HC Verma i Solutions for Class 12 Science Physics Chapter 18 Geometrical Optics are provided here with simple step-by-step explanations. These solutions for Geometrical Optics are extremely popular among class 12 Science students for Physics Geometrical Optics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the HC Verma i Book of class 12 Science Physics Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s HC Verma i Solutions. All HC Verma i Solutions for class 12 Science Physics are prepared by experts and are 100% accurate.

#### Page No 410:

#### Question 1:

Is the formula "Real depth/Apparent $\mathrm{depth}=\mathrm{\mu}$ valid if viewed from a position quite away from the normal?

#### Answer:

No, because this formula is only valid when the angles of incidence and refraction are small. When the angles of incidence and refraction are large, we cannot use relations sin*i** = *tan*i** *and sin*r** *= tan*r* to derive the formula for the apparent depth.

#### Page No 410:

#### Question 2:

Can you ever have a situation in which a light ray goes undeviated through a prism?

#### Answer:

There is only one such situation. Light rays go undeviated through a prism only when the prism is kept in a medium that has the same refractive index as that of the prism itself. In all other situations, there will always be some minimum deviation in the path of a light ray passing through the prism.

#### Page No 410:

#### Question 3:

Why does a diamond shine more than a glass piece cut to the same shape?

#### Answer:

A diamond shines more because of the phenomenon of total internal reflection (TIR). The refractive index of diamond is ≈2.4 and that of glass is ≈1.5.

By using the relation $\mu =\frac{1}{\mathrm{sin}C}$, where *C* is the critical angle, we get refractive index (*μ*).

For a large refractive index, the value of the critical angle is small. Thus, for the diamond, the critical angle for total internal reflection is much smaller than that of the glass. Therefore, a great percentage of incident light gets internally reflected several times before it emerges out.

#### Page No 410:

#### Question 4:

A narrow beam of light passes through a slab obliquely and is then received by an eye (figure 18-Q1). The index of refraction of the material in the slab fluctuates slowly with time. How will it appear to the eye? The twinkling of stars has a similar explanation.

Figure

#### Answer:

The beam of light will seem to appear and disappear, just like the twinkling of the stars. This is because the refractive index of the material in the slab fluctuates slowly with time. So, the light rays get refracted differently with time and this causes them to get concentrated during certain times, or get diffused at other times.

#### Page No 410:

#### Question 5:

Can a plane mirror ever form a real image?

#### Answer:

No, a plane mirror can never form a real image. This is because in all possible situations of image formation, the light rays never actually meet after getting reflected, but they only appear to meet behind the mirror, forming a virtual image always.

#### Page No 410:

#### Question 6:

If a piece of paper is placed at the position of a virtual image of a strong light source, will the paper burn after sufficient time? What happens if the image is real? What happens if the image is real but the source is virtual?

#### Answer:

No, the paper will not burn even after sufficient time. This is because the piece of paper is placed at the position of a virtual image of a strong light source, so light rays are not actually converging at that point, but seem to converge. As no light rays are concentrating on the point, the paper will never burn.

Instead, if the paper is placed at the position of a real image, it will burn as the light rays are actually concentrating on the point. Also, when we have the real image of a virtual source, it will also cause the burning of paper due to same reason; the light rays are actually meeting at the point.

#### Page No 410:

#### Question 7:

Can a virtual image be photographed by a camera?

#### Answer:

Yes, a virtual image can be photographed by a camera. This is because the light rays emitting from a virtual image and reaching the lens of the camera are real. A virtual image is formed from a reflecting surface after reflection of real incident rays, so these rays enter the camera to provide effects on the photographic film. Just like a virtual image can be seen by us in a mirror with our eyes, it can be photographed by a camera.

#### Page No 410:

#### Question 8:

In motor vehicles, a convex mirror is attached near the driver's seat to give him the view of the traffic behind. What is the special function of this convex mirror which a plane mirror can not do?

#### Answer:

The special function of a convex mirror is that it creates the image of a distant object that is reduced in size, is upright or erect and always lies within the virtual focal length of the mirror. A plane mirror cannot do this. Also, as the image is formed within the focal length, the image is close to the mirror as well as is small in size, enabling the driver to clearly view the nearer vehicles behind the motor vehicle.

#### Page No 410:

#### Question 9:

If an object far away from a convex mirror moves towards the mirror, the image also moves. Does it move faster, slower or at the same speed as compared to the object?

#### Answer:

The image of the object moves slower compared to the object. It can be explained using the mirror formula :$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

We know that for a convex mirror, the object distance (*u*) is positive, image distance (*v*) is negative and the focal length (*f*) is also negative. Thus mirror formula of a convex mirror is:$\frac{1}{u}-\frac{1}{v}=-\frac{1}{f}$

As *u* = +ve

$\frac{1}{v}-\frac{1}{f}>0\phantom{\rule{0ex}{0ex}}\frac{1}{v}>\frac{1}{f}\phantom{\rule{0ex}{0ex}}v<f$

Therefore, the image is always formed within the focal length of the mirror. Thus, the distance moved by the image is much slower than the distance moved by the object.

#### Page No 410:

#### Question 10:

Suppose you are inside the water in a swimming pool near an edge. A friends is standing on the edge. Do you find your friend taller or shorter than his usual height?

#### Answer:

When viewed from the water, the friend will seem taller than his usual height.

Let actual height be *h *and the apparent height be *h*'.

Here, the refraction is taking place from rarer to denser medium and a virtual image is formed.

Using

$\frac{{\mu}_{1}}{-u}+\frac{{\mu}_{2}}{v}=\frac{{\mu}_{2}-{\mu}_{1}}{R}$

Where refractive index of water is *μ*_{2} and refractive index of air is *μ*_{1}.

*u* and *v* are object and image distances, respectively.

*R *is the radius of curvature, here we will take it as ∞.

$\frac{{\mu}_{1}}{-u}+\frac{{\mu}_{2}}{v}=\frac{{\mu}_{2}-{\mu}_{1}}{\infty}\phantom{\rule{0ex}{0ex}}\frac{{\mu}_{1}}{u}=\frac{{\mu}_{2}}{v}\phantom{\rule{0ex}{0ex}}v=\frac{{\mu}_{2}}{{\mu}_{1}}\times u$

As *μ*_{1}_{ }= 1

*v* = *u** × **μ*_{2}

We know magnification is given by:

$m=\frac{v}{u}$

Putting the value of *v* in the above equation:

$m=\frac{u\times {\mu}_{2}}{u}\phantom{\rule{0ex}{0ex}}m={\mu}_{2}\phantom{\rule{0ex}{0ex}}$

As the magnification is greater than 1, so the apparent height seems to be greater than actual height.

#### Page No 410:

#### Question 11:

The equation of refraction at a spherical surface is

$\frac{{\mu}_{2}}{\nu}-\frac{{\mu}_{1}}{\mu}=\frac{{\mu}_{2}-{\mu}_{1}}{R}$.

Taking $R=\infty $, show that this equation leads to the equation

$\frac{\mathrm{Real}\mathrm{depth}}{\mathrm{Apparent}\mathrm{depth}}=\frac{{\mu}_{2}}{{\mu}_{1}}$

for refraction at a plane surface.

#### Answer:

Proof:

$\frac{{\mu}_{2}}{v}-\frac{{\mu}_{1}}{u}=\frac{{\mu}_{2}-{\mu}_{1}}{R}\phantom{\rule{0ex}{0ex}}\mathrm{Now}R=\infty \phantom{\rule{0ex}{0ex}}\frac{{\mu}_{2}}{v}-\frac{{\mu}_{1}}{u}=\frac{{\mu}_{2}-{\mu}_{1}}{\infty}\phantom{\rule{0ex}{0ex}}\frac{{\mu}_{2}}{v}-\frac{{\mu}_{1}}{u}=0\phantom{\rule{0ex}{0ex}}\frac{{\mu}_{2}}{v}=\frac{{\mu}_{1}}{u}\phantom{\rule{0ex}{0ex}}\frac{{\mu}_{1}}{{\mu}_{2}}=\frac{u}{v}\phantom{\rule{0ex}{0ex}}\mathrm{But}\frac{u}{v}=\frac{\mathrm{Real}\mathrm{depth}/\mathrm{height}}{\mathrm{Apparent}\mathrm{depth}/\mathrm{height}}\phantom{\rule{0ex}{0ex}}\therefore \frac{\mathrm{Real}\mathrm{depth}/\mathrm{height}}{\mathrm{Apparent}\mathrm{depth}/\mathrm{height}}=\frac{{\mu}_{\mathit{1}}}{{\mu}_{\mathit{2}}}$

#### Page No 410:

#### Question 12:

μA thin converging lens is formed with one surface convex and the other plane. Does the position of image depend on whether the convex surface or the plane surface faces the object?

#### Answer:

Yes. Using the lens maker formula we can show it. We know that the formula is:

$\frac{1}{f}=\left(\frac{{\mu}_{2}}{{\mu}_{1}}-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)$

Where, *f* is the focal length of the thin converging lens.

*μ*_{2} and *μ*_{1} are refractive indexes of lens and air respectively.

*R*_{1} and *R*_{2} are the radius of curvature of convex and plane surfaces respectively.

Here, *R*_{2 }= ∞ because of the plane surface.

__Case 1__

When the convex surface is facing the object, we have:

$\frac{1}{f}=\left(\frac{{\mu}_{2}}{{\mu}_{1}}-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{\infty}\right)\phantom{\rule{0ex}{0ex}}\frac{1}{f}=\left(\frac{{\mu}_{2}}{{\mu}_{1}}-1\right)\frac{1}{{R}_{1}}...\left(\mathrm{i}\right)$

__Case 2__

When the plane surface is facing the object, we have:

$\frac{1}{f}=\left(\frac{{\mu}_{2}}{{\mu}_{1}}-1\right)\left(\frac{1}{\infty}-\frac{1}{{R}_{1}}\right)\phantom{\rule{0ex}{0ex}}\frac{1}{f}=-\left(\frac{{\mu}_{2}}{{\mu}_{1}}-1\right)\frac{1}{{R}_{1}}...\left(\mathrm{ii}\right)$

For Case 1, the focal length is positive and for the Case 2 the focal length is negative. Thus, the image distance is different in both cases.

#### Page No 410:

#### Question 13:

A single lens is mounted in a tube. A parallel beam enters the tube and emerges out of the tube as a divergent beam. Can you say with certainty that there is a diverging lens in the tube?

#### Answer:

Yes, we can say so with certainty, as only a diverging lens can cause a parallel beam of light to diverge.

#### Page No 410:

#### Question 14:

An air bubble is formed inside water. Does it act as a converging lens or a diverging lens?

#### Answer:

It will act as a diverging lens. This is because the refractive index of air is less than that of water. Thus, light will diverge after passing through the bubble.

#### Page No 410:

#### Question 15:

Two converging lenses of unequal focal lengths can be used to reduce the aperture of a parallel beam of light without loosing the energy of the light. This increase the intensity. Describe how the converging lenses should be placed to do this.

#### Answer:

Let the two converging lenses be L_{1} and* *L_{2}, with focal lengths *f*_{1} and *f*_{2} respectively.

To reduce the aperture of a parallel beam of light without losing the energy of the light and also increase the intensity, we have to place the lens L_{2} within the focal range of L_{1.}

#### Page No 410:

#### Question 16:

If a spherical mirror is dipped in water, does its focal length change?

#### Answer:

No, the focal length of mirror will not change. This is because the focal length of a mirror does not depend on the refractive index of the medium in which the light rays travel.

#### Page No 410:

#### Question 17:

If a thin lens is dipped in water, does its focal length change?

#### Answer:

Yes, the focal length of the lens will change. This is because its focal length is dependent on the medium in which the light rays travel.

#### Page No 410:

#### Question 18:

Can mirrors give rise to chromatic aberration?

#### Answer:

No, mirrors cannot give rise to chromatic aberration. This is because chromatic aberration occurs due to the refraction of different colours of light. In case of mirrors, refraction of light does not take place.

#### Page No 410:

#### Question 19:

A laser light is focussed by a converging lens. Will there be a significant chromatic aberration?

#### Answer:

No, there will not be a significant chromatic aberration because a laser light is monochromatic in nature. Therefore, all light will get converged at the same point.

#### Page No 410:

#### Question 1:

A point source of light is placed in front of a plane mirror.

(a) All the reflected rays meet at a point when produced backward.

(b) Only the reflected rays close to the normal meet at a point when produced backward.

(c) Only the reflected rays making a small angle with the mirror meet at a point when produced backward.

(d) Light of different colours make different images.

#### Answer:

(a) All the reflected rays meet at a point when produced backward.

Here, the angle of reflection is equal to the angle of incidence. Therefore, all rays get reflected to converge at a single point to form the point image of the point source.

#### Page No 411:

#### Question 2:

Total internal reflection can take place only if

(a) light goes from optically rarer medium (smaller refractive index) to optically denser medium

(b) light goes from optically denser medium to rarer medium

(c) the refractive indices of the two media are close to each other

(d) the refractive indices of the two media are widely different.

#### Answer:

(b) light goes from optically denser medium to rarer medium

In this case the incident angle is greater than critical angle, so the light gets reflected back into the same denser medium, instead of being refracted .

#### Page No 411:

#### Question 3:

In image formation from spherical mirrors, only paraxial rays are considered because they

(a) are easy to handle geometrically

(b) contain most of the intensity of the incident light

(c) from nearly a point image of a point source

(d) show minimum dispersion effect.

#### Answer:

(c) form nearly a point image of a point source

Since, when reflected back, they meet at a single point forming a point image of a point source.

#### Page No 411:

#### Question 4:

A point object is placed at a distance of 30 cm from a convex mirror of focal length 30 cm. The image will form at

(a) infinity

(b) pole

(c) focus

(d) 15 cm behind the mirror.

#### Answer:

(d) 15 cm behind the mirror

Since* u* = − 30 cm

*v* = ?

*f* = 30 cm

From the mirror formula:

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{30}-\frac{1}{-30}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=\frac{1}{30}+\frac{1}{30}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=\frac{2}{30}\phantom{\rule{0ex}{0ex}}\therefore v=15\mathrm{cm}$

#### Page No 411:

#### Question 5:

Figure (18-Q2) shows two rays *A* and *B* being reflected by a mirror and going as *A*' and *B*'. The mirror

(a) is plane

(b) is convex

(c) is concave

(d) may be any spherical mirror.

Figure

#### Answer:

(a) is plane

This is because the reflected rays are still parallel, which is only possible if the mirror is a plane mirror. A spherical mirror will either converge or diverge the reflected rays.

#### Page No 411:

#### Question 6:

The image formed by a concave mirror

(a) is always real

(b) is always virtual

(c) is certainly real if the object is virtual

(d) is certainly virtual if the object is real.

#### Answer:

(c) is certainly real if the object is virtual

As a virtual object is the virtual image of the object, it is always formed beyond the focus of the concave mirror. Thus, the image formed by a concave mirror of the virtual object will always be real.

#### Page No 411:

#### Question 7:

Figure (18-Q3) shows three transparent media of refractive indices ${\mu}_{1},{\mu}_{2}\mathrm{and}{\mu}_{3}$. A point object *O* is placed in the medium ${\mu}_{2}$. If the entire medium on the right of the spherical surface has refractive index ${\mu}_{1}$, the image forms at *O*'. If this entire medium has refractive index ${\mu}_{3}$, the image forms at *O*". In the situation shown,

(a) the image forms between *O*' and *O*"

(b) the image forms to the left of *O*'

(c) the image forms to the right of *O*"

(d) two images form, one at *O*' and the other at *O*".

Figure

#### Answer:

(d) two images form, one at *O*' and the other at *O*"

Light will be refracted differently in both mediums on the right side. Thus, two images will be formed, one at* O*' due to refraction from medium *μ*_{1} and another at *O*" due to refraction from medium *μ*_{3}.μ1

#### Page No 411:

#### Question 8:

Four modifications are suggested in the lens formula to include the effect of the thickness *t* of the lens. Which one is likely to be correct?

(a) $\frac{1}{\nu}-\frac{1}{u}=\frac{t}{uf}$

(b) $\frac{t}{{\nu}^{2}}-\frac{1}{u}=\frac{1}{f}$

(c) $\frac{1}{\nu -t}-\frac{1}{u+t}=\frac{1}{f}$

(d) $\frac{1}{\nu}-\frac{1}{u}+\frac{t}{uv}=\frac{t}{f}$.

#### Answer:

(c) $\frac{1}{v-t}-\frac{1}{u+t}=\frac{1}{f}$

The thickness of lens will cause a shift in the position of object and image from the ideal condition, which will be:

${u}_{\mathrm{new}}=u+t(1+\frac{1}{\mu})\mathrm{and}{v}_{\mathrm{new}}=v-t(1+\frac{1}{\mu})\phantom{\rule{0ex}{0ex}}or{u}_{\mathrm{new}}=u+t+\frac{t}{\mu}\mathrm{and}{v}_{\mathrm{new}}=v-t+\frac{t}{\mu}$

As $\frac{t}{\mu}$ will be quite small, it can be ignored.

Thus, *u*_{new} =* u* + *t* and* **v*_{new} = *v* − *t*

We get the new lens formula:

$\frac{1}{v-t}-\frac{1}{u+t}=\frac{1}{f}$

#### Page No 411:

#### Question 9:

A double convex lens has two surfaces of equal radii *R* and refractive index $m=1\xb75$. We have,

(a) $f=R/2$

(b) $f=R$

(c) $f=-R$

(d) $f=2R$.

#### Answer:

(b) *f* = *R*

As from lens maker formula:

$\frac{1}{f}=(m-1)(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}})\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=(1.5-1)(\frac{1}{R}-\frac{1}{-R})(\mathrm{As}\mathrm{given}{R}_{1}={R}_{2}=R)\phantom{\rule{0ex}{0ex}}\frac{1}{f}=(0.5)(\frac{1}{R}+\frac{1}{R})\phantom{\rule{0ex}{0ex}}\frac{1}{f}=(0.5)\left(\frac{2}{R}\right)\phantom{\rule{0ex}{0ex}}f=R$

#### Page No 411:

#### Question 10:

A point source of light is placed at a distance of 2 *f* from a converging lens of focal length *f*. The intensity on the other side of the lens is maximum at a distance

(a) *f*

(b) between *f* and 2 *f*

(c) 2 *f*

(d) more than 2 *f*.

#### Answer:

(c) 2 *f*

Since the object is placed at 2 *f*, the image of the object will be formed at distance of 2 *f* from a converging lens.

It can also be shown from the lens formula:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Here, *u* = − 2 *f* and *f* = *f*

On putting the respective values we get:

$\frac{1}{v}-\frac{1}{-2f}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{2f}\phantom{\rule{0ex}{0ex}}=\frac{1}{2f}$

Therefore, image distance *v *= 2 *f*

#### Page No 411:

#### Question 11:

A parallel beam of light is incident on a converging lens parallel to its principal axis. As one moves away from the lens on the other side on its principal axis, the intensity of light

(a) remains constant

(b) continuously increases

(c) continuously decreases

(d) first increases then decreases.

