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#### Page No 430:

#### Question 1:

Can virtual image be formed on the retina in a seeing process?

#### Answer:

The retina acts as a screen; only real images can be obtained on the screen. In case of people having eye defects, the spectacles form the virtual image of the object and the eye lens form the real and inverted image on the retina.

#### Page No 430:

#### Question 2:

Can the image formed by a simple microscope be projected on a screen without using any additional lens or mirror?

#### Answer:

The image formed by a simple microscope is virtual and erect. So, it cannot be projected on a screen without using any additional lens or mirror.

#### Page No 430:

#### Question 3:

The angular magnification of a system is less than one. Does it mean that the image formed is inverted?

#### Answer:

No, angular magnification is the ratio of the angle subtended by the final image on the eye to the angle subtended by the object on the unaided eye. Its value less than one signifies reduction in the size of the image. It does not mean that the image is inverted.

#### Page No 430:

#### Question 4:

A simple microscope using a single lens often shows coloured image of a white source. Why?

#### Answer:

A simple microscope consists of a single convex lens. Sometimes due to chromatic and spherical aberrations, the image of a white source seems coloured at the corners of the lens and somewhere in between.

#### Page No 430:

#### Question 5:

A magnifying glass is a converging lens placed close to the eye. A farsighted person uses spectacles having converging lenses. Compare the functions of a converging lens used as a magnifying glass and as spectacles.

#### Answer:

A converging lens in a magnifying glass is of small focal length which is used to magnify an object which is placed close to the lens. On the other hand, converging lens used as spectacles is of varying focal length which depends upon the actual near point of the long-sighted person. It forms image at the near point of the defected eye which is further focussed by the eye lens.

#### Page No 430:

#### Question 6:

A person is viewing an extended object. If a converging lens is placed in front of his eyes, will he feel that the size has increased?

#### Answer:

If a person views an extended object through a converging lens, then it will appear larger and wider to him.

#### Page No 430:

#### Question 7:

The magnifying power of a converging lens used as a simple microscope is $\left(1+\frac{D}{f}\right).$ A compound microscope is a combination of two such converging lenses. Why don't we have magnifying power $\left(1+\frac{D}{{f}_{o}}\right)\left(1+\frac{D}{{f}_{e}}\right)?$ In other words, why can the objective not be treated as a simple microscope but the eyepiece can?

#### Answer:

In a simple microscope, the converging lens is used to magnify the object. It is done by the eyepiece in a compound microscope. But the purpose of the objective lens is the same, i.e., to form an enlarged, real and inverted image of the object at a distance less than the focal length of the eyepiece. So, its magnification power cannot be expressed in a way it is expressed for a simple microscope.

#### Page No 430:

#### Question 8:

By mistake, an eye surgeon puts a concave lens in place of the lens in the eye after a cataract operation. Will the patient be able to see clearly any object placed at any distance?

#### Answer:

The image formed by a concave lens is virtual and upright. It is smaller than the object and is formed between the object and the lens, irrespective of the position of the object. If, by mistake, an eye surgeon puts a concave lens in place of eye lens in a patient's eye, then the image will not be focused on the retina; this will lead to unclear vision.

#### Page No 430:

#### Question 9:

The magnifying power of a simple microscope is given by $1+\frac{D}{f},$ where *D* is the least distance for clear vision. For farsighted persons, *D* is greater than the usual. Does it mean that the magnifying power of a simple microscope is greater for a farsighted person as compared to a normal person? Does it mean that a farsighted person can see an insect more clearly under a microscope than a normal person?

#### Answer:

The magnifying power of a simple microscope depends on the ratio $\frac{D}{f}$ for a farsighted person. Here, *D* for a farsighted person is greater than that for a normal person, but the value of* f *remains the same. Therefore, the magnifying power of a simple microscope is greater for a farsighted person compared to that for a person with normal vision. Also, a farsighted person can see the insect more clearly under the microscope than a person with normal vision.

#### Page No 430:

#### Question 10:

Why are the magnification properties of microscopes and telescopes defined in terms of the ratio of angles and not in terms of the ratio of sizes of objects and images?

#### Answer:

Instruments like telescopes and microscopes deal with objects placed at different distances. Due to some physical factors, there is a relative change in heights not in the angle which the light emerging from them subtends on the lens. So, the magnification properties of instruments are defined in terms of the ratio of angles.

#### Page No 430:

#### Question 11:

An object is placed at a distance of 30 cm from a converging lens of focal length 15 cm. A normal eye (near point 25 cm, far point infinity) is placed close to the lens on the other side. (a) Can the eye see the object clearly? (b) What should be the minimum separation between the lens and the eye so that the eye can clearly see the object? (c) Can a diverging lens, placed in contact with the converging lens, help in seeing the object clearly when the eye is close to the lens?

#### Answer:

Object distance, *u*** = $-$**30 cm

Focal length, *f* = 15 cm

Image distance,* v* = ?

The lens formula is given by

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

(on the opposite side of the object)

(a) No, the eye placed close to the lens cannot see the object clearly.

(b) The eye should be 30 cm away from the lens to see the object clearly.

(c) The diverging lens will always form an image at a large distance from the eye; this image cannot be seen through the human eye.

#### Page No 431:

#### Question 1:

The size of an object as perceived by an eye depends primarily on

(a) actual size of the object

(b) distance of the object from the eye

(c) aperture of the pupil

(d) size of the image formed on the retina.

#### Answer:

(d) size of the image formed on the retina

An eye consists of a lens that works on the principle on which a glass lens works. It forms the image on the screen called retina. The magnification, in this case, depends on the ratio of the image to the object height.

#### Page No 431:

#### Question 2:

The muscles of a normal eye are least strained when the eye is focussed on an object

(a) far away from the eye

(b) very close to the eye

(c) at about 25 cm from the eye

(d) at about 1 m from the eye

#### Answer:

(a) far away from the eye

A normal eye can see from 25 cm to infinity; it faces least difficulty and strain focussing on the object as far as it could be.

#### Page No 431:

#### Question 3:

A normal eye is not able to see objects closer than 25 cm because

(a) the focal length of the eye is 25 cm

(b) the distance of the retina from the eye-lens is 25 cm

(c) the eye is not able to decrease the distance between the eye-lens and the retina beyond a limit

(d) the eye is not able to decrease the focal length beyond a limit.

#### Answer:

(d) the eye is not able to decrease the focal length beyond a limit

The ciliary muscles adjust the focal length to form an image on the retina, but the muscles cannot be strained beyond a limit. Hence, if the object is brought too close to the eye, the focal length cannot be adjusted to form the image on the retina.

#### Page No 431:

#### Question 4:

When objects at different distances are seen by the eye, which of the following remain constant?

(a) The focal length of the eye-lens.

(b) The object-distance from the eye-lens.

(c) The radii of curvature of the eye-lens.

(d) The image-distance from the eye-lens.

#### Answer:

(d) The image distance from the eye lens

In the human eye, the image is formed on the retina, which is at a fixed distance from the eye lens.

#### Page No 431:

#### Question 5:

A person *A* can clearly see objects between 25 cm and 200 cm. Which of the following may represent the range of clear vision for a person *B* having muscles stronger than *A*, but all other parameters of eye identical to that of *A*?

(a) 25 cm to 200 cm

(b) 18 cm to 200 cm

(c) 25 cm to 300 cm

(d) 18 cm to 300 cm

#### Answer:

(b) 18 cm to 200 cm

Person B has stronger ciliary muscles than person A. So, the muscles in his case can be strained more and the focal length of his eye can be reduced more compared to those of person A. While seeing far objects, the muscles are relaxed, so their strength will not affect the far point of the eye.

#### Page No 431:

#### Question 6:

The focal length of a normal eye-lens is about

(a) 1 mm

(b) 2 cm

(c) 25 cm

(d) 1m

#### Answer:

(b) 2 cm

Given:

Near point of the human eye,* u *= $-$25 cm

Distance between the retina and the eye lens,* v* = 2 cm (approximately)

Thus, we have the focal length,* f*.

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}\cong \frac{1}{2}-\frac{1}{-25}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}\cong \frac{27}{50}\phantom{\rule{0ex}{0ex}}\Rightarrow f\cong 2\mathrm{cm}$

#### Page No 431:

#### Question 7:

The distance of the eye-lens from the retina is *x*. For a normal eye, the maximum focal length of the eye-lens

(a) =* x*

(b) <* x*

(c) >* x*

(d) =* *2* x*

#### Answer:

(a) = *x*

For a normal eye, we have:

Far point at which the object can be placed, *u* = $\infty $

Distance between the eye lens and the retina, *v* = x

Thus, we have:

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{f}=\frac{1}{x}-\frac{1}{\infty}\phantom{\rule{0ex}{0ex}}\Rightarrow f=x$1f=1v−1u⇒1f≅12−1−25⇒1f≅2750⇒f≅2 cm1f=1v−1u⇒1f≅12−1−25⇒1f≅2750⇒f≅2 cm

#### Page No 431:

#### Question 8:

A man wearing glasses of focal length +1 m cannot clearly see beyond 1 m

(a) if he is farsighted

(b) if he is nearsighted

(c) if his vision is normal

(d) in each of these cases

#### Answer:

(d) in each of these cases

The man is wearing glasses of positive power (converging lens). Hence, he cannot see nearby objects clearly. In other words, he is farsighted. Since he cannot see beyond 1 m, he is nearsighted. If a person with normal vision wears glasses of focal length +1 m, then the person will not be able to see beyond 1 m.

#### Page No 431:

#### Question 9:

An object is placed at a distance *u* from a simple microscope of focal length *f*. The angular magnification obtained depends

(a) on *f* but not on *u*

(b) on *u* but not on *f*

(c) on *f* as well as *u*

(d) neither on *f* nor on *u*

#### Answer:

(c) on *f* as well as *u*

â€‹The angular magnification is the ratio of the angle subtended by the image to the angle subtended by the object on an unaided eye.

In a simple microscope,

$m=\frac{{\displaystyle \raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$u$}\right.}}{{\displaystyle \raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$D$}\right.}}\phantom{\rule{0ex}{0ex}}\mathrm{Here},\phantom{\rule{0ex}{0ex}}u=\mathrm{Object}\mathrm{distance}\mathrm{from}\mathrm{the}\mathrm{lens}\phantom{\rule{0ex}{0ex}}D=\mathrm{Image}\mathrm{distance}\mathrm{from}\mathrm{the}\mathrm{lens}\phantom{\rule{0ex}{0ex}}h=\mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{object}$

In normal adjustment, the object is placed at a distamce equal to â€‹focal length(f) from the lens and then magnification is given by *m = D/f.*

For *u < f*,* m = D/f *+1.

#### Page No 431:

#### Question 10:

To increase the angular magnification of a simple microscope, one should increase

(a) the focal length of the lens

(b) the power of the lens

(c) the aperture of the lens

(d) the object size

#### Answer:

(b) the power of the lens

For a simple microscope in normal adjustment, the object is placed at a distance equal to *f *(the â€‹focal length) from the lens, And the angular magnification is given by the relation

*m = D/f.*

For *u *< *f**, **m** = **D/f* + 1.

Power of the lens = 1/*f *

Angular magnification depends on power.