#### Answer:

(d) first increases then decreases

Since all beams of light are parallel to the principal axis, they will converge at the focus of a converging lens. Thus, the intensity of light increases till one reaches the focus and then starts decreasing as one moves beyond it.

#### Page No 411:

#### Question 12:

A symmetric double convex lens in cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens was 4 D, the power a cut-lens will be

(a) 2 D

(b) 3 D

(c) 4 D

(d) 5 D.

#### Answer:

(a) 2 D

The lens is cut into two equal parts by a plane perpendicular to the principal axis. Thus, the radius of curvature of both the lenses become half of its original value.

As,

$f=\frac{R}{2}\phantom{\rule{0ex}{0ex}}P=\frac{1}{f}$

Radius of curvature becomes half. Therefore, the focal also reduces to half its original value.

Thus, the power (*P*) of the two cut-lenses will be equal.

2 *P* = 4 D

*P* = 2 D

#### Page No 411:

#### Question 13:

A symmetric double convex lens is cut in two equal parts by a plane containing the principal axis. If the power of the original lens was 4 D, the power of a divided lens will be

(a) 2 D

(b) 3 D

(c) 4 D

(d) 5 D.

#### Answer:

(c) 4 D

As the lens is cut along the principal axis, the two new lens will act as two double convex lenses. Even after cutting the lens, the radius of the curvature of the two new lens will remain the same as that of the original lens, with the power equal to the original lens.

#### Page No 411:

#### Question 14:

Two concave lenses *L*_{1} and *L*_{2} are kept in contact with each other. If the space between the two lenses is filled with a material of smaller refractive index, the magnitude of the focal length of the combination

(a) becomes undefined

(b) remains unchanged

(c) increases

(d) decreases.

#### Answer:

(c) increases

The focal length of the combination will increase. For finding out the combination of lens we have the formula:

$\frac{1}{F}=\frac{1}{{f}_{1}}+\frac{1}{{f}_{2}}-\frac{\mathrm{d}}{{f}_{1}{f}_{2}}$

Where, *F *is the focal length for the combination

*d* is the separation between two lenses

Here, *d* = 0

$\frac{1}{F}=\frac{1}{{f}_{1}}+\frac{1}{{f}_{2}}\phantom{\rule{0ex}{0ex}}F=\frac{{f}_{1}{f}_{2}}{{f}_{1}+{f}_{2}}$

Hence, the focal length will increase.

#### Page No 411:

#### Question 15:

A thin lens is made with a material having refractive index $\mu =1\xb75$. Both the side are convex. It is dipped in water ($\mu =1\xb733$). It will behave like

(a) a convergent lens

(b) a divergent lens

(c) a rectangular slab

(d) a prism.

#### Answer:

(a) a convergent lens

Since the refractive index of lens is more than that of of water and both the sides are convex, it will act like a normal convergent lens (though its focal length will change).

#### Page No 411:

#### Question 16:

A convex lens is made of a material having refractive index $1\xb72$. Both the surfaces of the lens are convex. If it is dipped into water ($\mu =1\xb733$), it will behave like

(a) a convergent lens

(b) a divergent lens

(c) a rectangular slab

(d) a prism.

#### Answer:

(b) a divergent lens

Since the refractive index of lens is less than that of of water and both the sides are convex, it will act like a normal diverging lens (though its focal length will change to negative).

#### Page No 411:

#### Question 17:

A point object *O* is placed on the principal axis of a convex lens of focal length *f* = 20 cm at a distance of 40 cm to the left of it. The diameter of the lens is 10 cm. An eye is placed 60 cm to right of the lens and a distance *h* below the principal axis. The maximum value of *h* to see the image is

(a) 0

(b) 2.5 cm

(c) 5 cm

(d) 10 cm.

#### Answer:

(b) 2.5 cm

As the focal length of the lens is 20 cm and object distance is 40 cm from the lens, the image is formed at the centre of curvature at the right side of the lens.

From right angled triangle ABC,

$\mathrm{tan}\alpha =\frac{\mathrm{AB}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\alpha =\frac{5}{40}\phantom{\rule{0ex}{0ex}}\alpha ={\mathrm{tan}}^{-1}\left(\frac{5}{40}\right)$

and from right angled triangle, we have:

$\mathrm{tan}\alpha =\frac{\mathrm{DE}}{\mathrm{DC}}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\alpha =\frac{h}{20}\phantom{\rule{0ex}{0ex}}\mathrm{as}\mathrm{DB}-\mathrm{BC}=\mathrm{DC}=20\mathrm{cm}$

Putting the value of angle alpha, we get:

$\mathrm{tan}\left({\mathrm{tan}}^{-1}\left(\frac{5}{40}\right)\right)=\frac{h}{20}\phantom{\rule{0ex}{0ex}}\frac{5}{40}=\frac{h}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow h=2.5\mathrm{cm}$

#### Page No 411:

#### Question 18:

The rays of different colours fail to converge at a point after going through a converging lens. This defect is called

(a) spherical aberration

(b) distortion

(c) coma

(d) chromatic aberration.

#### Answer:

(d) chromatic aberration

When light rays of different colours do not converge at the same point after passing through a converging lens, it is called chromatic aberration. This happens because a lens has different refractive indices for different colours, i.e, for different wavelengths of light.

#### Page No 412:

#### Question 1:

A concave mirror having a radius of curvature 40 cm is placed in front of an illuminated point source at a distance of 30 cm from it. Find the location of the image.

#### Answer:

Using sign conventions, given,

Distance of object from mirror, *u* = − 30 cm,

Radius of curvature of concave mirror *R* = − 40 cm

Using the mirror equation,

$\frac{1}{v}+\frac{1}{u}=\frac{2}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{2}{R}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{2}{-40}-\frac{1}{-30}=\frac{1}{-20}+\frac{1}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{-30+20}{30\times 20}=\frac{-10}{30\times 20}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=-\frac{1}{60}$

or, *v* = − 60 cm

Hence, the required image will be located at a distance of 60 cm in front of the concave mirror.

#### Page No 412:

#### Question 2:

A concave mirror forms an image of 20 cm high object on a screen placed 5.0 m away from the mirror. The height of the image is 50 cm. Find the focal length of the mirror and the distance between the mirror and the object.

#### Answer:

Given,

Height of the object, *h*_{1} = 20 cm,

Distance of image from screen *v* = −5.0 m = −500 cm,

Since, we know that

$-\frac{v}{u}=\frac{{h}_{2}}{{h}_{1}}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{-(-500)}{u}=\frac{50}{20}$

Where '*u*' is the distance of object from screen.

(As the image is inverted)

$\mathrm{or}u=-500\times \frac{2}{5}\phantom{\rule{0ex}{0ex}}=-200\mathrm{cm}=-2.0\mathrm{m}$

Using mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\mathrm{or}\frac{1}{-5}+\frac{1}{-2}=\frac{1}{f}$

$\mathrm{or}-\frac{1}{f}=\frac{7}{10}\phantom{\rule{0ex}{0ex}}\mathrm{or}f=-\frac{10}{7}=-1.44\mathrm{m}$

Hence, the required focal length of the concave mirror is 1.44 m.

#### Page No 412:

#### Question 3:

A concave mirror has a focal length of 20 cm. Find the position or positions of an object for which the image-size is double of the object-size.

#### Answer:

Using sign conventions, given,

Focal length of the concave mirror:

*f* = −20 cm

As per the question,

Magnification (*m*) is:

$m=-\frac{v}{u}=2$

⇒ *v* = −2*u*

__ Case I (Virtual image):__

Using mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2u}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow u=\frac{f}{2}=10\mathrm{cm}$

__Case II (Real image)__

Using mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{1}{2u}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3}{2u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow u=\frac{3f}{2}=30\mathrm{cm}$

Hence, the required positions of objects are 10 cm or 30 cm from the concave mirror.

#### Page No 412:

#### Question 4:

A 1 cm object is placed perpendicular to the principal axis of a convex mirror of focal length 7.5 cm. Find its distance from the mirror if the image formed is 0.6 cm in size.

#### Answer:

Given,

Height of the object, *h*_{1} = 1 cm

Focal length of the concave mirror, *f* = 7.5 cm = $\frac{15}{2}\mathrm{cm}$

Magnification is given as,

$m=-\frac{v}{u}=\frac{{h}_{i}}{{h}_{0}}=0.6\phantom{\rule{0ex}{0ex}}\mathrm{Using}\mathrm{mirror}\mathrm{equation},\phantom{\rule{0ex}{0ex}}\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{0.6u}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5}{3u}-\frac{1}{u}=\frac{2}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow u=5\mathrm{cm}$

Hence, the distance of the object from the mirror is 5 cm.

#### Page No 412:

#### Question 5:

A candle flame 1.6 cm high is imaged in a ball bearing of diameter 0.4 cm. If the ball bearing is 20 cm away from the flame, find the location and the height of the image.

#### Answer:

Given,

Height (*h*_{1}) of the candle flame taken as object *AB* = 1.6 cm

Diameter of the ball bearing (*d*) = 0.4 cm

So, radius = 0.2 cm

Distance of object, *u* = 20 cm

Using mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{2}{R}$

Putting the values according to sign conventions, we get,

$\frac{1}{(-20)}+\frac{1}{v}=\frac{2}{0.2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{20}+10\phantom{\rule{0ex}{0ex}}\Rightarrow v=0.1\mathrm{cm}\mathrm{or}1.0\mathrm{mm}\mathrm{insi}\mathrm{de}\mathrm{the}\mathrm{ball}\mathrm{bearing}.\phantom{\rule{0ex}{0ex}}\mathrm{Magnification}=m\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}=-\frac{v}{u}=m\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{200}=\frac{1}{200}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{A}\text{'}\mathrm{B}\text{'}=\frac{\mathrm{AB}}{200}=+\frac{1.6}{200}=+0.08\mathrm{cm}(+0.008\mathrm{cm})$

Hence, the distance of the image is 1 cm.

Height of the image is 0.008 cm.

#### Page No 412:

#### Question 6:

A 3 cm tall object is placed at a distance of 7.5 cm from a convex mirror of focal length 6 cm. Find the location, size and nature of the image.

#### Answer:

Given,

Height of the object *AB*, *h*_{1} = 3 cm

Distance of the object from the convex mirror, *u* = −7.5 cm

Focal length of the convex mirror (*f*) = 6 cm

Using mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

Putting values according to sign convention,

$\frac{1}{v}=\frac{1}{6}-\frac{1}{(-7.5)}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=\frac{1}{6}+\frac{1}{7.5}=\frac{3}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{10}{3}\mathrm{cm}$

Magnification = *m* = $-\frac{v}{u}=\frac{10}{7.5\times 3}$

$\Rightarrow \frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}=\frac{10}{7.5\times 3}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{A}\text{'}\mathrm{B}\text{'}=\frac{100}{75}=\frac{4}{3}=1.33\mathrm{cm}$

Where A'B' is the height of the image.

Hence, the required location of the image is $\frac{10}{3}\mathrm{cm}$ from the pole and image height is 1.33 cm. Nature of the image is virtual and erect.

#### Page No 412:

#### Question 7:

A U-shaped wire is placed before a concave mirror having radius of curvature 20 cm as shown in figure. Find the total length of the image.

Figure

#### Answer:

Given,

Radius of curvature of concave mirror, *R* = 20 cm

So its focal length will be *f* = $\frac{R}{2}$ = −10 cm

For part AB of the U shaped wire, PB = 30 + 10 = 40 cm

Therefore, *u* = −40 cm

Using mirror formula,

$\Rightarrow \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}}=-\frac{1}{10}-\frac{1}{-40}=-\frac{3}{40$

$\Rightarrow v=\frac{-40}{3}=13.3\mathrm{cm}$

$\mathrm{So},\mathrm{PB}\text{'}=13.3\mathrm{cm}$

$m=\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}=\frac{-\left(v\right)}{u}$

$=\frac{-(-13.3)}{-40}=-\frac{1}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{A}\text{'}\mathrm{B}\text{'}=-\frac{10}{3}=-3.33\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

For part CD of the U shaped wire, PC = 30 cm

Therefore, *u* = −30 cm

Again, using mirror equation:

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$\frac{1}{v}}=-\frac{1}{10}-\frac{1}{-30}\phantom{\rule{0ex}{0ex}}{\displaystyle \frac{1}{v}}=-\frac{1}{10}+\frac{1}{30}=-\frac{1}{15$

⇒ *v* = −15 cm = PC'

$m=\frac{\mathrm{C}\text{'}\mathrm{D}\text{'}}{\mathrm{CD}}=-\frac{v}{u}$

$=-\frac{(-15)}{-30}=-\frac{1}{2}$

⇒ C'D' = 5 cm

⇒ B'C' = PC' − PB'

= 15 − 13.3 = 1.7 cm

Hence, the total length of the U- shaped wire is A'B' + B'C' + C'D'

= (3.3) + (1.7) + 5 = 10 cm

#### Page No 412:

#### Question 8:

A man uses a concave mirror for shaving. He keeps his face at a distance of 25 cm from the mirror and gets an image which is 1.4 times enlarged. Find the focal length of the mirror.

#### Answer:

Given,

Distance of the man's face (here, taken as object), *u* = −25 cm

According to the question, magnification, *m* = 1.4

$m=\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}=-\frac{v}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow 1.4=-\frac{\left(v\right)}{-25}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{14}{10}=\frac{v}{25}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{25\times 14}{10}=35\mathrm{cm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Using equation of mirror, we get:

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\Rightarrow \frac{1}{f}=\frac{1}{35}-\frac{1}{25}$

$=\frac{5-7}{175}=-\frac{2}{175}$

⇒ *f* = −87.5

Hence, the required focal length of the concave mirror is 87.5 cm.

#### Page No 412:

#### Question 1:

If the light moving in a straight line bends by a small but fixed angle, it may be a case of

(a) reflection

(b) refraction

(c) diffraction

(d) dispersion.

#### Answer:

(a) reflection

(b) refraction

When the light strikes on a surface nearly parallel to it, it then bends by a small and fixed angle after reflection. Also, when the light travels from one medium to another with slight differences in their refractive indices, it bends by a small angle. Thus, the bending of light by a small but fixed angle can be the case of either reflection or refraction.

#### Page No 412:

#### Question 2:

Mark the correct options.

(a) If the incident rays are converging, we have a real object.

(b) If the final rays are converging, we have a real image.

(c) The image of a virtual object is called a virtual image.

(d) If the image is virtual, the corresponding object is called a virtual object.

#### Answer:

(b) If the final rays are converging, we have a real image.

This is because a real image is formed by converging reflected/refracted rays from a mirror/lens.

#### Page No 412:

#### Question 3:

Which of the following (referred to a spherical mirror) do (does) not depend on whether the rays are paraxial or not?

(a) Pole

(b) Focus

(c) Radius of curvature

(d) Principal axis

#### Answer:

(a) Pole

(c) Radius of curvature

(d) Principal axis

Paraxial rays are the light rays close to the principal axis. The focus of the spherical mirror for paraxial rays are different from the focus for marginal rays. So, the focus depends on whether the rays are paraxial or marginal. The pole, radius of curvature and the principal axis of a spherical mirror do not depend on paraxial or marginal rays.

#### Page No 412:

#### Question 4:

The image of an extended object, placed perpendicular to the principal axis of a mirror, will be erect if

(a) the object and the image are both real

(b) the object and the image are both virtual

(c) the object is real but the image is virtual

(d) the object is virtual but the image is real.

#### Answer:

(c) the object is real but the image is virtual

(d) the object is virtual but the image is real

The virtual image of a real object and the real image of a virtual object are always erect.

#### Page No 412:

#### Question 5:

A convex lens forms a real image of a point object placed on its principals axis. If the upper half of the lens is painted black,

(a) the image will be shifted downward

(b) the image will be shifted upward

(c) the image will not be shifted

(d) the intensity of the image will decrease.

#### Answer:

(c) the image will not be shifted

(d) the intensity of the image will decrease

If the upper half portion of the convex lens is painted, then only the intensity of the image will decrease, as the amount of light passing through the lens will decrease. Also, there will be no shift in the position of the image because all the parameters remain the same.

#### Page No 412:

#### Question 6:

Consider three converging lenses *L*_{1}, *L*_{2} and *L*_{3} having identical geometrical construction. The index of refraction of *L*_{1} and *L*_{2} are ${\mu}_{1}\mathrm{and}{\mu}_{2}$ respectively. The upper half of the lens *L*_{3} has a refractive index ${\mu}_{1}$ and the lower half has ${\mu}_{2}$ (figure 18-Q3). A point object *O* is imaged at *O*_{1} by the lens *L*_{1} and at *O*_{2} by the lens *L*_{2} placed in same position. If *L*_{3} is placed at the same place,

(a) there will be an image at *O*_{1}

(b) there will be an image at *O*_{2}.

(c) the only image will form somewhere between *O*_{1} and *O*_{2}

(d) the only image will form away from *O*_{2}.

Figure

#### Answer:

(a) there will be an image at *O*_{1}

(b) there will be an image at *O*_{2}

The lens *L*_{1} converges the light at point *O*_{1} and lens *L*_{2} converges the light at *O*_{2}. As the upper half of lens *L*_{3} has a refractive index equal to that of L_{1}, it will converge the light at *O*_{1} and thus the image will be formed at O_{1}. Also, the lower half of lens *L*_{3} has a refractive index equal to that of lens *L*_{2}, it will converge the light at *O*_{2} and thus the image will be formed at O_{2}.

#### Page No 412:

#### Question 7:

A screen is placed a distance 40 cm away from an illuminated object. A converging lens is placed between the source and the screen and its is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens

(a) must be less than 10 cm

(b) must be greater than 20 cm

(c) must not be greater than 20 cm

(d) must not be less than 10 cm.

#### Answer:

(b) must be greater than 20 cm

Let the image be formed at a distance of *x* cm from the lens.

Therefore, the distance of the object from the lens, *u*, will be = (40 − x) cm

From lens formula:

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{1}{x}-\frac{1}{40-x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{40-x-x}{40x-{x}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow f=\frac{40x-{x}^{2}}{40-2x}\phantom{\rule{0ex}{0ex}}\Rightarrow f(40-2x)=40x-{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-2fx-40x+40f=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-(2f+40)x+40f=0$

Therefore, we get *x* as:

$x=\frac{(2f+40)\pm \sqrt{(2f+40{)}^{2}-160f}}{2}$

#### Page No 413:

#### Question 9:

Find the diameter of the image of the moon formed by a spherical concave mirror of focal length 7.6 m. The diameter of the moon is 3450 km and the distance between the earth and the moon is 3.8 × 10^{5} km.