#### Page No 431:

#### Question 11:

A man is looking at a small object placed at his near point. Without altering the position of his eye or the object, he puts a simple microscope of magnifying power 5 X before his eyes. The angular magnification achieved is

(a) 5

(b) 2.5

(c) 1

(d) 0.2

#### Answer:

(c) 1

$\mathrm{We}\mathrm{have}:\phantom{\rule{0ex}{0ex}}h=\mathrm{Object}\mathrm{height}\phantom{\rule{0ex}{0ex}}u=\mathrm{Object}\mathrm{distance}=25\mathrm{cm}\phantom{\rule{0ex}{0ex}}D=\mathrm{Near}\mathrm{point}=25\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}m=\frac{{\displaystyle \raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$u$}\right.}}{{\displaystyle \raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$D$}\right.}}\phantom{\rule{0ex}{0ex}}\Rightarrow m=\frac{{\displaystyle \raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$25$}\right.}}{{\displaystyle \raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$25$}\right.}}\phantom{\rule{0ex}{0ex}}\Rightarrow m=1\phantom{\rule{0ex}{0ex}}$

#### Page No 431:

#### Question 1:

When we see an object, the image formed on the retina is

(a) real

(b) virtual

(c) erect

(d) inverted

#### Answer:

(a) real

(d) inverted

The retina acts as a screen in the eye; only real, inverted images can be obtained on the screen.

#### Page No 431:

#### Question 2:

In which of the following the final image is erect?

(a) Simple microscope

(b) Compound microscope

(c) Astronomical telescope

(d) Galilean telescope

#### Answer:

(a) Simple microscope

(d) Galilean telescope

In a simple microscope, the image formed is virtual and above the axis on the same side of the object. Similarly, in a Galilean telescope, the image formed is virtual, erect and magnified and between the objective lens and the eyepiece. A compound microscope and an astronomical telescope produce inverted images.

#### Page No 431:

#### Question 3:

The maximum focal length of the eye-lens of a person is greater than its distance from the retina. The eye is

(a) always strained in looking at an object

(b) strained for objects at large distances only

(c) strained for objects at short distances only

(d) unstrained for all distances

#### Answer:

(a) always strained in looking at an object

The maximum focal length of a normal eye is equal to the distance of the lens from the retina. In case it is greater than the distance, the eye will be strained while focusing the objects on the retina that is at a fixed distance from the eye lens.

#### Page No 431:

#### Question 12:

A compound microscope forms an inverted image of an object. In which of the following cases it it likely to create difficulties? (a) Looking at small germs. (b) Looking at circular spots. (c) Looking at a vertical tube containing some water.

#### Answer:

(c) Looking at a vertical tube containing some water

If the experimentalist is looking at a vertical tube containing some water, he has to be careful, as the lower meniscus will appear as upper.

#### Page No 431:

#### Question 4:

Mark the correct options.

(a) If the far point goes ahead, the power of the divergent lens should be reduced.

(b) If the near point goes ahead, the power of the convergent lens should be reduced.

(c) If the far point is 1 m away from the eye, divergent lens should be used.

(d) If the near point is 1 m away from the eye, divergent lens should be used.

#### Answer:

(a) If the far point goes ahead, the power of the divergent lens should be reduced.

(c) If the far point is 1 m away from the eye, the divergent lens should be used.

As the far point (*x*) is shifted ahead, the focal length (*f*) will be increased.

Thus, we have:

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

As power (*P*) is equal to the reciprocal of the focal length, it will be reduced. Also, because the focal length is negative, the lens used will be divergent when the far point is 1 m away.

#### Page No 431:

#### Question 5:

The focal length of the objective of a compound microscope if *f _{o}* and its distance from the eyepiece is

*L*. The object is placed at a distance

*u*from the objective. For proper working of the instrument,

(a)

*L*<

*u*

(b)

*L*>

*u*

(c)

*f*<

_{o}*L*< 2

*f*

_{o}(d)

*L*> 2

*f*

_{o}#### Answer:

(b)* L* > *u*

(d) *L** *> 2*f*_{0}

In a compound microscope, the objective lens of a short focal length, *f _{o}*, is used. The focal length of the objective lens is less than the focal length of the eyepiece,

*f*, .The object is placed at a distance slightly greater than its focal length. The real, inverted image of the object forms somewhere in front of the eyepiece at a distance less than its focal length. This image acts as its object and the final image forms in between length, L, of the microscope. Please check and correct.

_{â€‹e}∴

*L > f*â€‹

_{o}â€‹+ f_{e }> 2f_{o}Also,

*L*>

*u*

#### Page No 432:

#### Question 4:

A child has near point at 10 cm. What is the maximum angular magnification the child can have with a convex lens of focal length 10 cm?

#### Answer:

Given:

Distance of near point, *D* = 10 cm

Focal length, *f* = 10 cm

When the image is formed at the near point, *D* = 10 cm, the magnifying power $\left(m\right)$ of a simple microscope is given as:

$m=1+\frac{D}{f}\phantom{\rule{0ex}{0ex}}=1+\frac{10}{10}=1+1=2$

Thus, the maximum angular magnification is 2.

#### Page No 432:

#### Question 11:

An optical instrument used for angular magnification has a 25 D objective and 20 D eyepiece. The tube length is 25 cm when the eye is least strained. (a) Whether it is a microscope or a telescope? (b) What is the angular magnification produced?

#### Answer:

Given:

Power of the objective lens = 25 D

The focal length of the objective lens is given by

${f}_{0}=\frac{1}{25\mathrm{D}}=0.04\mathrm{m}=4\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Power of the eyepiece = 5 D

The focal length of the eyepiece is given by

${f}_{e}=\frac{1}{20\mathrm{D}}=0.05\mathrm{m}=5\mathrm{cm}$

(a) As* **f*_{o} < *f*e, the instrument must be a microscope.

(b) Tube length,* l* = 25 cm

Image distance for the eye lens, *v _{e}* = 25 cm

We know:

*l*= ${v}_{o}+{f}_{e}$

$\Rightarrow {v}_{0}=l-{f}_{e}$=25 $-$ 5 = 20 cm

$\mathrm{On}\mathrm{applying}\mathrm{the}\mathrm{lens}\mathrm{formula}\mathrm{for}\mathrm{the}\mathrm{objective}\mathrm{lens},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\frac{1}{{v}_{0}}-\frac{1}{{u}_{0}}=\frac{1}{{f}_{0}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow -\frac{1}{{u}_{o}}=\frac{1}{4}-\frac{1}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{1}{{u}_{0}}=\frac{5-1}{20}=\frac{4}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow {u}_{0}=\frac{20}{4}=5\mathrm{cm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

The angular magnification of microscope $\left(m\right)$ is given by

$m=\frac{{v}_{0}}{{u}_{0}}\left(\frac{D}{{f}_{e}}\right)\phantom{\rule{0ex}{0ex}}=\frac{20}{5}\left(\frac{25}{5}\right)=20\mathrm{cm}$

So, the required angular magnification of the microscope is 20.

#### Page No 432:

#### Question 1:

A person looks at different trees in an open space with the following details. Arrange the trees in decreasing order of their apparent sizes.

Tree |
Height(m) |
Distance from the eye(m) |

A |
2.0 | 50 |

B |
2.5 | 80 |

C |
1.8 | 70 |

D |
2.8 | 100 |

#### Answer:

Let the tree make a visual angle *θ* with the eyes.

Thus, we have:

$\theta =\frac{\mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{tree}}{\mathrm{Distance}\mathrm{from}\mathrm{the}\mathrm{eye}}\phantom{\rule{0ex}{0ex}}$

For tree A, ${\theta}_{\mathrm{A}}=\frac{2}{50}=0.04$

For tree B, ${\theta}_{\mathrm{B}}=\frac{2.5}{80}=0.03125$

For tree C, ${\theta}_{\mathrm{C}}=\frac{1.8}{70}=0.0257$

For tree D, ${\theta}_{\mathrm{D}}=\frac{2.8}{100}=0.028$

Now,

*θ*_{A} > *θ*_{B} > *θ*_{D} > *θ*_{C}

The more the value of *θ*, the bigger the apparent size of the tree.

So, the arrangement in decreasing order is A, B, D and C.

#### Page No 432:

#### Question 2:

An object is to be seen through a simple microscope of focal length 12 cm. Where should the object be placed so as to produce maximum angular magnification? The least distance for clear vision is 25 cm.

#### Answer:

For the simple microscope, we have:

Focal length, *f* = 12 cm

Least distance of clear vision, *d* = 25 cm

For maximum angular magnification,

Image distance = Least distance of clear vision

*v* = *d* = − 25 cm

$\mathrm{The}\mathrm{lens}\mathrm{formula}\mathrm{is}\mathrm{given}\mathrm{by}\phantom{\rule{0ex}{0ex}}\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{u}=\frac{1}{v}-\frac{1}{f}$

On substituting the values, we get:

$\frac{1}{u}=\frac{1}{-25}-\frac{1}{12}\phantom{\rule{0ex}{0ex}}\frac{1}{u}=-\frac{25+12}{300}=-\frac{37}{300}$

$\Rightarrow u=-\frac{300}{37}=-8.1\mathrm{cm}$

So, to produce maximum angular magnification, the object should be placed 8.1 cm away from the lens.

#### Page No 432:

#### Question 3:

A simple microscope has a magnifying power of 3.0 when the image is formed at the near point (25 cm) of a normal eye. (a) What is its focal length? (b) What will be its magnifying power if the image is formed at infinity?

#### Answer:

For the simple microscope, we have:

Least distance of clear vision, *D* = 25 cm

Magnifying power, *m* = 3

Now,

Let the focal length be *f*.

(a) When the image is formed at *D* = 25 cm, the magnifying power of the simple microscope is given by

$m=1+\left(\frac{D}{f}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 3=1+\left(\frac{25}{f}\right)$

$\Rightarrow \frac{25}{f}=2\phantom{\rule{0ex}{0ex}}\Rightarrow f=\frac{25}{2}=12.5\mathrm{cm}$

(b) When the image is formed at infinity in normal adjustment, the magnifying power is given by

$m=\frac{D}{f}=\frac{25}{12.5}=2.0$

So, the magnifying power is 2 if the image is formed at infinity.

#### Page No 432:

#### Question 5:

A simple microscope is rated 5 X for a normal relaxed eye. What will be its magnifying power for a relaxed farsighted eye whose near point is 40 cm?

#### Answer:

For a simple microscope,

Magnification for a normal relaxed eye, *m = *5

Least distant of distinct vision, D = 25 cm

Now,

Let the focal length be *f.*

The image will form at infinity, as the eye is relaxed.

The magnifying power $\left(m\right)$ of a simple microscope in normal adjustment is given by

$m=\frac{D}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow 5=\frac{25}{f}\phantom{\rule{0ex}{0ex}}\Rightarrow f=5\mathrm{cm}$

For the relaxed far-sighted eye, *D* = 40 cm.

Magnification $\left(m\right)$:

$m=\frac{D}{f}=\frac{40}{5}=8$

The magnifying power for a simple microscope is 8X.

#### Page No 432:

#### Question 6:

Find the

maximum magnifying power of a compound microscope having a 25 diopter lens as the objective, a 5 diopter lens as the eyepiece and the separation 30 cm between the two lenses. The least distance for clear vision is 25 cm.#### Answer:

For the compound microscope, we have:

Power of the objective lens = 25 D

The focal length of the objective lens is given by

${f}_{0}=\frac{1}{25\mathrm{D}}=0.04\mathrm{m}=4\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Power of the eyepiece = 5 D

The focal length of the eyepiece is given by

${f}_{e}=\frac{1}{5\mathrm{D}}=0.2\mathrm{m}=20\mathrm{cm}$

Least distance of clear vision,* D* = 25 cm

Separation between the objective and the eyepiece, *L* = 30 cm

Magnifying power is maximum when the image is formed by the eyepiece at the least distance of clear vision, i.e., *D *= 25 cm.