#### Answer:

Given,

Focal length of the concave mirror,* f *= − 7.6 m

Distance between earth and moon taken as object distance, *u* = −3.8 × 10^{5} km

Diameter of moon = 3450 km

Using mirror equation:

$\Rightarrow \frac{1}{v}=\frac{1}{u}+\frac{1}{f}$

$\therefore \frac{1}{v}+\left(-\frac{1}{3.8\times {10}^{8}}\right)=\left(-\frac{1}{7.6}\right)$

As we know, the distance of moon from earth is very large as compared to focal length it can be taken as ∞.

Therefore, image of the moon will be formed at focus, which is inverted.

$\Rightarrow \frac{1}{v}=-\frac{1}{7.6}$

⇒ *v* = −7.6 m

We know that magnification is given by:

$m=-\frac{v}{u}=\frac{{d}_{\mathrm{image}}}{{d}_{\mathrm{object}}}$

$\Rightarrow \frac{-(-7.6)}{(-3.8\times {10}^{8})}=\frac{{d}_{image}}{3450\times {10}^{3}}$

${d}_{\mathrm{image}}=-\frac{3450\times 7.6\times {10}^{3}}{3.8\times {10}^{8}}$

= $-$0.069 m = $-$6.9 cm

Hence, the required diameter of the image of moon is 6.9 cm.

#### Page No 413:

#### Question 10:

A particle goes in a circle of radius 2.0 cm. *A* concave mirror of focal length 20 cm is placed with its principal axis passing through the centre of the circle and perpendicular to its plane. The distance between the pole of the mirror and the centre of the circle is 30 cm. Calculate the radius of the circle formed by the image.

#### Answer:

Given,

Distance of the circle from the mirror taken as object distance, *u* = −30 cm,

Focal length of the concave mirror, *f *= −20 cm

Using mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{v}+\left(-\frac{1}{30}\right)=-\frac{1}{20}$

$\Rightarrow \frac{1}{v}=\frac{1}{30}-\frac{1}{20}=\frac{1}{60}$

⇒ *v* = $-$60 cm

Therefore, image of the circle is formed at a distance of 60 cm in front of the mirror.

We know magnification (*m*) is given by:

$m=-\frac{v}{u}=\frac{{R}_{\mathrm{image}}}{{R}_{\mathrm{object}}}$

$\Rightarrow -\frac{(-60)}{(-30)}=\frac{{R}_{\mathrm{image}}}{2}$

Where *R*_{object} and *R*_{image} are radius of the object and radius of the image, respectively.

⇒ R_{image} = 4 cm

Hence, the required radius of the circle formed by the image is 4 cm.

#### Page No 413:

#### Question 11:

A concave mirror of radius *R* is kept on a horizontal table (figure). Water (refractive index = μ) is poured into it up to a height *h*. Where should an object be placed so that its image is formed on itself?

Figure

#### Answer:

Given,

A concave mirror of radius '*R*' kept on a horizontal table.

'*h*' is height up to which the water is poured into the concave mirror.

Let the object be placed at height '*x*' above the surface of water.

We know if we place the object at the centre of curvature of the mirror, then the image itself will be formed at the centre of curvature.

Therefore, the apparent position of the object with respect to the mirror should be at the centre of curvature so that the image is formed at the same position.

Since,

$\frac{\mathrm{Real}\mathrm{depth}}{\mathrm{Apparent}\mathrm{depth}}=\frac{1}{\mu}$

(with respect to mirror)

$\mathrm{Now},\frac{x}{R\mathit{-}h}\mathit{=}\frac{\mathit{1}}{\mu}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\mathit{}x\mathit{=}\frac{R\mathit{-}h}{\mu}$

Hence, the object should be placed at $\frac{R-h}{\mathrm{\mu}}$ above the water surface.

#### Page No 413:

#### Question 12:

A point source *S* is placed midway between two converging mirrors having equal focal length *f *as shown in figure. Find the values of *d* for which only one image is formed.

Figure

#### Answer:

Given,

Two converging mirrors having equal focal length '*f* '.

Both the mirrors will produce one image under two conditions:

__Case-1 __

When the point source is at the centre of curvature of the mirrors, i.e, at a distance of '2*f** *' from each mirror, the images will be produced at the same point 'S'. Therefore, *d* = 2*f* + 2*f* = 4*f*

__Case-II __

When the point source *'*S' is at focus, i.e., at a distance '*f* ' from each mirror, the rays from the source after reflecting from one mirror will become parallel and so these parallel rays, after the reflection from the other mirror the object itself. Thus, only one image is formed.

Here *d* = *f* + *f* = 2*f*

#### Page No 413:

#### Question 13:

A converging mirror *M*_{1}, a point source *S* and a diverging mirror *M*_{2} are arranged as shown in figure. The source is placed at a distance of 30 cm from *M*_{1}. The focal length of each of the mirrors is 20 cm. Consider only the images formed by a maximum of two reflections. It is found that one image is formed on the source itself. (a) Find the distance between the two mirrors. (b) Find the location of the image formed by the single reflection from *M*_{2}.

Firegu

#### Answer:

Given,

Converging mirror *M*_{1} with focal length (*f*_{1}) = 20 cm

Converging mirror *M*_{2} with focal length (*f*_{2}) = 20 cm

*f*_{1} = *f*_{2} = 20 cm = *f*

Point source is at a distance of 30 cm from *M*_{1}.

As the 1^{st} reflection is through the mirror *M*_{1},

*u = *−30 cm

*f* = −20 cm

Using mirror equation:

$\Rightarrow \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{v}+\frac{1}{-30}=-\frac{1}{20}$

$\Rightarrow \frac{1}{v}=-\frac{1}{20}+\frac{1}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=-\frac{1}{60}\phantom{\rule{0ex}{0ex}}\Rightarrow v=-60\mathrm{cm}$

and for the 2^{nd} reflection at mirror *M*_{2},

*u* = 60 − (30 + *x*) = 30 − *x*

*v* = − *x*, *f* = 20 cm

Again using the mirror equation:

$\Rightarrow \frac{1}{30-x}-\frac{1}{x}=\frac{1}{20}$

$\Rightarrow \frac{x-30+x}{x(30-x)}=\frac{1}{20}$

⇒ 40*x* − 600 = 30*x* − *x*^{2}

⇒ *x*^{2} + 10*x* − 600 = 0

$\Rightarrow x=\frac{10+50}{2}=\frac{40}{2}$

= 20 cm or − 30 cm

∴ Total distance between the two lines is 20 + 30 = 50 cm

(b) Location of the image formed by the single reflection from *M*_{2} = 60 $-$ 50 = 10

Thus, the image formed by the single reflection from *M*_{2} is at a distance of 10 cm from mirror *M*_{2}_{. }

#### Page No 413:

#### Question 14:

A light ray falling at an angle of 45° with the surface of a clean slab of ice of thickness 1.00 m is refracted into it at an angle of 30°. Calculate the time taken by the light rays to cross the slab. Speed of light in vacuum = 3 × 10^{8} m s^{−1}.

#### Answer:

Given,

Angle of incidence, *i* = 45°

Angle of refraction, *r* = 30°

Using Snell's law,

$\frac{\mathrm{sin}i}{\mathrm{sin}r}=\frac{3\times {10}^{8}}{v}$

$=\frac{\mathrm{sin}45\xb0}{\mathrm{sin}30\xb0}=\frac{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}{\left({\displaystyle \frac{1}{2}}\right)}$

$=\frac{2}{\sqrt{2}}=\sqrt{2}$

$\mathrm{So},\phantom{\rule{0ex}{0ex}}v=\frac{3\times {10}^{8}}{\sqrt{2}}\mathrm{m}/s$

Let *x* be the distance travelled by light in the slab.

Now,

$x=\frac{1\mathrm{m}}{\mathrm{cos}30\xb0}=\frac{2}{\sqrt{3}}\mathrm{m}$

We know:

Time taken $=\frac{\mathrm{Distance}}{\mathrm{Speed}}$

$=\frac{2}{\sqrt{3}}\times \frac{\sqrt{2}}{3\times {10}^{8}}$

= 0.54 × 10^{−8}

= 5.4 × 10^{−9} s

#### Page No 413:

#### Question 15:

A pole of length 1.00 m stands half dipped in a swimming pool with water level 50.0 cm higher than the bed. The refractive index of water is 1.33 and sunlight is coming at an angle of 45° with the vertical. Find the length of the shadow of the pole on the bed.

#### Answer:

Given,

Length of the pole = 1.00 m

Water level of the swimming pool is 50.0 cm higher than the bed.

Refractive index (*μ*) of water = 1.33

According to the figure, shadow length = BA' = BD + DA'= 0.5 + 0.5 tan *r*

Using Snell's law:

$\mathrm{Now}1.33=\frac{\mathrm{sin}45\xb0}{\mathrm{sin}r}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}r=\frac{1}{1.33\sqrt{2}}=0.53\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}r=\sqrt{1-{\mathrm{sin}}^{2}r}\phantom{\rule{0ex}{0ex}}=\sqrt{1-(0.53{)}^{2}}=0.85\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{tan}r=0.6235\phantom{\rule{0ex}{0ex}}$

Therefore, shadow length of the pole = (0.5)$\times $(1 + 0.6235) = 0.81175 m

= 81.2 cm

#### Page No 413:

#### Question 16:

A small piece of wood is floating on the surface of a 2.5 m deep lake. Where does the shadow form on the bottom when the sum is just setting? Refractive index of water = 4/3.

#### Answer:

Given,

Depth of the lake = 2.5 m

Refractive index (*μ*) of water = $\frac{4}{3}$

When the sun is just setting, *θ* is approximately = 90˚

Therefore, incidence angle is 90˚

Using Snell's law:

$\therefore \frac{\mathrm{sin}i}{\mathrm{sin}r}=\frac{{\mathrm{\mu}}_{2}}{{\mathrm{\mu}}_{1}}$

$\Rightarrow \frac{1}{\mathrm{sin}r}=\frac{{\displaystyle \frac{4}{3}}}{1}$

$\Rightarrow \mathrm{sin}r=\frac{3}{4}$

$\Rightarrow r=49\xb0$

From the figure, W'O = *x *is the distance of the shadow

$\mathrm{Thus},\left({\displaystyle \frac{x}{2.5}}\right)=\mathrm{tan}r=1.15$

$\Rightarrow x=2.5\times 1.15=2.8\mathrm{m}$

#### Page No 413:

#### Question 17:

An object *P* is focussed by a microscope *M*. *A* glass slab of thickness 2.1 cm is introduced between *P* and *M*. If the refractive index of the slab is 1.5, by what distance should the microscope be shifted to focus the object again?

#### Answer:

Given,

Thickness of the glass slab, *d* = 2.1 cm

Refractive index, μ = 1.5

Shift due to the glass slab is given by,

$\u2206t=\left[1-\left(\frac{1}{\mathrm{\mu}}\right)\right]d$

$=\left[1-\left(\frac{1}{1.5}\right)\right]\left(2.1\right)$

$=\frac{1}{3}\left(2.1\right)=0.7\mathrm{cm}$

Hence, the microscope should be shifted by 0.70 cm to focus the object '*P*' again.

#### Page No 413:

#### Question 18:

A vessel contains water up to a height of 20 cm and above it an oil up to another 20 cm. The refractive indices of the water and the oil are 1.33 and 1.30 respectively. Find the apparent depth of the vessel when viewed from above.

#### Answer:

Given,

Height of water, *d*_{w} = 20 cm

Height of oil, *d*_{O} = 20 cm

The refractive index of the water (*μ*_{w}) = 1.33

The refractive index of oil (*μ*_{O}) = 1.30

Shift due to water is given by,

$\u2206{t}_{w}=1-\left(\frac{1}{{\mathrm{\mu}}_{\mathrm{w}}}\right){d}_{w}$

$=\left[1-\left(\frac{1}{1.33}\right)\right]20$

$=\frac{1}{4}\left(20\right)=5\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Shift due to oil,

$\u2206{t}_{\mathrm{O}}=\left[1-\left(\frac{1}{1.3}\right)\right]20\phantom{\rule{0ex}{0ex}}=4.6\mathrm{cm}$

Therefore, total shift, Δ*t* = 5 + 4.6 = 9.6 cm

Hence, apparent depth = 40 − (9.6) = 30.4 cm between the surface.

#### Page No 413:

#### Question 19:

Locate the image of the point *P* as seen by the eye in the figure.

Figure

#### Answer:

Given,

From the figure we can infer that the air is present in between the sheet, so it does not affect the shift. Therefore, the shift is only due to 3 sheets of different refractive indices, which is given by:

$\u2206t=\left[1-\frac{1}{{\mu}_{1}}\right]{t}_{1}+\left[1-\frac{1}{{\mu}_{2}}\right]{t}_{2}+\left[1-\frac{1}{{\mu}_{3}}\right]{t}_{3}$

$=\left[1-\left(\frac{1}{12}\right)\right]\left(0.2\right)+\left[1-\left(\frac{1}{1.3}\right)\right]\left(0.3\right)+\left[1-\left(\frac{1}{1.4}\right)\right]\left(0.4\right)$

= 0.2 cm

Hence, location of image of point P is located 0.2 cm above point P.

#### Page No 413:

#### Question 20:

*k* transparent slabs are arranged one over another. The refractive indices of the slabs are μ_{1}, μ_{2}, μ_{3}, ... μ_{k} and the thicknesses are *t*_{1} *t*_{2}, *t*_{3}, ... *t _{k}*. An object is seen through this combination with nearly perpendicular light. Find the equivalent refractive index of the system which will allow the image to be formed at the same place.

#### Answer:

*k* number of transparent slabs are arranged one over the other.

Refractive indices of the slabs = μ_{1}, μ_{2}, μ_{3}, ..., μ_{k}

Thickness of the slabs = *t*_{1}, *t*_{2}, *t*_{3},.., *t _{k}*

Shift due to one slab:

$\u2206t=\left[1-\frac{1}{\mathrm{\mu}}\right]t$

For the combination of multiple slabs, the shift is given by,

$\u2206t=\left[1-\frac{1}{{\mathrm{\mu}}_{1}}\right]{t}_{1}+\left[1-\frac{1}{{\mathrm{\mu}}_{2}}\right]{t}_{2}+...+\left[1-\frac{1}{{\mathrm{\mu}}_{k}}\right]{t}_{k}...\left(\mathrm{i}\right)$

Let

*μ*be the refractive index of the combination of slabs.

The image is formed at the same place.

So, the shift will be:

$\mathrm{\Delta}t=\left[1-\left(\frac{1}{\mathrm{\mu}}\right)\right]({t}_{1}+{t}_{2}...+{t}_{k})...\left(\mathrm{ii}\right)$

Equating (i) and (ii), we get:

$\left[1-\left(\frac{1}{\mathrm{\mu}}\right)\right]({t}_{1}+{t}_{2}...+{t}_{k})=\left[1-\frac{1}{{\mathrm{\mu}}_{1}}\right]{t}_{1}+\left[1-\frac{1}{{\mathrm{\mu}}_{2}}\right]{t}_{2}+\left[1-\frac{1}{{\mathrm{\mu}}_{k}}\right]{t}_{k}\phantom{\rule{0ex}{0ex}}=({t}_{1}+{t}_{2}...+{t}_{k})-\left(\frac{{t}_{1}}{{\mathrm{\mu}}_{1}}+\frac{{t}_{2}}{{\mathrm{\mu}}_{2}}+\frac{{t}_{k}}{{\mathrm{\mu}}_{k}}\right)\phantom{\rule{0ex}{0ex}}=-\frac{1}{\mathrm{\mu}}\sum _{\mathrm{i}=1}^{\mathrm{k}}{t}_{\mathrm{i}}=-\sum _{\mathrm{i}=1}^{\mathrm{k}}\left(\frac{{t}_{\mathrm{i}}}{{\mathrm{\mu}}_{\mathrm{i}}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\mu}=\frac{\sum _{\mathrm{i}=1}^{\mathrm{k}}{t}_{\mathrm{i}}}{-\sum _{\mathrm{i}=1}^{\mathrm{k}}\left({\displaystyle \raisebox{1ex}{${t}_{\mathrm{i}}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{\mu}}_{\mathrm{i}}$}\right.}\right)}$

Hence, the required equivalent refractive index is $\frac{\sum _{\mathrm{i}=1}^{\mathrm{k}}{t}_{\mathrm{i}}}{\sum _{\mathrm{i}=1}^{\mathrm{k}}\left({\displaystyle \raisebox{1ex}{${t}_{\mathrm{i}}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{\mu}}_{\mathrm{i}}$}\right.}\right)}$.

#### Page No 413:

#### Question 21:

A cylindrical vessel of diameter 12 cm contains 800π cm^{3} of water. A cylindrical glass piece of diameter 8.0 cm and height 8.0 cm is placed in the vessel. If the bottom of the vessel under the glass piece is seen by the paraxial rays (see figure), locate its image. The index of refraction of glass is 1.50 and that of water is 1.33.

Figure

#### Answer:

Given,

Diameter of the cylindrical vessel = 12 cm

∴ radius *r* = 6 cm

Diameter of the cylindrical glass piece = 8 cm

∴ radius *r*^{' }= 4 cm and its height, *h*_{1} = 8 cm

Refractive index of glass, *μ*_{g} = 1.5

Refractive index of water, *μ*_{w} = 1.3

Let *h* be the final height of the water column.

The volume of the cylindrical water column after the glass piece is put will be:

$\pi $*r*^{2}*h* = 800π + $\pi $*r'*^{2}*h*_{1}

or *r*^{2}*h* = 800 + *r*_{'}^{2}*h*_{1}

or (6)^{2} *h* = 800 + (4)^{2} 8

$\mathrm{or}h=\frac{800+128}{36}=\frac{928}{36}=25.7\mathrm{cm}$

There will be two shifts; due to the glass block as well as water:

$\u2206{t}_{1}=\left[1-\frac{1}{{\mu}_{g}}\right]{t}_{g}=\left[1-\frac{1}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]\times 8=2.26\mathrm{cm}\phantom{\rule{0ex}{0ex}}\u2206{t}_{2}=\left[1-\frac{1}{{\mu}_{\mathrm{w}}}\right]{t}_{\mathrm{w}}=\left[1-\frac{1}{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right]\times \left(25.7-8\right)=4.44\mathrm{cm}$

Hence, the total shift = (Δ*t*_{1}+ Δ*t*_{2}) = (2.66 + 4.44) cm = 7.1 cm above the bottom.

#### Page No 414:

#### Question 22:

Consider the situation in figure. The bottom of the pot is a reflecting plane mirror, *S* is a small fish and *T* is a human eye. Refractive index of water is μ. (a) At what distance(s) from itself will the fish see the image(s) of the eye? (b) At what distance(s) from itself will the eye see the image(s) of the fish.

Figure

#### Answer:

Given,

Refractive index of water = *μ.*

Height of the pot = *H*

Let us take *x *as the distance of the image of the eye formed above the surface of the water as seen by the fish.