For the eyepiece, we have:

*v*_{e} = − 25 cm and *f*_{e} = 20 cm

The lens formula is given by

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

$\Rightarrow \frac{1}{-25}-\frac{1}{20}=-\frac{(4+5)}{100}$

$\Rightarrow {u}_{e}=\frac{-100}{9}=11.11\mathrm{cm}$

Let *u*_{o} and *v*_{o} be the object and image distance for the objective lens.

So, for the objective lens, the image distance will be

${V}_{0}=L-{u}_{0}$

$\phantom{\rule{0ex}{0ex}}{v}_{0}=30-11.11\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{o}=18.89\mathrm{cm}\dots \left(1\right)$

As the image produced is real, ${v}_{o}=+18.89\mathrm{cm}$.

The lens formula is given by

$\frac{1}{{u}_{o}}=\frac{1}{{v}_{o}}-\frac{1}{{f}_{o}}\phantom{\rule{0ex}{0ex}}=\frac{1}{18.89}-\frac{1}{4}=-0.197\phantom{\rule{0ex}{0ex}}{u}_{o}=-5.07\mathrm{cm}\dots \left(2\right)\phantom{\rule{0ex}{0ex}}$

Maximum magnifying power of the compound microscope:

$m=\frac{{v}_{0}}{{u}_{0}}\left(1+\frac{\mathrm{D}}{{f}_{e}}\right)\phantom{\rule{0ex}{0ex}}=-\left(\frac{18.89}{-5.07}\right)\left(1+\frac{25}{20}\right)$

= 3.7225 × 2.25 = 8.376

So, the maximum magnifying power of the compound microscope is 8.376.

#### Page No 432:

#### Question 7:

The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9.8 cm to 11.8 cm. If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if the image is always needed at 24 cm from the eye.

#### Answer:

For the compound microscope, we have:

Focal length of the objective, *f*_{o} = 1.0 cm

Focal length of the eyepiece, *f*_{e} = 6 cm

Image distance from the eyepiece, *v*_{e} = $-$24 cm

Separation between the objective and the eyepiece = 9.8 cm to 11.8 cm

The lens formula is given by

$\frac{1}{{v}_{e}}-\frac{1}{{u}_{e}}=\frac{1}{{f}_{e}}\phantom{\rule{0ex}{0ex}}\frac{1}{-24}-\frac{1}{{u}_{e}}=\frac{1}{6}$

$\Rightarrow \frac{1}{{u}_{e}}=-\frac{1}{6}-\frac{1}{24}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{u}_{e}}=-\frac{4+1}{24}=-\frac{5}{24}\phantom{\rule{0ex}{0ex}}\therefore {u}_{e}=-\frac{24}{5}\mathrm{cm}=-4.8\mathrm{cm}$

(a) Separation between the objective and the eyepiece = 9.8 cm

So, for the objective lens, the image distance will be

*v*_{o} = 9.8 − 4.8 = 5 cm

The lens formula for the objective lens is given by

$\frac{1}{{v}_{0}}-\frac{1}{{u}_{0}}=\frac{1}{{f}_{0}}\phantom{\rule{0ex}{0ex}}\frac{1}{5}-\frac{1}{{u}_{0}}=\frac{1}{1}$

$\Rightarrow -\frac{1}{{u}_{0}}=1-\frac{1}{5}=\frac{5-1}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow {u}_{0}=-\frac{5}{4}\mathrm{cm}=-1.25\mathrm{cm}$

Magnifying power of the compound microscope:

$m=\frac{{v}_{0}}{{u}_{0}}\left(1+\frac{D}{{f}_{e}}\right)\phantom{\rule{0ex}{0ex}}$

$=\frac{5}{1.25}\left(1+\frac{24}{6}\right)=20$

(b) Separation between the objective and the eyepiece = 11.8 cm

We have:

*m* = 30

So, the required range of the magnifying power is 20–30.

#### Page No 432:

#### Question 8:

An eye can distinguish between two points of an object if they are separated by more than 0.22 mm when the object is placed at 25 cm from the eye. The object is now seen by a compound microscope having a 20 D objective and 10 D eyepiece separated by a distance of 20 cm. The final image is formed at 25 cm from the eye. What is the minimum separation between two points of the object which can now be distinguished?

#### Answer:

For the compound microscope, we have:

Power of the objective lens = 20 D

The focal length of the objective lens is given by

${f}_{0}=\frac{1}{20\mathrm{D}}=0.05\mathrm{m}=5\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Power of the eyepiece = 10 D

The focal length of the eyepiece is given by

${f}_{e}=\frac{1}{10\mathrm{D}}=0.1\mathrm{m}=10\mathrm{cm}$

Least distance for clear vision,* D* = 25 cm,

To distinguish the two points having minimum separation, the magnifying power should be maximum.

For the eyepiece, we have:

Image distance, *v _{e}* = − 25 cm

Focal length,

*f*= 10 cm

_{e}The lens formula is given by

$\frac{1}{{u}_{e}}=\frac{1}{{v}_{e}}-\frac{1}{{f}_{e}}\phantom{\rule{0ex}{0ex}}$

_{}

$=\frac{1}{-25}-\frac{1}{10}\phantom{\rule{0ex}{0ex}}=\frac{-2-5}{50}=\frac{-7}{50}$

Separation between the objective and the eyepiece = 20 cm

So, the image distance for the objective lens $\left({v}_{0}\right)$ can be obtained as:

${v}_{0}=20-\frac{50}{7}=\frac{90}{7}\mathrm{cm}$

The lens formula for the objective lens is given by

$\frac{1}{{u}_{0}}=\frac{1}{{v}_{0}}-\frac{1}{{f}_{0}}\phantom{\rule{0ex}{0ex}}=\frac{7}{90}-\frac{1}{5}=\frac{7-18}{90}=-\frac{11}{90}$

$\Rightarrow {u}_{0}=\frac{-90}{11}\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Magnification of the compound microscope:

$m=\frac{{v}_{0}}{{u}_{0}}\left(1+\frac{D}{{f}_{e}}\right)$

$=-\frac{{\displaystyle \frac{90}{7}}}{{\displaystyle \frac{-90}{11}}}\left(1+\frac{25}{10}\right)\phantom{\rule{0ex}{0ex}}=\frac{11}{7}\times \left(3.5\right)=5.5$

∴ Minimum separation that the eye can distinguish = $\frac{0.22}{5.5}$ mm = 0.04 mm

#### Page No 432:

#### Question 9:

A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. Find the focal length of the eyepiece.

#### Answer:

For the compound microscope, we have:

Magnifying power, *m* = 100

Focal length of the objective, *f*_{o} = 0.5 cm

Tube length, *l* = 6.5 cm

Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eyepiece.

∴ *v*_{o} +* **f*_{e} = 6.5 cm …(1)

The magnifying power for normal adjustment is given by

$m=\frac{{v}_{o}}{{u}_{o}}\times \frac{D}{{f}_{e}}\phantom{\rule{0ex}{0ex}}=-\left[1-\frac{{v}_{o}}{{f}_{o}}\right]\frac{D}{{f}_{e}}\phantom{\rule{0ex}{0ex}}\Rightarrow 100=-\left[1-\frac{{v}_{o}}{0.5}\right]\times \frac{25}{{f}_{e}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2{v}_{o}-4{f}_{e}=1\dots \left(2\right)$

On solving equations (1) and (2), we get:

*V*_{o} = 4.5 cm and *f*_{e} = 2 cm

Thus, the focal length of the eyepiece is 2 cm.

#### Page No 432:

#### Question 10:

A compound microscope consists of an objective of focal length 1 cm and an eyepiece of focal length 5 cm. An object is placed at a distance of 0.5 cm from the objective. What should be the separation between the lenses so that the microscope projects an inverted real image of the object on a screen 30 cm behind the eyepiece?

#### Answer:

For the compound microscope, we have:

Focal length of the objective, *f*_{o} = 1.0 cm

Focal length of the eyepiece, *f*_{e} = 5 cm

Distance of the object from the objective, *u*_{0} = 0.5 cm

Distance of the image from the eyepiece, *v*_{e} = 30 cm

The lens formula for the objective lens is given by

$\frac{1}{{v}_{0}}-\frac{1}{{u}_{0}}=\frac{1}{{f}_{0}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{v}_{0}}+\frac{1}{0.5}=\frac{1}{1}$

$\Rightarrow \frac{1}{{v}_{0}}=1-\frac{10}{5}=-1\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{0}=-1\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

The objective will form a virtual image at the side same as that of the object at a distance of 1 cm from the objective lens. The image formed by the objective will act as a virtual object for the eyepiece.

$\mathrm{The}\mathrm{lens}\mathrm{formula}\mathrm{for}\mathrm{the}\mathrm{eyepiece}\mathrm{is}\mathrm{given}\mathrm{by}\phantom{\rule{0ex}{0ex}}\frac{1}{{v}_{e}}-\frac{1}{{u}_{e}}=\frac{1}{{f}_{e}}$

$\Rightarrow \frac{1}{30}-\frac{1}{{u}_{e}}=\frac{1}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{1}{{u}_{e}}=\frac{1}{5}-\frac{1}{30}=\frac{6-1}{30}=\frac{1}{6}$

⇒ *u _{e}* = − 6 cm

∴ Separation between the objective and the eyepiece = 6 − 1 = 5 cm

#### Page No 432:

#### Question 12:

An astronomical telescope is to be designed to have a magnifying power of 50 in normal adjustment. If the length of the tube is 102 cm, find the powers of the objective and the eyepiece.

#### Answer:

For the astronomical telescope,

Magnifying power, *m* = 50

Length of the tube, *L* = 102 cm

Let the focal length of objective and eye piece be *f*_{0} and *f*_{e}, respectively.

Now, using $m=\frac{{f}_{o}}{{f}_{e}}$, we get:

*f*_{o}_{ }= 50*f*_{e}_{ }…(1)

And,

*L* = *f*_{o} + *f*_{e} = 102 cm …(2)

On substituting the value of *f*_{o} from (1) in (2), we get:

50*f*e + *f*e = 102

⇒ 51*f*e = 102

⇒ *f*e = 2 cm = 0.02 m

And,

*f*_{o} = 50$\times $0.02 = 1 m

Power of the objective lens = $\frac{1}{{f}_{o}}$ = 1 D

And,

Power of the eye piece lens = $\frac{1}{{f}_{e}}=\frac{1}{0.02}$ = 50 D

#### Page No 432:

#### Question 13:

The eyepiece of an astronomical telescope has a focal length of 10 cm. The telescope is focussed for normal vision of distant objects when the tube length is 1.0. m. Find the focal length of the objective and the magnifying power of the telescope.

#### Answer:

In the astronomical telescope (in normal adjustment),

Focal length of the eyepiece, *f*e = 10 cm

Length of the tube, *L* = 1 m = 100 cm

Focal length of the objective, *f*_{o} = ?