We can infer from from the diagram,

$\frac{H}{x}=\frac{\mathrm{Real}\mathrm{depth}}{\mathrm{Apparent}\mathrm{depth}}=\frac{1}{\mu}\phantom{\rule{0ex}{0ex}}\mathrm{or}x=\mu H$

The distance of the direct image $=\frac{H}{2}+\mu H=H\left(\mu +\frac{1}{2}\right)$

Similarly, image through mirror = $\frac{H}{2}+(H+x)=\frac{3H}{2}+\mu H=H\left(\frac{3}{2}+\mu \right)$

b) We know that:

$\frac{{\displaystyle \raisebox{1ex}{$H$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{y}=\mu ,Soy=\frac{H}{2\mu}$

Where, *y* is the distance of the image of the fish below the surface as seen by the eye.

Direct image = $H+y=H+\frac{H}{2\mu}=H\left(1+\frac{1}{2\mu}\right)$

Again another image of fish will be formed *H*/2 below the mirror.

Real depth for that image of fish becomes *H* + *H*/2 = 3*H*/2

So, apparent depth from the surface of water = 3*H*/2*μ*

So, distance of the image from the eye $=\frac{H}{2}+\frac{3H}{2\mu}=H\left(1+\frac{3}{2\mu}\right)$

#### Page No 414:

#### Question 23:

A small object is placed at the centre of the bottom of a cylindrical vessel of radius 3 cm and height 4 cm filled completely with water. Consider the ray leaving the vessel through a corner. Suppose this ray and the ray along the axis of the vessel are used to trace the image. Find the apparent depth of the image and the ratio of real depth to the apparent depth under the assumptions taken. Refractive index of water = 1.33.

#### Answer:

Given,

Refractive index of water *μ = *1.33

Radius of the cylindrical vessel = 3 cm

Height of the cylindrical vessel = 4 cm

Let *x* be the length of BD

According to the diagram,

$\frac{x}{3}=cotr....\left(\mathrm{i}\right)$

Using Snell's law,

$\frac{\mathrm{sin}i}{\mathrm{sin}r}=\frac{1}{1.33}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}r=\frac{4}{3}\mathrm{sin}i=\frac{4}{3}\times \frac{3}{5}=\frac{4}{5}\phantom{\rule{0ex}{0ex}}\mathrm{As}\mathrm{sin}i=\frac{BC}{AC}=\frac{3}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cot}r=\frac{3}{4}.....\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}$

From (i) and (ii),

$\frac{x}{3}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{9}{4}=2.25\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Hence, the ratio of real and apparent depth of the image will be $4:2.25=1.78$

#### Page No 414:

#### Question 24:

A cylindrical vessel, whose diameter and height both are equal to 30 cm, is placed on a horizontal surface and a small particle *P* is placed in it at a distance of 5.0 cm from the centre. An eye is placed at a position such that the edge of the bottom is just visible (see figure). The particle *P* is in the plane of drawing. Up to what minimum height should water be poured in the vessel to make the particle *P* visible?

Figure

#### Answer:

Given,

Diameter and height (*h*) of the cylindrical vessel = 30 cm

Therefore, its radius (*r*) = 15 cm

We know the refractive index of water (μ_{w}) = 1.33 $=\frac{4}{3}$

Using Snell's law,

$\frac{\mathrm{sin}i}{\mathrm{sin}r}=\frac{1}{{\mathrm{\mu}}_{\mathrm{w}}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{sin}i}{\mathrm{sin}r}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\mathrm{As}r=45\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}i=\frac{3}{4\sqrt{2}}$

Point P will be visible when the refracted ray makes an angle of 45˚ at the point of refraction.

Let *x *be the distance of point P from X.

$\mathrm{tan}45\xb0=\frac{x+10}{d}\phantom{\rule{0ex}{0ex}}\Rightarrow d=x+10...\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\mathrm{i}=\frac{x}{d}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3}{\sqrt{23}}=\frac{d-10}{d}\left(\because \mathrm{sin}\mathrm{i}=\frac{3}{4\sqrt{2}}\Rightarrow \mathrm{tan}\mathrm{i}=\frac{3}{\sqrt{23}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3}{\sqrt{23}}-1=\frac{-10}{d}\phantom{\rule{0ex}{0ex}}\Rightarrow d\mathit{=}\frac{\sqrt{23}\times 10}{\sqrt{23}-3}\phantom{\rule{0ex}{0ex}}d=26.7\mathrm{cm}$

Hence, the required minimum height of water = 26.7 cm

#### Page No 414:

#### Question 25:

A light ray is incident at an angle of 45° with the normal to a √2 cm thick plate (μ = 2.0). Find the shift in the path of the light as it emerges out from the plate.

#### Answer:

Given,

Angle of incidence = 45˚

Thickness of the plate = $\sqrt{2}\mathrm{cm}$

Refractive index (μ) of the plate = 2.0

Applying Snell's law,

$\frac{\mathrm{sin}i}{\mathrm{sin}r}=\frac{1}{\mu}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{sin}i}{\mathrm{sin}r}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}r=\frac{\mathrm{sin}45\xb0}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}r=\frac{1}{2\sqrt{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow r=21\xb0$

Therefore, $\theta =\left(45\xb0-21\xb0\right)\phantom{\rule{0ex}{0ex}}=24\xb0$

From figure, we know that BD is the shift in path which is equal to (AB)sin 24˚

$=0.406\times \mathrm{AB}=\frac{\mathrm{AE}}{\mathrm{cos}21\xb0}\times 0.406\phantom{\rule{0ex}{0ex}}=0.62\mathrm{cm}$

Hence, the required shift in the path of light is 0.62 cm.

#### Page No 414:

#### Question 26:

An optical fibre (μ = 1.72) is surrounded by a glass coating (μ = 1.50). Find the critical angle for total internal reflection at the fibre-glass interface.

#### Answer:

Given,

Refractive index of the optical fibre is represented by μ_{o} = 1.72

Refractive index of glass coating is represented by μ_{g}= 1.50

Let the critical angle for glass be *θ*_{c}

Using the Snell's law,

$\frac{\mathrm{sin}i}{\mathrm{sin}r}=\frac{\mathrm{sin}{\theta}_{\mathrm{c}}}{\mathrm{sin}90\xb0}=\frac{{\mathrm{\mu}}_{\mathrm{g}}}{{\mathrm{\mu}}_{0}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{sin}{\theta}_{\mathrm{c}}}{\mathrm{sin}90\xb0}=\frac{1.50}{1.72}=\frac{75}{86}\phantom{\rule{0ex}{0ex}}\Rightarrow {\theta}_{\mathrm{c}}={\mathrm{sin}}^{-1}\left(\frac{75}{86}\right)$

Hence, the required critical angle is ${\mathrm{sin}}^{-1}\left(\frac{75}{86}\right)$

#### Page No 414:

#### Question 27:

A light ray is incident normally on the face *AB* of a right-angled prism *ABC* (μ = 1.50) as shown in figure. What is the largest angle ϕ for which the light ray is totally reflected at the surface *AC*?

Figure

#### Answer:

Given,

Refractive index (*μ*) of prism = 1.50

Let us take *θ*_{c} as the critical angle for the glass.

So, According to Snell's law,

$\frac{\mathrm{sin}{\mathrm{\theta}}_{c}}{\mathrm{sin}90\xb0}=\frac{1}{\mu}$

$\Rightarrow \mathrm{sin}\mathrm{\theta}c=\frac{1}{1.5}=\frac{2}{3}$

$\Rightarrow {\theta}_{\mathrm{c}}={\mathrm{sin}}^{-1}\frac{2}{3}$

Condition for total internal reflection: 90° − $\varphi $> θ_{c}

⇒ $\varphi $ < 90° $-$ θ_{c}

$\Rightarrow \varphi 90\xb0-{\mathrm{sin}}^{-1}\left(\frac{2}{3}\right)$

$\Rightarrow \varphi {\mathrm{cos}}^{-1}\left(\frac{2}{3}\right)$

Hence, the largest angle for which light is totally reflected at the surface *AC* is ${\mathrm{cos}}^{-1}\left(\frac{2}{3}\right)$.

#### Page No 414:

#### Question 28:

Find the maximum angle of refraction when a light ray is refracted from glass (μ = 1.50) to air.

#### Answer:

Given,

Refractive index (*μ*) of the glass = 1.5

We know from the definition of a critical angle (*θ*_{c}) that if a refracted angle measures more than 90°, then total internal reflection takes place.

Hence, the maximum angle of refraction is 90°.

#### Page No 414:

#### Question 29:

Light is incident from glass (μ = 1.5) to air. Sketch the variation of the angle of deviation δ with the angle of incident *i* for 0 < *i* < 90°.

#### Answer:

Given,

Refractive index of glass, *μ*_{g} = 1.5

Refractive index of air, *μ*_{a}= 1.0

Angle of incidence 0° < *i* < 90

Let us take *θ*_{c} as the Critical angle

$\Rightarrow \frac{\mathrm{sin}\theta \mathrm{c}}{\mathrm{sin}r}=\frac{{\mu}_{\mathrm{a}}}{{\mu}_{\mathrm{g}}}$

$\Rightarrow \frac{\mathrm{sin}{\theta}_{\mathrm{c}}}{\mathrm{sin}90\xb0}=\frac{1}{1.5}=0.66\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}{\theta}_{\mathrm{c}}=0.66\phantom{\rule{0ex}{0ex}}\Rightarrow {\theta}_{\mathrm{c}}=\mathrm{sin}{}^{-1}\left(0.66\right)$

⇒ *θ*_{c} = 40°48"

The angle of deviation (*δ*) due to refraction from glass to air increases as the angle of incidence increases from 0° to 40°48". The angle of deviation (*δ*) due to total internal reflection further increases from 40°48" to 45° and then it decreases, as shown in the graph.

#### Page No 414:

#### Question 30:

Light is incident from glass (μ = 1.50) to water (μ = 1.33). Find the range of the angle of deviation for which there are two angles of incidence.

#### Answer:

Given,

Refractive index of glass $:{\mu}_{\mathrm{g}}=1.5=\frac{3}{2}$

Refractive index of water$:{\mu}_{\mathrm{w}}=1.33=\frac{4}{3}$

As per the question,

For two angles of incidence,

1. When light passes straight through the Normal,

⇒ Angle of incidence = 0°

⇒ Angle of refraction = 0°

⇒ Angle of deviation = 0°

2. When light is incident at critical angle *θ*_{c},

$\frac{\mathrm{sin}{\theta}_{\mathrm{c}}}{\mathrm{sin}r}=\frac{{\mu}_{\mathrm{w}}}{{\mu}_{\mathrm{g}}}$

(since the light is passing from glass to water)

$\Rightarrow \mathrm{sin}{\theta}_{\mathrm{c}}=\frac{8}{9}$

$\Rightarrow {\theta}_{\mathrm{c}}={\mathrm{sin}}^{-1}\left(\frac{8}{9}\right)=62.73\xb0$

⇒ Angle of deviation

=90° − *θ*_{c}

$=90-{\mathrm{sin}}^{-1}\frac{8}{9}$

$={\mathrm{cos}}^{-1}\frac{8}{9}$

=37.27°

Here, if the angle of incidence increased beyond the critical angle, total internal reflection occurs and deviation decreases.

Therefore, the range of angle of deviation is in between 0 to 37.27° or ${\mathrm{cos}}^{-1}\left(\frac{8}{9}\right)$.

#### Page No 414:

#### Question 31:

Light falls from glass (μ = 1.5) to air. Find the angle of incidence for which the angle of deviation is 90°.

#### Answer:

Given,

Light falls from glass to air.

Refractive index (*μ*) of glass = 1.5

Critical angle (*θ*_{c}) $={\mathrm{sin}}^{-1}\left(\frac{1}{\mathrm{\mu}}\right)$

$={\mathrm{sin}}^{-1}\left(\frac{1}{1.5}\right)=41.80$

We know that the maximum attainable angle of deviation in refraction is (90° − 41.8°)

= 47.2°

In this case, total internal reflection must have taken place.

In reflection,

Deviation = 180° − 2*i* = 90°

⇒ 2*i* = 90°

⇒ *i* = 45°

Hence, the required angle of incidence is 45°.

#### Page No 414:

#### Question 32:

A point source is placed at a depth *h* below the surface of water (refractive index = μ). (a) Show that light escapes through a circular area on the water surface with its centre directly above the point source. (b) Find the angle subtended by a radius of the area on the source.

#### Answer:

Given,

Refractive index is *μ*

(a)

Let the point source be P, which is placed at a depth of *h *from the surface of water.

Let us take *x* as the radius of the circular area.

and let *θ*_{c} be the critical angle.

Thus,

$\frac{x}{h}=\mathrm{tan}{\theta}_{c}\phantom{\rule{0ex}{0ex}}\frac{x}{h}=\frac{\mathrm{sin}{\theta}_{c}}{\sqrt{1-{\mathrm{sin}}^{2}{\theta}_{c}}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{\mu}$}\right.}}{\sqrt{1-{\displaystyle \frac{1}{{\mathrm{\mu}}^{2}}}}}\left(\because \mathrm{sin}{\theta}_{c}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{\mu}$}\right.\right)\phantom{\rule{0ex}{0ex}}\frac{x}{h}=\frac{{\displaystyle 1}}{\sqrt{{\mathrm{\mu}}^{2}-1}}\phantom{\rule{0ex}{0ex}}x=\frac{h}{\sqrt{{\mathrm{\mu}}^{2}-1}}$

Clearly from figure, the light escapes through a circular area at a fixed distance *r* on the water surface, directly above the point source.

That makes a circle, the centre of which is just above P.

(b)

The angle subtended by the radius of the circular area on the point source P:

$\Rightarrow \mathrm{sin}{\mathrm{\theta}}_{\mathrm{c}}=\frac{1}{\mathrm{\mu}}$

$\Rightarrow {\theta}_{\mathrm{c}}={\mathrm{sin}}^{-1}\left(\frac{1}{\mathrm{\mu}}\right)$

#### Page No 414:

#### Question 33:

A container contains water up to a height of 20 cm and there is a point source at the centre of the bottom of the container. *A* rubber ring of radius *r* floats centrally on the water. The ceiling of the room is 2.0 m above the water surface. (a) Find the radius of the shadow of the ring formed on the ceiling if *r* = 15 cm. (b) Find the maximum value of *r* for which the shadow of the ring is formed on the ceiling. Refractive index of water = 4/3.

#### Answer:

Given,

Height (*h*) of the water in the container = 20 cm

Ceiling of the room is 2.0 m above the water surface.

Radius of the rubber ring = *r*

Refractive index of water = 4/3

(a)

From the figure, we can infer:

$\mathrm{sin}i=\frac{15}{25}\phantom{\rule{0ex}{0ex}}$

Using Snell's law, we get:

$\frac{\mathrm{sin}i}{\mathrm{sin}r}=\frac{1}{\mathrm{\mu}}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}i=\frac{4}{5}$

From the figure, we have:

$\mathrm{tan}r=\frac{x}{2}\phantom{\rule{0ex}{0ex}}\mathrm{So},\phantom{\rule{0ex}{0ex}}\mathrm{sin}r=\frac{\mathrm{tan}r}{\sqrt{1+{\mathrm{tan}}^{2}r}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\sqrt{1+{\displaystyle \frac{{x}^{2}}{4}}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\displaystyle x}}{\sqrt{4+{\displaystyle {x}^{2}}}}=\frac{4}{5}$

$\Rightarrow 25{x}^{2}=16(4+{x}^{2})\phantom{\rule{0ex}{0ex}}\Rightarrow 9{x}^{2}=64\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{8}{3}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Total radius of the shadow = $\frac{8}{3}+0.15=2.81\mathrm{m}$

(b)

Condition for the maximum value of *r*:

Angle of incidence should be equal to the critical angle, i.e., $i={\theta}_{\mathrm{c}}$.

Let us take *R* as the maximum radius.

Now,

$\mathrm{sin}{\theta}_{\mathrm{c}}=\frac{\mathrm{sin}{\theta}_{\mathrm{c}}}{\mathrm{sin}r}\phantom{\rule{0ex}{0ex}}=\frac{R}{\sqrt{{R}^{2}+20}}=\frac{3}{4}(\mathrm{sin}r\mathit{}=1)\phantom{\rule{0ex}{0ex}}\Rightarrow 16{R}^{2}=9{R}^{2}+9\times 400\phantom{\rule{0ex}{0ex}}\Rightarrow 7{R}^{2}=9{\mathrm{R}}^{2}+9\times 400\phantom{\rule{0ex}{0ex}}\Rightarrow R=22.67\mathrm{cm}$

#### Page No 414:

#### Question 34:

Find the angle of minimum deviation for an equilateral prism made of a material of refractive index 1.732. What is the angle of incidence for this deviation?

#### Answer:

Given,

Refractive index (*μ*) of the material from which prism is made = 1.732

We know refractive index is given by:

$\mu =\frac{\mathrm{sin}\left[{\displaystyle \frac{{\delta}_{\mathrm{min}}+A}{2}}\right]}{\mathrm{sin}\left[{\displaystyle \frac{A}{2}}\right]}$

Where *δ*_{min} is the angle of minimum deviation and *A *is the angle of prism = 60˚

$\Rightarrow 1.732\times \mathrm{sin}(30\xb0)=\mathrm{sin}\left(\frac{{\delta}_{\mathrm{min}}+60\xb0}{2}\right)$

$\Rightarrow \frac{1.732}{2}=\mathrm{sin}\left(\frac{{\delta}_{\mathrm{min}}+60\xb0}{2}\right)$

$\Rightarrow \left(\frac{{\delta}_{\mathrm{min}}+60\xb0}{2}\right)=60\xb0$

δ_{min} = 60°

δ_{min} = 2*i* − A

2*i** = *120°

*i* = 60°

Hence, the required angle of deviation is 60°.

#### Page No 414:

#### Question 35:

Find the angle of deviation suffered by the light ray shown in figure. The refractive index μ = 1.5 for the prism material.

Figure

#### Answer:

Given,

The refractive index of the prism material (μ) = 1.5

Angle of prism form the figure = 4˚

We know that,

$\mu =\frac{\mathrm{sin}\left({\displaystyle \frac{\mathrm{A}+{\delta}_{\mathrm{m}}}{2}}\right)}{\mathrm{sin}{\displaystyle \frac{A}{2}}}\phantom{\rule{0ex}{0ex}}\mu =\frac{\left({\displaystyle \frac{\mathrm{A}+{\delta}_{\mathrm{m}}}{2}}\right)}{{\displaystyle \frac{A}{2}}}\left(\because forsmallangle\mathrm{sin}\theta \approx \theta \right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{A}+{\delta}_{\mathrm{m}}}{2\xb0}\phantom{\rule{0ex}{0ex}}1.5=\frac{4\xb0+{\delta}_{\mathrm{m}}}{2\xb0}\phantom{\rule{0ex}{0ex}}{\delta}_{\mathrm{m}}=2\xb0$

Hence, the angle of deviation is 2˚

#### Page No 414:

#### Question 36:

A light ray, going through a prism with the angle of prism 60°, is found to deviate by 30°. What limit on the refractive index can be put from these data?