We know:

*f _{o} + f_{e} = L*

$\therefore $

*f*

_{o}=

*L –*

*f*

_{e}= 100 – 10 = 90 cm

Magnifying power $\left(m\right)$ in normal adjustment:

*m*= $\frac{{f}_{o}}{{f}_{e}}=\frac{90}{10}=9$

#### Page No 432:

#### Question 14:

A Galilean telescope is 27 cm long when focussed to form an image at infinity. If the objective has a focal length of 30 cm, what is the focal length of the eyepiece?

#### Answer:

The image will be formed at infinity.

Given:

Focal length of the objective,* **f*_{o} = 30 cm

Length of the tube,* L =* 27 cm

Now,

*L* = *${f}_{o}-\left|{f}_{e}\right|$*

In a Galilean telescope, the eyepiece lens is concave.

Thus, we have:

*f*_{e} = *f*_{o} – *L* = 30 – 27 = 3 cm

#### Page No 432:

#### Question 15:

A farsighted person cannot see objects placed closer to 50 cm. Find the power of the lens needed to see the objects at 20 cm.

#### Answer:

Here,

*u* = – 20 cm and *v* = – 50 cm

The lens formula is given by

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

On putting the values, we get:

$\frac{1}{f}=\frac{1}{\left(-50\right)}-\frac{1}{\left(-20\right)}\phantom{\rule{0ex}{0ex}}=\frac{3}{100}$

$\Rightarrow f=\frac{100}{3}\mathrm{cm}=\frac{1}{3}\mathrm{m}$

∴ Power of the lens = $\frac{1}{f}$ = 3 D

#### Page No 432:

#### Question 16:

A nearsighted person cannot clearly see beyond 200 cm. Find the power of the lens needed to see objects at large distances.

#### Answer:

For a near-sighted person, *u* = ∞.

Now, we have:

*v* = – 200 cm = – 2 m

The lens formula is given by

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

On putting the respective values, we get:

$\frac{1}{f}=\frac{1}{\left(-2\right)}-\frac{1}{\infty}=-\frac{1}{2}$

∴ Power of the lens = $\frac{1}{f}$ = – 0.5 D

#### Page No 432:

#### Question 17:

A person wears glasses of power − 2.5 D. Is the person farsighted or nearsighted? What is the far point of person without the glasses?

#### Answer:

The person must be near-sighted because the person wears glasses of power –2.5 D.

For a near-sighted person, *u* = ∞.

Focal length, *f* = $\frac{1}{\mathrm{Power}}$ = $\frac{1}{\left(-2.5\right)}$ = − 0.4 m = − 40 cm

The lens formula is given by

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

Putting the values, we get:

$\frac{1}{v}=\frac{1}{\infty}+\frac{1}{\left(-40\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow v=-40$

So, the far point of the near-sighted person is 40 cm.

#### Page No 432:

#### Question 18:

A professor reads a greeting card received on his 50th birthday with + 2.5 D glasses keeping the card 25 cm away. Ten years later, he reads his farewell letter with the same glasses but he has to keep the letter 50 cm away. What power of lens should he now use?

#### Answer:

Given:

The professor can read the card at a distance of 25 cm using a glass of + 2.5 D.

After 10 years, he can read the farewell letter at a distance (*u*) of – 50 cm.

Focal length, *f* = $\frac{1}{\mathrm{Power}}$ = $\frac{1}{2.5}$ = 0.4 m = 40 cm

The lens formula is given by

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

Putting the values, we get:

$\frac{1}{v}=\frac{1}{\left(-50\right)}+\frac{1}{40}\phantom{\rule{0ex}{0ex}}\Rightarrow v=200\mathrm{cm}$

∴ His near point,* v* = 200 cm

To read the farewell letter at a distance of 25 cm,

*u* = – 25 cm

The lens formula is given by

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

Putting the values, we get:

$\frac{1}{f}=\frac{1}{\left(200\right)}-\frac{1}{\left(-25\right)}=+\frac{9}{200}\phantom{\rule{0ex}{0ex}}\Rightarrow f=\frac{200}{9}\mathrm{cm}=\frac{2}{9}\mathrm{m}$

∴ Power of the lens = $\frac{1}{f}$ = 4.5 D

He should use a lens of power + 4.5 D.

#### Page No 432:

#### Question 19:

A normal eye has retina 2 cm behind the eye-lens. What is the power of the eye-lens when the eye is (a) fully relaxed, (b) most strained?

#### Answer:

(a) When the eye lens is fully relaxed, we have:

*u = ∞*

Distance of the retina from the eye lens,* v* = 2 cm = 0.02 m

The lens formula is given by

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

Putting the respective values, we get:

$\frac{1}{f}=\frac{1}{0.02}-\frac{1}{\infty}=\frac{1}{0.02}$

∴ Power of the lens = $\frac{1}{f}$ = 50 diopters

So, in a fully relaxed condition, the power of the eye lens is 50 D.

(b) When the eye lens is most strained:

*u* = – 25 cm = – 0.25 m

The lens formula is given by

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

Putting the values, we get:

$\frac{1}{f}=\frac{1}{0.02}-\frac{1}{\left(-0.25\right)}\phantom{\rule{0ex}{0ex}}=50+4=54$

∴ Power of the lens = $\frac{1}{f}$ = 54 diopters

So, in the most strained condition, the power of the eye lens is 54 D.

#### Page No 432:

#### Question 20:

The near point and the far point of a child are at 10 cm and 100 cm. If the retina is 2.0 cm behind the eye-lens, what is the range of the power of the eye-lens?

#### Answer:

Given:

Near point of the child = 10 cm

Far point of the child = 100 cm

The retina is at a distance of 2 cm behind the eye lens.

Thus, we have:

*v =* 2 cm = 0.02 m

When the near point is 10 cm,

*u* = − 10 cm = − 0.1 m

*v =* 2 cm = 0.02 m

The lens formula is given by

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

On putting the respective values, we get:

$\frac{1}{f}=\frac{1}{0.02}-\frac{1}{\left(-0.1\right)}\phantom{\rule{0ex}{0ex}}=50+10=60\mathrm{m}$

∴ Power of the lens,* P* = $\frac{1}{f}$ = 60 D

Now, consider the near point 100 cm.