#### Answer:

Given,

The angle of the prism (*A*) = 60˚

The angle of deviation (*δ*_{m}) = 30˚

$\mathrm{Refractive}\mathrm{index},\phantom{\rule{0ex}{0ex}}\mathrm{\mu}\le \frac{\mathrm{sin}\left({\displaystyle \frac{A+{\delta}_{\mathrm{m}}}{2}}\right)}{\mathrm{sin}\left({\displaystyle \frac{A}{2}}\right)}\phantom{\rule{0ex}{0ex}}\mathrm{\mu}\le \frac{\mathrm{sin}\left({\displaystyle \frac{60\xb0+{\delta}_{\mathrm{m}}}{2}}\right)}{\mathrm{sin}\left({\displaystyle \frac{60\xb0}{2}}\right)}\phantom{\rule{0ex}{0ex}}\mathrm{\mu}\le 2\mathrm{sin}\left(\frac{60\xb0+{\delta}_{\mathrm{m}}}{2}\right)$

As there is one ray that has been found which has deviated by 30˚, the angle of minimum deviation should be either equal to or less than 30˚ but it can not be more than 30˚.

Therefore,

$\mathrm{\mu}\le 2\mathrm{sin}\left(\frac{60\xb0+{\delta}_{\mathrm{m}}}{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{\mu}\le 2\mathrm{sin}\left(\frac{60\xb0+30\xb0}{2}\right)=2\mathrm{sin}\left(45\xb0\right)$

Refractive index (μ) will be more if angle of deviation (*δ*_{m}) is more.

$\mathrm{\mu}\le 2\times \frac{1}{\sqrt{2}}$

$\Rightarrow \mathrm{\mu}\le \sqrt{2}$

Hence, the required limit of refractive index is $\sqrt{2}$

#### Page No 414:

#### Question 37:

Locate the image formed by refraction in the situation shown in figure.

Figure

#### Answer:

Given,

Let the refractive indices of two mediums be μ_{1}=1.0 and μ_{2} =1.5

Point C is the centre of curvature, the distance between C and the pole is 20 cm.

Therefore, radius of curvature (*R)* = 20 cm

Distance between source S and pole is 25 cm.

Therefore, object distance (*u)* = −25

Using lens equation,

$\frac{{\mathrm{\mu}}_{2}}{v}-\frac{{\mathrm{\mu}}_{1}}{u}=\frac{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}}{\mathrm{R}}$

$\Rightarrow \frac{1.5}{v}-\frac{1}{(-25)}=\frac{0.5}{20}$

$\Rightarrow \frac{1.5}{v}=\frac{-3}{200}$

$\therefore v=-\frac{200\times 1.5}{0.3\times 10}=-100$

Hence, the required location of the image is 100 cm from the pole and on the side of S.

#### Page No 414:

#### Question 38:

A spherical surface of radius 30 cm separates two transparent media *A *and *B* with refractive indices 1.33 and 1.48 respectively. The medium *A* is on the convex side of the surface. Where should a point object be placed in medium *A* so that the paraxial rays become parallel after refraction at the surface?

#### Answer:

Given,

Spherical surface of radius (*R*) = 30 cm

Medium A has refractive index (*μ _{1}*) = 1.33

Medium B has refractive index (

*μ*) = 1.48

_{2}Medium A is the convex side of surface.

Since,

We know that paraxial rays become parallel after refraction

i.e, the image of the point object will be formed at infinity.

Therefore

*v = ∞*

Using the lens equation,

$\frac{{\mathrm{\mu}}_{2}}{v}-\frac{{\mathrm{\mu}}_{1}}{u}=\frac{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}}{\mathrm{R}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1.48}{\infty}-\frac{1.33}{u}=\frac{1.48-1.33}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{1.33}{u}-\frac{0.15}{30}\phantom{\rule{0ex}{0ex}}\therefore u=-266.0\mathrm{cm}$

Hence, the object is placed at a distance of 266.0 cm from the convex surface on side A.

#### Page No 415:

#### Question 39:

Figure shows a transparent hemisphere of radius 3.0 cm made of a material of refractive index 2.0. (a) A narrow beam of parallel rays is incident on the hemisphere as shown in the figure. Are the rays totally reflected at the plane surface? (b) Find the image formed by the refraction at the first surface. (c) Find the image formed by the reflection or by the refraction at the plane surface. (d) Trace qualitatively the final rays as they come out of the hemisphere.

Figure

#### Answer:

Given,

The radius of the transparent hemisphere (*R*) = 3.0 cm

Refractive index of the material (*μ _{2}*) = 2.0

Let the critical angle be

*θ*

_{c}

∴ critical angle is given by

*θ*

_{c}=${\mathrm{sin}}^{-1}\left(\frac{1}{{\mathrm{\mu}}_{2}}\right)={\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)=30\xb0$

(a)

From the figure it is seen that the angle of incidence is greater than the critical angle, so the rays are totally reflected at the plane surface.

(b)

Using the lens equation:

$\frac{{\mathrm{\mu}}_{2}}{v}-\frac{{\mathrm{\mu}}_{1}}{u}=\frac{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{v}-\left(-\frac{1}{\infty}\right)=\frac{2-1}{3}(\mathrm{for}\mathrm{parallel}\mathrm{rays}u=\infty )\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{v}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow v=6\mathrm{cm}$

If we complete the sphere then the image will be formed diametrically opposite to A.

(c)

By internal reflection, the image is formed in front of A.

#### Page No 415:

#### Question 40:

A small object is embedded in a glass sphere (μ = 1.5) of radius 5.0 cm at a distance 1.5 cm left to the centre. Locate the image of the object as seen by an observer standing (a) to the left of the sphere and (b) to the right of the sphere.

#### Answer:

Given,

Radius of the sphere = 5.0 cm

Refractive index of the sphere (μ_{1}) = 1.5

An object is embedded in the glass sphere 1.5 cm left to the centre.

(a)

When the image is seen by observer from left of the sphere,

from surface the object distance (*u*) = − 3.5 cm

μ_{1} = 1.5

μ_{2}_{ }= 1

*v*_{1} = ?

Using lens equation:

$\frac{{\mathrm{\mu}}_{2}}{{v}_{1}}-\frac{{\mathrm{\mu}}_{1}}{u}=\frac{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}}{\mathrm{R}}$

$\frac{1}{{v}_{1}}-\frac{1.5}{-(3.5)}=\frac{-0.5}{-5}$

$\frac{1}{{v}_{1}}=\frac{1}{10}-\frac{3}{7}$

$=\frac{7-30}{70}$

$=\frac{-23}{70}$

${v}_{1}=\frac{-70}{23}\simeq -3\mathrm{cm}$

So the image will be formed at 2 cm (5 cm $-$ 3cm) left to centre.

(b)

When the image is seen by observer from the right of the sphere,

*u* = −(5.0 + 1.5) = − 6.5,

*R* = −5.00 cm

μ_{1} = 1.5, μ_{2} = 1, *v* = ?

Using lens equation:

$\frac{{\mathrm{\mu}}_{2}}{{v}_{1}}-\frac{{\mathrm{\mu}}_{1}}{u}=\frac{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}}{R}$

$\frac{1}{v}-\frac{1.5}{6.3}=\frac{1}{10}$

$\frac{1}{v}=\frac{1}{10}-\frac{3}{13}$

$=\frac{13-30}{130}$

$=\frac{-17}{130}$

$v=\frac{-130}{17}=-7.65\mathrm{cm}$

Therefore, the image will be formed 7.6 − 5 = 2.6 towards left from centre.

#### Page No 415:

#### Question 41:

A biconvex thick lens is constructed with glass (μ = 1.50). Each of the surfaces has a radius of 10 cm and the thickness at the middle is 5 cm. Locate the image of an object placed far away from the lens.

#### Answer:

Given,

Biconvex lens with each surface has a radius (*R*_{1}_{ }= *R*_{2} *= R*) = 10 cm,

and thickness of the lens (*t*) = 5 cm

Refractive index of the lens (μ) = 1.50

Object is at infinity (∴ u = ∞ )

First refraction takes place at A.

We know that,

$\frac{{\mathrm{\mu}}_{g}}{v}-\frac{{\mathrm{\mu}}_{\mathrm{a}}}{u}=\frac{{\mathrm{\mu}}_{\mathrm{g}}-{\mathrm{\mu}}_{\mathrm{a}}}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1.5}{v}-\left(-\frac{1}{\infty}\right)=\frac{1.5-1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1.5}{v}=\frac{0.5}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow v=30\mathrm{cm}$

Now, the second refraction is at B.

For this, a virtual object is the image of the previous refraction.

Thus, *u* = (30 − 5) = 25 cm

$\frac{{\mathrm{\mu}}_{\mathrm{a}}}{v}-\frac{{\mathrm{\mu}}_{\mathrm{g}}}{u}=\frac{{\mathrm{\mu}}_{\mathrm{a}}-{\mathrm{\mu}}_{\mathrm{g}}}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}-\frac{1.5}{25}=\frac{1-1.5}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}-\frac{1.5}{25}=\frac{-0.5}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow v=9.1\mathrm{cm}$

Hence, the image is formed 9.1 cm far from the second refraction or second surface of the lens.

#### Page No 415:

#### Question 42:

A narrow pencil of parallel light is incident normally on a solid transparent sphere of radius *r*. What should be the refractive index is the pencil is to be focussed (a) at the surface of the sphere, (b) at the centre of the sphere.

#### Answer:

Given,

The radius of the transparent sphere = *r*

Refraction at convex surface.

As per the question,

*u* = −∞, μ_{1} = 1, μ_{2} = ?

(a)

When image is to be focused on the surface,

Image distance (*v*) = 2*r*, Radius of curvature (*R*) = *r*

We know that,

$\frac{{\mu}_{2}}{v}-\frac{{\mu}_{1}}{u}=\frac{{\mu}_{2}-{\mu}_{1}}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\mu}_{2}}{2r}-\left(\frac{1}{-\infty}\right)=\frac{{\mu}_{2}-1}{r}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\mu}_{2}}{2r}=\frac{{\mu}_{2}-1}{r}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mu}_{2}=2{\mu}_{2}-2\phantom{\rule{0ex}{0ex}}\Rightarrow {\mu}_{2}=2$

(b)

When the image is to be focused at the centre,

Image distance (*v*) = *r*, Radius of curvature (*R*) = *r*

$\frac{{\mu}_{2}}{v}-\frac{{\mu}_{1}}{u}=\frac{{\mu}_{2}-{\mu}_{1}}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\mu}_{2}}{r}-\left(\frac{1}{-\infty}\right)=\frac{{\mu}_{2}-1}{r}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\mu}_{2}}{r}=\frac{{\mu}_{2}-1}{r}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mu}_{2}={\mu}_{2}-1\phantom{\rule{0ex}{0ex}}$

The above equation is impossible.

Hence, the image cannot be focused at centre.

#### Page No 415:

#### Question 43:

One end of a cylindrical glass rod (μ = 1.5) of radius 1.0 cm is rounded in the shape of a hemisphere. The rod is immersed in water (μ = 4/3) and an object is placed in the water along the axis of the rod at a distance of 8.0 cm from the rounded edge. Locate the image of the object.

#### Answer:

Given,

Radius (*R*) of the cylindrical rod = 1.0 cm

Refractive index (μ_{g}) of the rod = 1.5 = $\frac{3}{2}$

Refractive index (μ_{w}) of water = 4/3

We know:

$\frac{{\mu}_{g}}{v}-\frac{{\mu}_{w}}{u}=\frac{{\mu}_{g}-{\mu}_{w}}{R}$

As per the question, *u* = −8 cm.

Now,

$\frac{3}{2v}-\left(-\frac{4}{3\times 8}\right)=\frac{{\displaystyle \frac{3}{2}}-{\displaystyle \frac{4}{3}}}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3}{2v}+\frac{1}{6}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\infty $

Hence, the image will be formed at infinity (∞).

#### Page No 415:

#### Question 44:

A paperweight in the form of a hemisphere of radius 3.0 cm is used to hold down a printed page. An observer looks at the page vertically through the paperweight. At what height above the page will the printed letters near the centre appear to the observer?

#### Answer:

Given:

Radius of the paperweight (*R*) = 3 cm

Refractive index of the paperweight (μ_{2}) = 3/2

Refractive index of the air (μ_{1}) = 1

In the first case, the refraction is at A.

*u =* 0 and *R =* ∞

We know:

$\frac{{\mathrm{\mu}}_{2}}{v}-\frac{{\mathrm{\mu}}_{1}}{u}=\frac{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{v}-\frac{1}{0}=\frac{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}}{\infty}\phantom{\rule{0ex}{0ex}}\Rightarrow v=0$

Therefore, the image of the letter is formed at the point.

For the second case, refraction is at point B.

Here,

Object distance, *u *= −3 cm

*R *= − 3 cm

μ_{1} = 3/2

μ_{2} = 1

Thus, we have:

$\frac{{\mathrm{\mu}}_{2}}{v}-\frac{{\mathrm{\mu}}_{1}}{u}=\frac{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}}{R}\phantom{\rule{0ex}{0ex}}\frac{1}{v}-\frac{3}{2\times (-3)}=\frac{1-{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{-3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}+\frac{3}{2\times 3}=\frac{{\displaystyle 1}}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{{\displaystyle 1}}{6}-\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow v=-3\mathrm{cm}$

Hence, there will be no shift in the final image.

#### Page No 415:

#### Question 45:

Solve the previous problem if the paperweight is inverted at its place so that the spherical surface touches the paper.

#### Answer:

Given,

Taking the radius of the paperweight as its thickness = 3 cm

Refractive index of the paperweight (μ_{g}) = 3/2

Refractive index of the air (μ_{1}) = 1

Image shift is given by:

$\u2206t=\left(1-\frac{1}{\mathrm{\mu}}\right)t\phantom{\rule{0ex}{0ex}}=\left(1-\frac{2}{3}\right)3\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{3}\right)3\phantom{\rule{0ex}{0ex}}=1\mathrm{cm}$

The upper surface of the paperweight is flat and the spherical spherical surface is in contact with the printed letter.

Therefore, we will take it as a simple refraction problem.

Hence, the image will appear 1 cm above point A.

#### Page No 415:

#### Question 46:

A hemispherical portion of the surface of a solid glass sphere (μ = 1.5) of radius *r* is silvered to make the inner side reflecting. An object is placed on the axis of the hemisphere at a distance 3*r* from the centre of the sphere. The light from the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted at the unsilvered part. Locate the final image formed.

#### Answer:

As shown in the figure, OQ = 3*r** *and OP = r

Thus, PQ = 2*r*

For refraction at APB,

we know that,

$\frac{{\mu}_{2}}{v}-\frac{{\mu}_{1}}{u}=\frac{{\mu}_{2}-{\mu}_{1}}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1.5}{v}-\left(\frac{1}{-2r}\right)=\frac{0.5}{r}=\frac{1}{2r}\left(\because u=-2r\right)\phantom{\rule{0ex}{0ex}}\Rightarrow v=\infty \phantom{\rule{0ex}{0ex}}$

For the reflection in the concave mirror,

*u* = ∞

Thus, *v *= focal length of mirror = *r*/2

For the refraction of APB of the reflected image,

*u* = −3*r*/2

$\Rightarrow \frac{1}{v}-\frac{1.5}{3r/2}=\frac{-0.5}{-r}\left(\mathrm{Here},{\mathrm{\mu}}_{1}=1.5,{\mathrm{\mu}}_{2}=1andR=-r\right)\phantom{\rule{0ex}{0ex}}\Rightarrow v=-2r\phantom{\rule{0ex}{0ex}}$

As negative sign indicates images formed inside APB, so image should be at C.

Therefore, the final image is formed on the reflecting surface of the sphere.

#### Page No 415:

#### Question 47:

The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface as shown in figure. (a) Where should a pin be placed on the axis so that its image is formed at the same place? (b) If the concave part is filled with water (μ = 4/3), find the distance through which the pin should be moved so that the image of the pin again coincides with the pin.

Figure

#### Answer:

(a) Let *F* be the the focal length of the given concavo-convex lens. Then,

$\frac{1}{F}=\frac{1}{{f}_{1}}+\frac{1}{{f}_{m}}+\frac{1}{{f}_{1}}$

$=\frac{2}{{f}_{1}}+\frac{1}{{f}_{m}}$ $\left[\because \frac{1}{{f}_{m}}=(\mathrm{\mu}-1)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)\right]$

$\frac{1}{F}=\frac{2}{30}+\frac{1}{15}=\frac{2}{15}$

⇒ *F *= 7.5 cm

Hence, *R* = 15 cm

Therefore, the pin should be placed at a distance of 15 cm from the lens.

(b) If the concave part is filled with water,

For focal length *F*'

$\frac{1}{F\mathit{\text{'}}}=\frac{2}{{f}_{w}}+\frac{2}{{f}_{1}}+\frac{1}{{f}_{m}}$

$=\frac{2}{90}+\frac{2}{30}+\frac{1}{15}$ $\left[\because \frac{1}{{f}_{w}}=\left(\frac{4}{3}-1\right)\left(+\frac{1}{30}\right)\right]$

$\therefore F\text{'}=\frac{45}{7}\mathrm{cm}$

Thus, pin should be placed at a distance of $\frac{90}{7}$ cm from the lens.

#### Page No 415:

#### Question 48:

A double convex lens has focal length 25 cm. The radius of curvature of one of the surfaces is double of the other. Find the radii, if the refractive index of the material of the lens is 1.5.

#### Answer:

Given,

Double convex lens of focal length, *f* = 25 cm

Refractive index of the material, *μ* = 1.5

As per the question, the radius of curvature of one surface is twice that of the other.

i.e. *R*_{1}*= R* and *R*_{2}= −2*R*

(according to sign conventions)

Using the lens maker formula $\frac{1}{f}=(\mathrm{\mu}-1)\left[\frac{\mathit{1}}{{\mathit{R}}_{\mathit{1}}}\mathit{-}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{2}}}\right]$, we have:

$\frac{1}{25}=\left(1.5-1\right)\left[\frac{1}{R}-\left(\frac{1}{-2R}\right)\right]$

$\Rightarrow \frac{1}{25}=0.5\left[\frac{3R}{2}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{25}=\left[\frac{3R}{4}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow R=18.75\mathrm{cm}$

*R*_{1} = 18.75 cm and* **R*_{2} = 37.5 cm

Hence, the required radii of the double convex lens are 18.75 cm and 37.5 cm.

#### Page No 415:

#### Question 49:

The radii of curvature of a lens are + 20 cm and + 30 cm. The material of the lens has a refracting index 1.6. Find the focal length of the lens (a) if it is placed in air, and (b) if it is placed in water (μ = 1.33).

#### Answer:

Given,

Radii of curvature of a lens, (R_{1}_{)} = +20 cm and R_{2} = +30 cm

Refractive index of the material of the lens, (μ) = 1.6

Refractive index of water, (μ_{water}) = 1.33

(a)

When the lens is placed in air,

Using lens maker formula:

$\frac{1}{f}=(\mathrm{\mu}-1)\left[\frac{\mathit{1}}{{\mathit{R}}_{\mathit{1}}}\mathit{-}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{2}}}\right]$

$\frac{1}{f}=0.6\left[\frac{1}{20}-\frac{1}{30}\right]\phantom{\rule{0ex}{0ex}}$

$f=\frac{0.6}{1060\times 10}$

* f* = 100 cm

Thus, the focal length of the lens is 100 cm when it is placed in air.

(b)When the lens is placed in water

$\frac{1}{f}=\left[\frac{{\mathrm{\mu}}_{\mathrm{lens}}}{{\mathrm{\mu}}_{\mathrm{water}}}-1\right]\left[\frac{\mathit{1}}{{R}_{\mathit{1}}}\mathit{-}\frac{\mathit{1}}{{R}_{\mathit{2}}}\right]$

$=\left(\frac{1.60}{1.33}-1\right)\left[\frac{1}{60}\right]$

$=\frac{28}{133\times 60}\simeq \frac{1}{300}$

* $\Rightarrow $f* = 300 cm.

Thus, the focal length of the lens is 300 cm when it is placed in water.

#### Page No 415:

#### Question 50:

Lenses are constructed by a material of refractive index 1.50. The magnitude of the radii of curvature are 20 cm and 30 cm. Find the focal lengths of the possible lenses with the above specifications.

#### Answer:

Given,

Refractive index of the material, (μ) = 1.50

Magnitudes of the radius of curvature:

*R*_{1} = 20 cm and *R*_{2} = 30 cm

From the given data we can make four possible lenses, using the lens maker formula.

$\frac{1}{f}=(\mathrm{\mu}-1)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)$

Four possible lenses:

(a) 1^{st} lens is double convex, in which *R*_{1} = +20 cm and *R*_{2} = −30 cm

$=0.5\left[\frac{1}{20}-\frac{1}{\left(-30\right)}\right]=\frac{0.5\times 5}{60}$

*f* = +24 cm

(b) 2^{nd} lens is double concave, in which *R*_{1} = −20 cm and *R*_{2} = +30 cm

$=0.5\left[\frac{-1}{20}-\frac{1}{30}\right]$

*f* = −24 cm

(c) 3^{rd} lens is concave concave in which *R*_{1} = −20 cm and *R*_{2} = −30 cm

$=0.5\left[\frac{-1}{20}-\frac{1}{\left(-30\right)}\right]$

*f* = −120 cm

(d) 4^{th} lens is concave convex, in which *R*_{1} = +20 cm and *R*_{2} = +30 cm

$=0.5\left[\frac{1}{20}-\frac{1}{30}\right]$

*f* = +120 cm

#### Page No 415:

#### Question 51:

A thin lens made of a material of refractive index μ_{2} has a medium of refractive index μ_{1} on one side and a medium of refractive index μ_{3} on the other side. The lens is biconvex and the two radii of curvature have equal magnitude *R*. *A* beam of light travelling parallel to the principal axis is incident on the lens. Where will the image be formed if the beam is incident from (a) the medium μ_{1}_{ }and (b) from the medium μ_{3}?

#### Answer:

Given,

A biconvex lens with two radii of curvature that have equal magnitude *R.*

Refractive index of the material of the lens is μ_{2}.

First medium of refractive index is μ_{1}_{.}

Second medium of refractive index is μ_{3}.

As per the question, the light beams are travelling parallel to the principal axis of the lens.

i.e., *u* (object distance) = ∞

(a) The light beam is incident on the lens from first medium μ_{1}.

Thus, refraction takes place at first surface

Using equation of refraction,

$\frac{{\mathrm{\mu}}_{2}}{v}-\frac{{\mathrm{\mu}}_{1}}{u}=\frac{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}}{R}$

Where, *v* is the image distance.

Applying sign convention, we get:

$\frac{{\mathrm{\mu}}_{2}}{v}-\frac{{\mathrm{\mu}}_{1}}{\left(-\infty \right)}=\frac{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}}{{\mathrm{\mu}}_{2}R}\phantom{\rule{0ex}{0ex}}\Rightarrow \Rightarrow v=\frac{{\mathrm{\mu}}_{2}R}{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}}\phantom{\rule{0ex}{0ex}}$

Now, refraction takes place at 2^{nd} surface

Thus,$\frac{{\mathrm{\mu}}_{3}}{v}-\frac{{\mathrm{\mu}}_{2}}{u}=\frac{{\mathrm{\mu}}_{3}-{\mathrm{\mu}}_{2}}{R}\phantom{\rule{0ex}{0ex}}$

Here, the image distance of the previous case becomes object distance:

$\frac{{\mathrm{\mu}}_{3}}{v}=-\left[\frac{{\mathrm{\mu}}_{3}-{\mathrm{\mu}}_{2}}{R}-\frac{{\mathrm{\mu}}_{2}}{{\mathrm{\mu}}_{2}R}\left({\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}\right)\right]\phantom{\rule{0ex}{0ex}}=-\left[\frac{{\mathrm{\mu}}_{3}-{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{2}+{\mathrm{\mu}}_{1}}{R}\right]\phantom{\rule{0ex}{0ex}}v=-\left[\frac{{\mathrm{\mu}}_{3}R}{{\mathrm{\mu}}_{3}-2{\mathrm{\mu}}_{2}+{\mathrm{\mu}}_{1}}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Therefore, the image is formed at $\frac{{\mathrm{\mu}}_{3}R}{2{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}-{\mathrm{\mu}}_{3}}$

(b) The light beam is incident on the lens from second medium μ_{3}.

Thus, refraction takes place at second surface.

Using equation of refraction,

$\frac{{\mathrm{\mu}}_{2}}{v}-\frac{{\mathrm{\mu}}_{3}}{u}=\frac{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{3}}{R}$

Where, *v* is the image distance

Applying sign convention, we get:

$\frac{{\mathrm{\mu}}_{2}}{v}-\frac{{\mathrm{\mu}}_{3}}{\left(-\infty \right)}=\frac{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{3}}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{3}}{{\mathrm{\mu}}_{2}R}\phantom{\rule{0ex}{0ex}}\Rightarrow \Rightarrow v=\frac{{\mathrm{\mu}}_{2}R}{{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{3}}\phantom{\rule{0ex}{0ex}}$

Now, refraction takes place at 2nd surface.

Thus,$\frac{{\mathrm{\mu}}_{1}}{v}-\frac{{\mathrm{\mu}}_{2}}{u}=\frac{{\mathrm{\mu}}_{1}-{\mathrm{\mu}}_{2}}{R}\phantom{\rule{0ex}{0ex}}$

Here, the image distance of the previous case becomes object distance.

$\frac{{\mathrm{\mu}}_{1}}{v}=-\left[\frac{{\mathrm{\mu}}_{1}-{\mathrm{\mu}}_{2}}{R}-\frac{{\mathrm{\mu}}_{2}}{{\mathrm{\mu}}_{2}R}\left({\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{3}\right)\right]\phantom{\rule{0ex}{0ex}}=-\left[\frac{{\mathrm{\mu}}_{1}-{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{2}+{\mathrm{\mu}}_{3}}{R}\right]\phantom{\rule{0ex}{0ex}}v=-\left[\frac{{\mathrm{\mu}}_{1}R}{{\mathrm{\mu}}_{3}-2{\mathrm{\mu}}_{2}+{\mathrm{\mu}}_{1}}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Therefore, the image is formed at $\frac{{\mathrm{\mu}}_{1}R}{2{\mathrm{\mu}}_{2}-{\mathrm{\mu}}_{1}-{\mathrm{\mu}}_{3}}$

#### Page No 415:

#### Question 52:

A convex lens has a focal length of 10 cm. Find the location and nature of the image if a point object is placed on the principal axis at a distance of (a) 9.8 cm, (b) 10.2 cm from the lens.

#### Answer:

Given:

Focal length (*f*) of the convex lens = 10 cm

(a) As per the question, the object distance (*u*) is 9.8 cm.

The lens equation is given by:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

=$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{10}-\frac{1}{9.8}$

=$\frac{1}{v}=\frac{9.8-10}{98}=\frac{-0.2}{98}$

*= v* = − 98 × 5

= − 490 cm (Same side of the object)

*v* = 490 cm (Virtual and on on the side of object)

Magnification of the image $=\frac{v}{u}$

$=\frac{-490}{-9.8}\phantom{\rule{0ex}{0ex}}=50$

Therefore, the image is erect and virtual.

(b) Object distance, *u* = 10.2 cm

The lens equation is given by:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

=$\frac{1}{v}=\frac{1}{10}-\frac{1}{10.2}$

$=\frac{10.2-10}{102}=\frac{0.2}{102}$

*= v* = 102 × 5 = 510 cm (Real and on the opposite side of the object)

Magnification of the image $=\frac{v}{u}$

$=\frac{510}{-9.8}\phantom{\rule{0ex}{0ex}}=-52.04$

Therefore, the image is real and inverted.

#### Page No 415:

#### Question 53:

A slide projector has to project a 35 mm slide (35 mm × 23 mm) on a 2 m × 2 m screen at a distance of 10 m from the lens. What should be the focal length of the lens in the projector?

#### Answer:

Given,

We are projecting a slide of 35 mm $\times $ 23 mm on a 2 m $\times $ 2 m screen using projector.

Therefore, the magnification required by the projector is:

$m=\frac{v}{u}$

Here,

*v* = Image distance

*u* = Object distance

We will take 35 mm as the object size

∵ 35 mm > 23 mm

$\therefore m=\frac{v}{u}=\frac{{h}_{i}}{{h}_{o}}\phantom{\rule{0ex}{0ex}}\frac{v}{u}=\frac{2\times {10}^{3}}{35}\phantom{\rule{0ex}{0ex}}\Rightarrow u=\frac{35}{2\times {10}^{3}}\times v\phantom{\rule{0ex}{0ex}}\Rightarrow u=\frac{35}{2\times {10}^{3}}\times 10\phantom{\rule{0ex}{0ex}}\Rightarrow u=0.175\mathrm{mm}$

The lens formula is given by

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{10}+\frac{1}{0.175}=\frac{1}{f}$

$\Rightarrow \frac{10175}{1750}=\frac{1}{f}$

$\Rightarrow f=\frac{1750}{10175}\phantom{\rule{0ex}{0ex}}f=0.172\mathrm{mm}$

Hence, the required focal length is 0.172 mm.

#### Page No 415:

#### Question 54:

A particle executes a simple harmonic motion of amplitude 1.0 cm along the principal axis of a convex lens of focal length 12 cm. The mean position of oscillation is at 20 cm from the lens. Find the amplitude of oscillation of the image of the particle.

#### Answer:

When the particle is at point B,

$\frac{1}{{v}_{\mathrm{B}}}=\frac{1}{f}+\frac{1}{u}$

$\frac{1}{{v}_{B}}=\frac{1}{12}-\frac{1}{19}$

$\Rightarrow {v}_{\mathrm{B}}=\frac{12\times 19}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{\mathrm{B}}=32.57\mathrm{cm}$

When particle is at point A,

$\frac{1}{{v}_{\mathrm{A}}}=\frac{1}{f}+\frac{1}{u}$

$\frac{1}{{v}_{\mathrm{A}}}=\frac{1}{12}-\frac{1}{21}$

$\Rightarrow {v}_{\mathrm{A}}=\frac{12\times 21}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{\mathrm{A}}=28\mathrm{cm}$

$\mathrm{Amplitude}\mathrm{of}\mathrm{image}=\frac{{v}_{\mathrm{A}}-{v}_{\mathrm{B}}}{2}$

$=\frac{4.5}{2}=2.2\mathrm{cm}$

#### Page No 415:

#### Question 55:

An extended object is placed at a distance of 5.0 cm from a convex lens of focal length 8.0 cm. (a) Draw the ray diagram (to the scale) to locate the image and from this, measure the distance of the image from the lens. (b) Find the position of the image from the lens formula and see how close the drawing is to the correct result.

#### Answer:

Given,

Object distance, (*u*) = 5.0 cm

Focal length (*f*) of convex lens = 8.0 cm

(a)

(b)

Using lens formula:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Where *v* is the image distance,

on putting the given values we get:

$\frac{1}{v}-\frac{1}{\left(-5\right)}=\frac{1}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{8}-\frac{1}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{-3}{40}\phantom{\rule{0ex}{0ex}}\therefore v=-13.3\mathrm{cm}$

Hence, the position of the image from the lens is 13.3 cm.

#### Page No 415:

#### Question 56:

A pin of length 2.00 cm is placed perpendicular to the principal axis of a converging lens. An inverted image of size 1.00 cm is formed at a distance of 40.0 cm from the pin. Find the focal length of the lens and its distance from the pin.

#### Answer:

Given,

Height of the object, (*h*_{0}) = 2 cm,

Height of the image, (*h*_{1}) = 1 cm

distance between image and object, ($-$*u* + *v*) = 40 cm

We know that,

Magnification is given by:

$m=\frac{{h}_{i}}{{h}_{0}}=\frac{v}{u},\frac{1}{2}=\frac{v}{-u}$

*u* = −2*v*

Using lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}-\frac{1}{-2v}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}+\frac{1}{2v}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow f=\frac{2v}{3}$

As per the question,

= −*u* + *v* = 40 cm,

= −(−2v) + *v* = 40 cm,

= 3*v* = 40 cm

$v=13.3=\frac{40}{3}\mathrm{cm},u=\frac{80}{3}$, *u* = 26.66 cm.

*∴ f* = 8.9 cm

Hence, the required focal length is 8.9 cm and object distance is 26.66 cm.

#### Page No 416:

#### Question 57:

A convex lens produces a double size real image when an object is placed at a distance of 18 cm from it. Where should the object be placed to produce a triple size real image?

#### Answer:

Given,

Object distance, *u* = $-$18 cm

Size of the image is two times the size of the object

i.e., *h*_{1} = 2*h*_{0},

We know that,

magnification is given by, *m *= $\frac{v}{u}$

$\Rightarrow \frac{{h}_{1}}{{h}_{0}}=\frac{v}{u}=2$

∴ *v* = 2*u* = 36 cm.

Using lens formula:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{36}+\frac{1}{18}=\frac{1}{f}$

$\Rightarrow \frac{1}{f}=\frac{3}{36}$

⇒ *f* = 12 cm.

For*, **h*_{1} = 3*h*_{0}

$\frac{{h}_{i}}{{h}_{0}}=\frac{v}{u}$

$\Rightarrow \frac{v}{u}=3\Rightarrow v=3u$

$-\frac{1}{3u}-\frac{1}{u}=\frac{1}{12}\phantom{\rule{0ex}{0ex}}$

*u* = $-$16 cm.

Therefore, the object should be placed at a distance of 16 cm in front of the lens.

#### Page No 416:

#### Question 58:

A pin of length 2.0 cm lies along the principal axis of a converging lens, the centre being at a distance of 11 cm from the lens. The focal length of the lens is 6 cm. Find the size of the image.

#### Answer:

Given,

Length of the pin = 2.0 cm

Focal length (*f*) of the lens = 6 cm

As per the question, the centre of the pin is 11 cm away from the lens.

i.e., the object distance (*u*) = 10 cm

Since, we have to calculate the image of A and B, Let the image be A' and B'

So, the length of the A'B' = size of the image.

Using lens formula:

$\frac{1}{{v}_{\mathrm{A}}}-\frac{1}{{u}_{\mathrm{A}}}=\frac{1}{f}$

Where* **v*_{A} and *u*_{A} are the image and object distances from point A.

$\Rightarrow \frac{1}{{v}_{\mathrm{A}}}-\frac{1}{-10}=\frac{1}{6}$

$\frac{1}{{v}_{\mathrm{A}}}=\frac{1}{6}-\frac{1}{10}=\frac{1}{15}$

${v}_{\mathrm{A}}=15\mathrm{cm}$

Similarly for point B,

Lens formula:

$\frac{1}{{v}_{\mathrm{B}}}-\frac{1}{{u}_{\mathrm{B}}}=\frac{1}{f}$

Where* **v*_{A} and *u*_{A} are the image and object distances from point B.

$\frac{1}{{v}_{\mathrm{B}}}-\frac{1}{-12}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{\mathrm{B}}=12\mathrm{cm}$

Length of image = *v*_{A} − *v*_{B} = 15 − 12 = 3 cm.

#### Page No 416:

#### Question 59:

The diameter of the sun is 1.4 × 10^{9} m and its distance from the earth is 1.5 × 10^{11} m. Find the radius of the image of the sun formed by a lens of focal length 20 cm.

#### Answer:

Given,

Diameter of the sun = 1.4 × 10^{9} m

Distance between sun and earth is taken as object distance (*u*) = − 150 × 10^{11} cm,

Focal length (*f*) of the lens = 20 cm

Using lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

The object distance is very large as compared to focal length of the lens.

Hence, the image is formed at the focus.

$\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{150\times {10}^{11}}$

$=\frac{750\times {10}^{9}-1}{150\times {10}^{11}}$

$\simeq \frac{750\times {10}^{9}}{150\times {10}^{11}}$

$\Rightarrow v=\frac{150\times {10}^{11}}{750\times {10}^{9}}\phantom{\rule{0ex}{0ex}}$

We know, Magnification (*m*) is given by:

$\left(m\right)=\frac{v}{u}=\frac{{h}_{2}}{{h}_{1}}$

$\Rightarrow {h}_{2}=\frac{v}{u}\times {h}_{1}$

$=\frac{150\times {10}^{11}}{750\times {10}^{9}\times 150\times {10}^{11}}\times 0.7\times {10}^{12}\mathrm{mm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=0.93\mathrm{mm}$

Hence, the required radius of the image of the sun is 0.93 mm.

#### Page No 416:

#### Question 60:

A 5.0 diopter lens forms a virtual image which is 4 times the object placed perpendicularly on the principal axis of the lens. Find the distance of the object from the lens.

#### Answer:

Given,

Power of the lens (*P*) = $\frac{1}{\mathrm{Focal}\mathrm{length}}$ = 5.0 D

The height of the image is four times the height of the object.

i.e.

${h}_{i}=4{h}_{0},\frac{{h}_{i}}{{h}_{0}}=4$

We know magnification (*m*) is also given by

$m=\frac{{h}_{i}}{{h}_{0}}=\frac{v}{u}\Rightarrow \frac{v}{u}=4,v=4u$

The lens maker formula is given by

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Here, *v* is the image distance and *u* is the object distance.

Now,

$\frac{1}{f}=P=5\mathrm{and}f=\frac{1}{5}\mathrm{m}=20\mathrm{cm}\phantom{\rule{0ex}{0ex}}\frac{1}{4u}-\frac{1}{u}=\frac{1}{20}\phantom{\rule{0ex}{0ex}}-\frac{3}{4u}=\frac{1}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow u=-\frac{60}{4}=-15\mathrm{cm}$

Hence, the required object distance is 15 cm.

#### Page No 416:

#### Question 61:

A diverging lens of focal length 20 cm and a converging mirror of focal length 10 cm are placed coaxially at a separation of 5 cm. Where should an object be placed so that a real image is formed at the object itself?

#### Answer:

Let the object be placed at a distance x cm from the lens (away from the mirror).

For the concave lens (I^{st} refraction) *u* = − *x*, *f* = − 20 cm

From lens formula:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}=\frac{1}{(-20)}+\frac{1}{(-x)}\Rightarrow v=-\left({\displaystyle \frac{20x}{x+20}}\right)\phantom{\rule{0ex}{0ex}}$

Thus, the virtual image due to the first refraction lies on the same side as that of object (A'B').

This image becomes the object for the concave mirror,

For the mirror,

$u=-\left(5+\frac{20x}{x+20}\right)\phantom{\rule{0ex}{0ex}}=-\left(\frac{25x+100}{x+20}\right)\phantom{\rule{0ex}{0ex}}f=-10\mathrm{cm}$

From mirror equation,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{-10}+\frac{x+20}{25x+100}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{10x+200-25x-100}{250(x+4)}$

$\Rightarrow v=\frac{250(x+4)}{100-15x}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{250(x+4)}{15x-100}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{50(x+4)}{(3x-20)}$

Thus, this image is formed towards left of the mirror.

Again for second refraction in concave lens,

$u=-\left[\frac{5-50(x+4)}{3x-20}\right]$

(assuming that image of mirror is formed between the lens and mirror 3*x* − 20),

*v* = + *x* (since the final image is produced on the object A"B")

using lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{x}+\frac{1}{{\displaystyle \frac{\left[5-50(x\times 4)\right]}{3x-20}}}=\frac{1}{-20}$

⇒ 25*x*^{2} − 1400*x* − 6000 = 0

⇒ *x*^{2} − 56*x* − 240 = 0

⇒ (*x* − 60) (*x* + 4) = 0

So, *x* = 60 m

The object should be placed at a distance 60 cm from the lens farther away from the mirror, so that the final image is formed on itself.

#### Page No 416:

#### Question 62:

A converging lens of focal length 12 cm and a diverging mirror of focal length 7.5 cm are placed 5.0 cm apart with their principal axes coinciding. Where should an object be placed so that its image falls on itself?

#### Answer:

Let the object be placed at a distance *x *cm from the lens (away from the mirror).

For the convex lens (1^{st} refraction) *u* = − *x*, *f* = − 12 cm

From the lens formula:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}=\frac{1}{(-12)}+\frac{1}{(-x)}\Rightarrow v=-\left({\displaystyle \frac{12x}{x+12}}\right)$

Thus, the virtual image due to the first refraction lies on the same side as that of object A'B'.

This image becomes the object for the convex mirror,

For the mirror,

$u=-\left(5+\frac{12x}{x+12}\right)\phantom{\rule{0ex}{0ex}}=-\left(\frac{17x+60}{x+12}\right)\phantom{\rule{0ex}{0ex}}f=-7.5\mathrm{cm}$

From mirror equation,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{-7.5}+\frac{x+12}{17x+60}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{17x+60-7.5}{7.5(17x+60)}$

$\Rightarrow v=\frac{7.5(17x+60)}{52.5-127.5x}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{250(x+4)}{15x-100}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{50(x+4)}{(3x-20)}$

Thus, this image is formed towards the left of the mirror.

Again for second refraction in concave lens,

$u=-\left[\frac{5-50(x+4)}{3x-20}\right]$

(assuming that the image of mirror formed between the lens and mirror is 3*x* − 20),

*v* = + *x* (since, the final image is produced on the object A"B")

Using lens formula:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{x}+\frac{1}{{\displaystyle \frac{\left[5-50(x\times 4)\right]}{3x-20}}}=\frac{1}{-20}$

⇒ 25*x*^{2} − 1400*x* − 6000 = 0

⇒ *x*^{2} − 56*x* − 240 = 0

⇒ (*x* − 60) (*x* + 4) = 0

Thus, *x* = 60 m

The object should be placed at a distance 60 cm from the lens farther away from the mirror, so that the final image is formed on itself.

#### Page No 416:

#### Question 63:

A converging lens and a diverging mirror are placed at a separation of 15 cm. The focal length of the lens is 25 cm and that of the mirror is 40 cm. Where should a point source be placed between the lens and the mirror so that the light, after getting reflected by the mirror and then getting transmitted by the lens, comes out parallel to the principal axis?

#### Answer:

Given,

Distance between convex lens and convex mirror is 15 cm.

Focal length (*f*_{l}) of the lens is 25 cm.

Focal length (*f*_{2}) of the mirror is 40 cm.

Let *x* cm be the object distance from the mirror.

Therefore*,
u* = −

*x*cm

*v*= 25 − 15 = + 10 cm (∵ focal length of lens = 25 cm)

∴

*f*

_{l }= + 40 cm

Using lens formula:

$\Rightarrow \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{x}=\frac{1}{10}-\frac{1}{40}\Rightarrow x=\frac{40}{3}$

Thus, the object distance is $\left(15-{\displaystyle \frac{40}{3}}\right)=\frac{5}{3}$

= 1.67 cm from the lens

#### Page No 416:

#### Question 64:

A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart with common principal axis. A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens. Find the locations of the two images formed.

#### Answer:

Given:

Convex lens of focal length (*f*_{l}) = 15 cm

Concave mirror of focal length (*f*_{2}) = 10 cm

Distance between the lens and the mirror = 50 cm

Point source is placed at a distance of 40 cm from the lens.

It means the point source is at the focus of the mirror.

Thus, two images will be formed:

(a) One due to direct transmission of light through the lens.

(b) One due to reflection and then transmission of the rays through the lens.

__Case 1__:

(S') For the image by direct transmission, we have:

Object distance (*u*) = − 40 cm

*f*_{l} = 15 cm

Using the lens formula, we get:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{15}+\frac{1}{(-40)}\phantom{\rule{0ex}{0ex}}=\frac{40-15}{40\times 15}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{40\times 15}{40-15}\phantom{\rule{0ex}{0ex}}$

Therefore, *v* is 24 cm to the left from the lens.

__Case II__:

(S') Since the object is placed at the focus of the mirror, the rays become parallel to the lens after reflection.

∴ Object distance (*u*) = ∞ ⇒ *f*_{l} = 15 cm

$\Rightarrow \frac{1}{v}-\frac{1}{u}=\frac{1}{15}$

Thus, *v* is 15 cm to the left of the lens.

#### Page No 416:

#### Question 65:

Consider the situation described in the previous problem. Where should a point source be placed on the principal axis so that the two images form at the same place?

#### Answer:

Given,

Convex lens of focal length (*f*_{l}) = 15 cm

Concave mirror of focal length (*f*_{m}) = 10 cm

Distance between lens and mirror = 50 cm

Thus, two images will be formed,

(a) One due to direct transmission of light through lens.

(b) One due to reflection and then transmission of the rays through lens.

Let the point source be placed at a distance of '*x*' from the lens as shown in the Figure, so that images formed by lens and mirror coincide.

For lens,

We use lens formula:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{15x}{x-15}...(\mathrm{i})$

For mirror,

Object distance will be (50 − *x*)

We use the formula:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{15}$

*u* = − (50 − *x*)

*f*_{m} = − 10 cm

$\mathrm{So},\frac{1}{{v}_{\mathrm{m}}}+\frac{1}{-(50-x)}=-\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{v}_{\mathrm{m}}}=\frac{1}{50-x}-\frac{1}{10}\phantom{\rule{0ex}{0ex}}=\frac{10-50+x}{10(50-x)}\phantom{\rule{0ex}{0ex}}{v}_{\mathrm{m}}=\frac{x-40}{10(50-x)}...\left(\mathrm{ii}\right)$

Since the distance between lens and mirror is 50 cm,

*v* − *v*_{m} = 50

from equation (i) and (ii):

$\Rightarrow \frac{15x}{x-15}-\frac{10(50-x)}{(x-40)}=50$

⇒(3*x*^{2} − 120*x*) − (100*x* − 2*x*^{2} − 1500 + 30*x*)

= 10 (*x*^{2} − 55*x* + 600)

⇒ 5*x*^{2}^{ }− 250*x* − 1500 = 10*x*^{2} − 550*x* + 6000

⇒ 5*x*^{2} − 300*x* + 4500 = 0

⇒ *x*^{2} − 60*x* + 900 = 0

⇒ (*x* − 30)^{2} = 0

*x* = 30 cm.

∴ Point source should be placed at a distance of 30 cm from the lens on the principal axis, so that the two images form at the same place.

#### Page No 416:

#### Question 66:

A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart. If a pin of length 2.0 cm is placed 30 cm from the lens farther away from the mirror, where will the final image form and what will be the size of the final image?

#### Answer:

Given,

Convex lens of focal length *f*_{l} = 15 cm

Concave mirror of focal length f_{m} = 10 cm

Distance between mirror and lens = 50 cm

Length of the pin (object length) *h*_{0} = 2.0 cm

As per the question

The pin (object) is placed at a distance of 30 cm from the lens on the principle axis.

Using lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{{f}_{\mathrm{l}}}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{u{f}_{\mathrm{l}}}{u+{f}_{\mathrm{l}}}$

Since, *u* = −30 cm and *f*_{l} = 15 cm

$\mathrm{So},v=\frac{(-30)\times 15}{(-30+15)}=\frac{-450}{-15}=30\mathrm{cm}$

From the figure it can be seen that image of the object (AB) is real and inverted (A'B') and it is of the same size as the object. This image (A'B') is at a distance of 20 cm from the concave mirror, which is formed at the centre of curvature of the mirror. Thus, mirror will form the image (A'B') at the same place as (A''B'') and will be of the same size. Now, due to the refraction from the lens, the final image (A''B'') will be formed at AB and will be of the same size as the object (AB).

#### Page No 416:

#### Question 67:

A point object is placed on the principal axis of a convex lens (*f* = 15 cm) at a distance of 30 cm from it. A glass plate (μ = 1.50) of thickness 1 cm is placed on the other side of the lens perpendicular to the axis. Locate the image of the point object.

#### Answer:

Given,

Convex lens of focal length (*f*) = 15 cm

Object distance, (*u*) = −30 cm

Using lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{15}-\frac{1}{30}=\frac{1}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow v=30$

Thus ,*v* = 30 cm

Therefore, the image of the object will be formed at a distance of 30 cm to the right side of the lens.

[since, μ_{g} = 1.5 and thickness (*t*) = 1 cm]

$\mathrm{\Delta}t=\left(1-{\displaystyle \frac{1}{{\mathrm{\mu}}_{\mathrm{g}}}}\right)t$

$=1-\frac{2}{3}=\frac{1}{3}=0.33\mathrm{cm}$

Hence, the image of the object will be formed at 30 + 0.33 = 30.33 cm from the lens on the right side, due to the glass having thickness 1 cm.

#### Page No 416:

#### Question 68:

A convex lens of focal length 20 cm and a concave lens of focal length 10 cm are placed 10 cm apart with their principal axes coinciding. A beam of light travelling parallel to the principal axis and having a beam diameter 5.0 mm, is incident on the combination. Show that the emergent beam is parallel to the incident one. Find the beam diameter of the emergent beam.

#### Answer:

Given,

Focal length of the convex lens, *f*_{d} = 20 cm

Focal length of the concave lens, *f*_{c} = 10 cm

Beam diameter of the incident light, *d *= 5.0 mm

Distance between both the lenses is 10 cm.

As per the question, the incident beam of light is parallel to the principal axis.

Let it be incident on the convex lens.

Now, let B be the focus of the convex lens where the image by the convex lens should be formed.

For the concave lens,

Object distance (*u*) = + 10 cm (Virtual object is on the right of concave lens.)

Focal length, *f*_{c} = − 10 cm

Using the lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{-10}+\frac{1}{+10}=\infty \phantom{\rule{0ex}{0ex}}\Rightarrow v=\infty $

Thus, after refraction in the concave lens, the emergent beam becomes parallel.

As shown, in triangles XYB and PQB,

$\frac{\mathrm{PQ}}{\mathrm{XY}}=\frac{\mathrm{RB}}{\mathrm{ZB}}=\frac{10}{20}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{PQ}=\frac{1}{2}\times 5=2.5\mathrm{mm}$

Thus, the beam diameter of the emergent light is 2.5 mm.

Similarly, we can prove that if the beam of light is incident on the side of the concave lens, the beam diameter (*d*) of the emergent light will be 1 cm.

#### Page No 416:

#### Question 69:

A diverging lens of focal length 20 cm and a converging lens of focal length 30 cm are placed 15 cm apart with their principal axes coinciding. Where should an object be placed on the principal axis so that its image is formed at infinity?

#### Answer:

Given,

Focal length of convex lens, *f*_{c} = 30 cm

Focal length of concave lens, *f*_{d} = 15 cm

Distance between both the lenses,* d* = 15 cm

Let (*f*) be the equivalent focal length of both the lenses.

$\frac{1}{f}=\frac{1}{{f}_{1}}+\frac{1}{{f}_{2}}-\frac{d}{{f}_{1}{f}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{30}+\left(\frac{1}{20}\right)-\left(\frac{5}{30(-20)}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{120}\phantom{\rule{0ex}{0ex}}\Rightarrow f=120\mathrm{cm}$

As focal length is positive, so it will be a converging lens.

Let '*d*_{1}' be the distance from diverging lens, so that the emergent beam is parallel to the principal axis and the image will be formed at infinity.

${d}_{1}=\frac{df}{{f}_{c}}=\frac{15\times 120}{30}=60\mathrm{cm}$

It should be placed 60 cm left to the diverging lens. The object should be placed

(120 − 60) = 60 cm from the diverging lens

Let *d*_{2} be the distance from the converging lens. Then,

${d}_{2}=\frac{df}{{f}_{d}}=\frac{15\times 120}{20}$

*d*_{2} = 90 cm

Thus, it should be placed (120 + 90) cm = 210 cm right to converging lens.

#### Page No 416:

#### Question 70:

A 5 mm high pin is placed at a distance of 15 cm from a convex lens of focal length 10 cm. A second lens of focal length 5 cm is placed 40 cm from the first lens and 55 cm from the pin. Find (a) the position of the final image, (b) its nature and (c) its size.

#### Answer:

Given:

Length of the high pin = 5.00 mm

Focal length of the first convex lens, *f* = 10 cm

Distance between the first lens and the pin = 15 cm

Focal length of the second convex lens, *f*_{1} = 5 cm

Distance between the first lens and the second lens = 40 cm

Distance between the second lens and the pin = 55 cm

(a) Image formed by the first lens:

Here,

Object distance, *u* = − 15 cm

Focal length, *f* = 10 cm

The lens formula is given by

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=\frac{1}{10}-\frac{1}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow v=30\mathrm{cm}$

Now,

This will be object for the second lens.

∴ Object distance for the second lens, ${u}_{1}$ = − (40 − 30)

$\Rightarrow {u}_{1}$ = − 10 cm

Focal length of the second lens, *f*_{1}_{ }= 5 cm

The lens formula is given by

$\frac{1}{{v}_{1}}=\frac{1}{{f}_{1}}+\frac{1}{{u}_{1}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{v}_{1}}=\frac{1}{5}-\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}=10\mathrm{cm}$

Therefore, the final position of the image is 10 cm right from the second lens.

(b) Magnification $\left(m\right)$ by the first lens is given by

$m=\frac{{h}_{i}}{{h}_{0}}=\frac{v}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow {h}_{i}=-\frac{5\times 30}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow {h}_{i}=-10\mathrm{mm}$

Magnification by the second lens:

$\frac{{h}_{\mathrm{final}}}{{h}_{i}}=\frac{v}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{10}{-10}=\frac{{h}_{\mathrm{final}}}{-10}\phantom{\rule{0ex}{0ex}}\Rightarrow {h}_{\mathrm{final}}=10\mathrm{mm}$

Thus, the image will be erect and real.

(c) Size of the final image is 10 mm.

#### Page No 416:

#### Question 71:

A point object is placed at a distance of 15 cm from a convex lens. The image is formed on the other side at a distance of 30 cm from the lens. When a concave lens is placed in contact with the convex lens, the image shifts away further by 30 cm. Calculate the focal lengths of the two lenses.

#### Answer:

Given,

Distance between point object and convex lens, *u* = 15 cm

Distance between the image of the point object and convex lens, *v* = 30 cm

Let* **f*_{c} be the focal length of the convex lens.

Then, using lens formula, we have:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{{f}_{c}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{f}_{\mathrm{c}}}=\frac{1}{30}-\frac{1}{(-15)}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{f}_{\mathrm{c}}}=\frac{1}{30}+\frac{1}{15}=\frac{3}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow {f}_{\mathrm{c}}=10\mathrm{cm}$

Now, as per the question, the concave lens is placed in contact with the convex lens. So the image is shifted by a distance of 30 cm.

Again, let *v*_{f} be the final image distance from concave lens, then:

${v}_{\mathrm{f}}$ = + (30 + 30) = + 60 cm

Object distance from the concave lens, *v *= 30 cm

If *f*_{d} is the focal length of concave lens then

Using lens formula, we have:

$\frac{1}{{v}_{\mathrm{f}}}-\frac{1}{v}=\frac{1}{{f}_{\mathrm{d}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{f}_{\mathrm{d}}}=\frac{1}{60}-\frac{1}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{f}_{d}}=\frac{30-60}{60\times 30}=\frac{-30}{60\times 30}\phantom{\rule{0ex}{0ex}}\Rightarrow {f}_{d}=-60\mathrm{cm}$

Hence, the focal length (*f*_{c}_{ }) of convex lens is 10 cm and that of the concave lens (*f*_{d} ) is 60 cm.

#### Page No 416:

#### Question 72:

Two convex lenses, each of focal length 10 cm, are placed at a separation of 15 cm with their principal axes coinciding. (a) Show that a light beam coming parallel to the principal axis diverges as it comes out of the lens system. (b) Find the location of the virtual image formed by the lens system of an object placed far away. (c) Find the focal length of the equivalent lens. (Note that the sign of the focal length is positive although the lens system actually diverges a parallel beam incident on it.)

#### Answer:

Given,

Focal length of each convex lens, *f = *10 cm

Distance between both the lens,* d *= 15 cm

(a) As shown in the figure, the light rays are falling parallel to the principal axis of the first lens, therefore the rays will converge within the focal length of the second lens. Hence, all the rays emerging from the lens system are diverging.

(b) Using lens formula for first convex lens,

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{10}+\frac{1}{(-\infty )}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow v=10\mathrm{cm}$

For the second convex lens,

Object distance, *u* = −(15 − 10) = −5 cm

If *v*_{1} is the image distance for the second convex lens, then,

Applying lens formula we have:

$\frac{1}{{v}_{1}}=\frac{1}{f}+\frac{1}{u}$

$\Rightarrow \frac{1}{{v}_{1}}=\frac{1}{10}+\frac{1}{-5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{v}_{1}}=\frac{-1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}=-10\mathrm{cm}$

Thus, the virtual image of the object will be at 5 cm from the first convex lens.

(c) The equivalent focal length of the lens system ^{$\left(F\right)$} is given by,

$\frac{1}{F}=\frac{1}{{f}_{1}}+\frac{1}{{f}_{2}}-\frac{d}{{f}_{1}{f}_{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Here},f={f}_{1}={f}_{2}=10\mathrm{cm}\phantom{\rule{0ex}{0ex}}\therefore \frac{1}{F}=\frac{1}{10}+\frac{1}{10}-\frac{15}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{F}=\frac{5}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow F=20\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Therefore the equivalent focal length (*F*) is 20 cm

#### Page No 416:

#### Question 73:

A ball is kept at a height *h* above the surface of a heavy transparent sphere made of a material of refractive index μ. The radius of the sphere is *R*. At *t* = 0, the ball is dropped to fall normally on the sphere. Find the speed of the image formed as a function of time for $t<\sqrt{\frac{2h}{g}}$. Consider only the image by a single refraction.

#### Answer:

Given,

A transparent sphere with refractive index* μ* and a of radius *R*.

A ball is kept at height *h* above the sphere.

As per the question, at *t* = 0 the ball is dropped normally on the sphere.

Let take the time taken to travel from point A to B as *t* .

Thus, the distance travelled by the ball is given by:

=$\frac{1}{2}g{t}^{2}$

Where *g* is acceleration due to gravity.

Therefore, the distance BC is given by:

$=h-\frac{1}{2}g{t}^{2}$

We are assuming this distance of the object from lens at any time* t.*

So here,

*u* = $-\left(h-\frac{1}{2}g{t}^{2}\right)$

Taking:

Refractive index of air, *μ*_{1} = 1

Refractive index of sphere, *μ*_{2} = * μ* (given)

Thus,

$\frac{\mathrm{\mu}}{v}-\frac{1}{-\left(h-{\displaystyle \frac{1}{2}}g{t}^{2}\right)}=\frac{\mathrm{\mu}-1}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{\mu}}{v}=\frac{\mathrm{\mu}-1}{R}-\frac{1}{\left(h-{\displaystyle \frac{1}{2}}g{t}^{2}\right)}=\frac{\left(\mathrm{\mu}-1\right)\left(h-\frac{1}{2}g{t}^{2}\right)-R}{R\left(h-\frac{1}{2}g{t}^{2}\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Let *v *be the image distance at any time *t.* Then,

$v=\frac{\mathrm{\mu}R\left(h-\frac{1}{2}g{t}^{2}\right)}{\left(\mathrm{\mu}-1\right)\left(h-\frac{1}{2}g{t}^{2}\right)-R}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Therefore, velocity of the image $\left(V\right)$ is given by,

$V=\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{\mathrm{\mu}R\left(h-\frac{1}{2}g{t}^{2}\right)}{\left(\mathrm{\mu}-1\right)\left(h-\frac{1}{2}g{t}^{2}\right)-R}\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\mu}{R}^{2}gt}{{\left[\left(\mathrm{\mu}-1\right)\left(h-\frac{1}{2}g{t}^{2}\right)-R\right]}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 416:

#### Question 74:

A particle is moving at a constant speed *V* from a large distance towards a concave mirror of radius *R* along its principal axis. Find the speed of the image formed by the mirror as a function of the distance *x* of the particle from the mirror.

#### Answer:

Given,

Radius of the concave mirror is *R*

Therefore focal length of the mirror, $f=\frac{R}{2}$

Velocity of the particle, *$V=\frac{\mathrm{d}x}{\mathrm{d}t}$*

Object distance, *u* = −x

Using mirror equation,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}$

On putting the respective values we get,

$\frac{1}{v}+\frac{1}{-x}=-\frac{2}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=-\frac{2}{R}+\frac{1}{x}=\frac{R-2x}{Rx}\phantom{\rule{0ex}{0ex}}\therefore v=\frac{Rx}{R-2x}\phantom{\rule{0ex}{0ex}}$

Velocity of the image is given by *V*_{1}

*${V}_{1}=\frac{\mathrm{dv}}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{Rx}{R-2x}\right]\phantom{\rule{0ex}{0ex}}=\frac{\left[\frac{\mathrm{d}}{\mathrm{d}t}\left(Rx\right)\left(R-2x\right)\right]-\left[\frac{\mathrm{d}}{\mathrm{d}t}\left(R-2x\right)\left(Rx\right)\right]}{{\left(R-2x\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{R\left[\left(\frac{\mathrm{dx}}{\mathrm{d}t}\right)\left(R-2x\right)\right]-\left[2\frac{\mathrm{dx}}{\mathrm{d}t}x\right]}{{\left(R-2x\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{R\left[\left(V\right)\left(R-2x\right)\right]-\left[2V\times 0\right]}{{\left(R-2x\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{V{R}^{2}}{(2x-R{)}^{2}}$*

#### Page No 416:

#### Question 75:

A small block of mass *m* and a concave mirror of radius *R* fitted with a stand lie on a smooth horizontal table with a separation *d* between them. The mirror together with its stand has a mass *m*. The block is pushed at *t* = 0 towards the mirror so that it starts moving towards the mirror at a constant speed *V* and collides with it. The collision is perfectly elastic. Find the velocity of the image (a) at a time *t* < *d*/*V*, (b) at a time *t* > *d*/*V*.

#### Answer:

Note :

(a) At time *t* = *t*,

Object distance, *u* = −(*d* − V*t*)

Here *d* > V*t*, $f=-\frac{R}{2}$

Mirror formula is given by:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{2}{\mathrm{R}}+\frac{1}{d-\mathrm{V}t}\phantom{\rule{0ex}{0ex}}$

$=\frac{-2(d-\mathrm{V}t)+\mathrm{R}}{\mathrm{R}(d-\mathrm{V}t)}$

$v=\frac{-\mathrm{R}(d-\mathrm{V}t)}{\mathrm{R}-2(d-\mathrm{V}t)}$

Differentiating w.r.t. '*t*':

$\frac{dv}{dt}=\frac{-\mathrm{RV}\left[\mathrm{R}-2(d-\mathrm{V}t)\right]-2\mathrm{V}\left[\mathrm{R}(d-\mathrm{V}t)\right]}{{\left[\mathrm{R}-2(d-\mathrm{V}t)\right]}^{2}}\phantom{\rule{0ex}{0ex}}=-\frac{{\mathrm{R}}^{2}\mathrm{V}}{{\left[2(d-\mathrm{V}t)-R\right]}^{2}}$

This is the required speed of mirror.

(b) When $t>\frac{d}{2}$ the collision between the mirror and mass will take place. As the collision is elastic, the object will come to rest and the mirror will start to move with the velocity V.

*u* = (*d* − V*t*).

At any time, $t>\frac{d}{2}$

The distance of mirror from the mass will be:

$x=V\left(t-\frac{d}{V}\right)=Vt-d$

Here,

Object distance, *u = $-\left(Vt-d\right)$ = $\left(d-Vt\right)$*

Focal length, *f* = $-\frac{R}{2}$

Now, mirror formula is given by:

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ = $\frac{1}{-{\displaystyle \frac{R}{2}}}-\frac{1}{d-Vt}$

$v\Rightarrow -\left[\frac{R\left(d-Vt\right)}{R-2\left(d-Vt\right)}\right]$

Velocity of image $\left({V}_{image}\right)$ is given by,

${V}_{image}=\frac{\mathrm{d}v}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{R\left(d-Vt\right)}{R+2\left(d-Vt\right)}\right]$

If* y *=* **d$-$**Vt*

$\frac{\mathrm{d}y}{\mathrm{dt}}$ = $-V$

Velocity of image:

${V}_{image}=\frac{d}{dt}\left[\frac{Ry}{R+2y}\right]=-{V}_{r}\left[\frac{R+2y-2y}{{\left(R+2y\right)}^{2}}\right]\phantom{\rule{0ex}{0ex}}=\frac{-V{R}^{2}}{{\left(R+2y\right)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

As the mirror is moving with velocity *V*,

Therefore,

*V*=*V*_{image} +* **V*_{mirror}

Absolute velocity of image = $V\left[1-\frac{{R}^{2}}{{\left(R+2y\right)}^{2}}\right]$

= $V\left[1-\frac{{R}^{2}}{2\left(Vt-d\right)-{R}^{2}}\right]$

#### Page No 416:

#### Question 76:

A gun of mass *M* fires a bullet of mass *m* with a horizontal speed *V*. The gun is fitted with a concave mirror of focal length *f* facing towards the receding bullet. Find the speed of separation of the bullet and the image just after the gun was fired.

#### Answer:

Given,

The focal length of the concave mirror is *f* and *M* is the mass of the gun. Horizontal speed of the bullet is* V.*

Let the recoil speed of the gun be *V*_{g}

Using the conservation of linear momentum we can write,

*MV*_{g} = *mV*

$\Rightarrow {V}_{\mathrm{g}}=\frac{m}{M}V$

Considering the position of gun and bullet at time *t* = *t*,

For the mirror, object distance, *u* = − (*Vt* + *V*_{g}*t*)

Focal length, *f* = − *f*

Image distance, *v* = ?

Using Mirror formula, we have:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{-f}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=-\frac{1}{f}+\frac{1}{(Vt+{V}_{\mathrm{g}}t)}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{-(Vt+{V}_{\mathrm{g}}t)+f}{(Vt+{V}_{\mathrm{g}}t)f}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{-(V+{V}_{\mathrm{g}})tf}{f-(V+{V}_{\mathrm{g}})t}$

The separation between image of the bullet and bullet at time *t *is given by:

$v-u=\frac{-(V+{V}_{\mathrm{g}})tf}{f-(V+Vg)t}+(V+{V}_{\mathrm{g}})t\phantom{\rule{0ex}{0ex}}=(V+{V}_{\mathrm{g}})t\left[\frac{f}{f-(V+{V}_{\mathrm{g}})t}+1\right]\phantom{\rule{0ex}{0ex}}=2\left(1+\frac{m}{M}\right)Vt$

Differentiating the above equation with respect to '*t*' we get,

$\frac{d(v-u)}{dt}=2\left(1+\frac{m}{M}\right)V$

Therefore, the speed of separation of the bullet and image just after the gun was fired is $2\left(1+\frac{m}{M}\right)V$.

#### Page No 416:

#### Question 77:

A mass *m* = 50 g is dropped on a vertical spring of spring constant 500 N m^{−1} from a height *h* = 10 cm as shown in figure. The mass sticks to the spring and executes simple harmonic oscillations after that. A concave mirror of focal length 12 cm facing the mass is fixed with its principal axis coinciding with the line of motion of the mass, its pole being at a distance of 30 cm from the free end of the spring. Find the length in which the image of the mass oscillates.

#### Answer:

Given,

Mass = 50 g

Spring constant, *k *= 500 Nm^{−1}

Height from where the mass is dropped on the spring, *h =* 10 cm

Focal length of concave mirror, *f* = 12 cm

Distance between the pole and the free end of the spring is 30 cm.

As per the question, when the mass is released it will stick to the spring and execute SHM.

At equilibrium position, weight of the mass is equal to force applied by the spring.

∴ *mg *= *kx*

where *g* is acceleration due to gravity

⇒ *x *=* mg/k*

$=\frac{50\times {10}^{-3}\times 10}{500}\phantom{\rule{0ex}{0ex}}={10}^{-3}\mathrm{m}=0.1\mathrm{cm}$

Therefore, the mean position of the SHM is (30 + 0.1 = 30.1) cm away from the pole of the mirror.

From the work energy principle,

final kinetic energy − initial kinetic energy = work done

⇒ 0 − 0 = *mg*(*h + δ*) − $\frac{1}{2}k{\delta}^{2}$

Where *δ *is the maximum compression of the spring.

*mg*(*h + δ*) = $\frac{1}{2}k{\delta}^{2}$

$\Rightarrow 50\times {10}^{-3}\times 10(0.1+\delta )=\frac{1}{2}500{\delta}^{2}\phantom{\rule{0ex}{0ex}}\delta =\frac{0.5\pm \sqrt{0.25+50}}{2\times 250}\phantom{\rule{0ex}{0ex}}=0.015\mathrm{m}=1.5\mathrm{cm}$

From the above figure,

Position of point B, (30 + 1.5) = 31.5 cm from the pole of the mirror

Therefore, amplitude of vibration of SHM, (31.5 − 30.1) = 1.4 cm

Position of point A from the pole of the mirror, (30.1 − 30.1) = 28.7 cm

For point A,

Object distance (*u*_{a}) = − 31.5

*f *= − 12 cm

By using the lens formula:

$\frac{1}{{v}_{\mathrm{a}}}-\frac{1}{{u}_{\mathrm{a}}}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{v}_{\mathrm{a}}}=\frac{1}{f}+\frac{1}{{u}_{\mathrm{a}}}\phantom{\rule{0ex}{0ex}}=\frac{1}{-12}+\frac{1}{-31.5}\phantom{\rule{0ex}{0ex}}={v}_{\mathrm{a}}=-19.38\mathrm{cm}$

For point B,

Object distance, (*u*_{b}) = − 28.7

*f *= − 12 cm

By using the lens formula:

$\frac{1}{{v}_{\mathrm{b}}}-\frac{1}{{u}_{\mathrm{b}}}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{v}_{\mathrm{b}}}=\frac{1}{f}+\frac{1}{{u}_{\mathrm{b}}}\phantom{\rule{0ex}{0ex}}=\frac{1}{-12}+\frac{1}{-28.7}\phantom{\rule{0ex}{0ex}}={v}_{\mathrm{b}}=-20.62\mathrm{cm}$

Hence, the image of the mass oscillates in length (20.62 − 19.38) = 1.24 cm

#### Page No 417:

#### Question 78:

Two concave mirrors of equal radii of curvature *R* are fixed on a stand facing opposite directions. The whole system has a mass *m* and is kept on a frictionless horizontal table (figure). Two blocks *A* and *B*, each of mass *m*, are placed on the two sides of the stand. At *t* = 0, the separation between *A* and the mirrors is 2 *R* and also the separation between *B* and the mirrors is 2 *R*. The block *B* moves towards the mirror at a speed *v*. All collisions which take place are elastic. Taking the original position of the mirrors-stand system to be *x* = 0 and *X*-axis along *AB*, find the position of the images of *A* and *B* at *t* = (a) $\frac{R}{v}$ (b) $\frac{3R}{v}$ (c) $\frac{5R}{v}$.

Figure

#### Answer:

Given,

*R* is the radii of curvature of two concave mirrors and *M* is the mass of the whole system.

Mass of the two blocks A and B is *m*.

As per the question,

At *t* = 0,

distance between block A and B is 2*R*

Block B is moving at a speed *v* towards the mirror.

Original position of the whole system at *x* = 0

(a)

At time *t* = $\frac{R}{v}$

The block B moved $\left(v\times \frac{R}{v}=R\right)$ *R *distance towards the mirror.

For block A,

object distance, *u* = − 2*R*

focal length of the mirror, *f* = − $\frac{R}{2}$

Using the mirror formula:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}=-\frac{2}{R}+\frac{1}{2R}\phantom{\rule{0ex}{0ex}}=-\frac{3}{2R}$

Therefore, *v* = − $\frac{2R}{3}$

Position of the image of block A is at $\frac{2R}{3}$ with respect to the given coordinate system.

For block B,

Object distance, *u* = − *R*

Focal length of the mirror, *f *= − $\frac{R}{2}$

Using the mirror formula:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=-\frac{2}{R}+\frac{1}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=-\frac{1}{R}$

Therefore, *v* = − *R*

Position of the image of block B is at the same place.

Similarly,

(b)

At time $\frac{3R}{v}$

Block B, after colliding with the mirror must have come to rest because the collision is elastic. Due to this, the mirror has travelled a distance *R *towards the block A, i.e., towards left from its initial position.

So, at this time

For block A

Object distance, *u* = − *R*

Focal length of the mirror, *f* = − $\frac{R}{2}$

Using the mirror formula:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}=-\frac{2}{R}+\frac{1}{R}\phantom{\rule{0ex}{0ex}}=-\frac{1}{R}$

Therefore, *v* = − *R*

Position of the image of block A is at − 2*R** *with respect to the given coordinate system.

For Block B,

Image of the block B is at the same place as it is at a distance of *R *from the mirror.

Therefore, the image of the block B is zero with respect to the given coordinate system.

(c)

At time $\frac{5R}{v}$

In a similar manner, we can prove that the position of the image of block A and B will be at − 3*R** *and $\frac{-4R}{3}$ respectively.

#### Page No 417:

#### Question 79:

Consider the situation shown in figure. The elevator is going up with an acceleration of 2.00 m s^{−2} and the focal length of the mirror is 12.0 cm. All the surfaces are smooth and the pulley is light. The mass-pulley system is released from rest (with respect to the elevator) at *t* = 0 when the distance of *B* from the mirror is 42.0 cm. Find the distance between the image of the block *B *and the mirror at *t* = 0.200 s. Take *g* = 10 m s^{−2}.

Figure

#### Answer:

Given,

Acceleration of the elevator, *a *= 2.00 m/s^{2}

Focal length of the mirror M, *f* = 12.00 cm

Acceleration due to gravity, *g* = 10 m/s^{2}

Mass of blocks A and B = *m*

As per the question, the mass–pulley system is released at time *t* = 0.

Let the acceleration of the masses A and B with respect to the elevator be *a.*

Using the free body diagram,

*T *−* mg + ma* − 2*m** *= 0 ...(i)

Also,

*T *− *ma *= 0 ...(ii)

From (i) and (ii), we get:

2*ma** = m*(*g + *2)

$\Rightarrow a=\frac{10+2}{2}\phantom{\rule{0ex}{0ex}}=\frac{12}{2}=6{\mathrm{ms}}^{-2}$

Now, the distance travelled by block B of mass *m *in time *t* = 0.2 s is given by

$s=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{As}(u=0)\phantom{\rule{0ex}{0ex}}s=\frac{1}{2}a{t}^{2}$

On putting the respective values, we get:

$s=\frac{1}{2}\times 6\times {\left(0.2\right)}^{2}\phantom{\rule{0ex}{0ex}}=0.12\mathrm{m}=12\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

As given in the question, the distance of block B from the mirror is 42 cm.

Object distance *u* from the mirror = − (42 − 12) = − 30 cm

Using the mirror equation,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}$

On putting the respective values, we get:

$\frac{1}{v}+\frac{1}{-30}=\frac{1}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{v}=\frac{1}{-30}=\frac{1}{12}+\frac{1}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow v=8.57\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Hence, the distance between block B and mirror M is 8.57 cm.

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