Here,

*u* = − 100 cm = − 1 m

*v =* 2 cm = 0.02 m

The lens formula is given by

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

Putting the values, we get:

$\frac{1}{f}=\frac{1}{0.02}-\frac{1}{\left(-1\right)}\phantom{\rule{0ex}{0ex}}=50+1=51$

∴ Power of the lens = $\frac{1}{f}$ = 51 D

So, the range of the power of the eye lens is from + 60 D to + 51 D.

#### Page No 432:

#### Question 21:

A nearsighted person cannot see beyond 25 cm. Assuming that the separation of the glass from the eye is 1 cm, find the power of lens needed to see distant objects.

#### Answer:

Given:

As the person wears spectacles at a distance of 1 cm from the eyes.

Distance of the image from the glass, *v* = Distance of the image from the eye − Separation between the glass and the eye

*v *= 25 − 1 = 24 cm = 0.24 m

For the near-sighted person, *u* = ∞.

The lens formula is given by

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

On substituting the values in the above formula, we get:

$\frac{1}{f}=\frac{1}{\left(-0.24\right)}-\frac{1}{\infty}=-42\phantom{\rule{0ex}{0ex}}$

Power of the lens,* P* = $\frac{1}{f}$ = − 4.2 D

Thus, the power of the lens required to see distant objects is − 4.2 D.

#### Page No 432:

#### Question 22:

A person has near point at 100 cm. What power of lens is needed to read at 20 cm if he/she uses (a) contact lens, (b) spectacles having glasses 2.0 cm separated from the eyes?

#### Answer:

Given: A person has near point of 100 cm.

According to the question, it is needed to read at a distance of 20 cm.

(a) When the contact lens is used:

* u* = – 20 cm = – 0.2 m,

* v* = – 100 cm = – 1 m

The lens formula is given by

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

On substituting the values in the above formula, we get:

$\frac{1}{f}=\frac{1}{\left(-1\right)}-\frac{1}{\left(-0.2\right)}=4\phantom{\rule{0ex}{0ex}}$

∴ Power of the lens,* P* = $\frac{1}{f}$ = 4 D

(b) When the person wears spectacles at a distance of 2 cm from the eyes:

*u* = – (20 – 2) = – 18 cm = – 0.18 m,

*v* = – 100 cm = –1 m

The lens formula is given by

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

On substituting the values, we get:

$\frac{1}{f}=\frac{1}{\left(-1\right)}-\frac{1}{\left(-0.18\right)}=4.53\phantom{\rule{0ex}{0ex}}$

∴ Power of the lens,* P* = $\frac{1}{f}$ = 4.53 D

#### Page No 432:

#### Question 23:

A lady uses + 1.5 D glasses to have normal vision from 25 cm onwards. She uses a 20 D lens as a simple microscope to see an object. Find the maximum magnifying power if she uses the microscope (a) together with her glass (b) without the glass. Do the answers suggest that an object can be more clearly seen through a microscope without using the correcting glasses?

#### Answer:

Given:

The lady uses glasses of +1.5 D to have normal vision from 25 cm onwards.

Least distance of clear vision, *D* = 25 cm

Focal length of the glasses, *f *= $\frac{1}{\mathrm{power}}=\frac{1}{1.5}$ m

She should have greater least distance of distinct vision without the glasses.

Take:

*u* = – 25 cm = $-$ 0.25 m

The lens formula is given by

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

Putting the values, we get:

$\frac{1}{v}=1.5-\frac{1}{0.25}=2.5\phantom{\rule{0ex}{0ex}}\Rightarrow v=0.4\mathrm{m}=40\mathrm{cm}$

Near point without glasses = 40 cm

Focal length of magnifying glass, *f* = $\frac{1}{20}$ = 0.05 m = 5 cm

(a) The maximum magnifying power $\left(m\right)$ if she uses the microscope together with her glass is given by

$m=1+\frac{D}{f}\phantom{\rule{0ex}{0ex}}$

Here,

D = 25 cm

*f* = 5 cm

On substituting the values, we get:

$m=1+\frac{25}{5}=6$

(b) Without the glasses,* D *= 40 cm.

$\therefore m=1+\frac{D}{f}=1+\frac{40}{5}=9\phantom{\rule{0ex}{0ex}}$

#### Page No 432:

#### Question 24:

A lady cannot see objects closer than 40 cm from the left eye and closer than 100 cm from the right eye. While on a mountaineering trip, she is lost from her team. She tries to make an astronomical telescope from her reading glasses to look for her teammates. (a) Which glass should she use as the eyepiece? (b) What magnification can she get with relaxed eye?

#### Answer:

Given:

For the left glass lens of the lady, we have:

*v* = – 40 cm and *u* = – 25 cm

The lens formula is given by

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

Putting the values, we get:

$\frac{1}{f}=\frac{1}{\left(-40\right)}-\frac{1}{\left(-25\right)}=\frac{3}{200}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow f=\frac{200}{3}=66.6\mathrm{cm}$

For the right glass lens of the lady, we have:

*v* = – 100 cm, *u* = – 25 cm

The lens formula is given by

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

Putting the values, we get:

$\frac{1}{f}=\frac{1}{\left(-100\right)}-\frac{1}{\left(-25\right)}=\frac{3}{100}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow f=\frac{100}{3}=33.3\mathrm{cm}$

(a) An astronomical telescope consists of two lenses: the objective lens having a large focal length and the eyepiece lens having a smaller focal length. So, she should use the right lens of focal length 33.3 cm as the eyepiece lens.

(b) With relaxed eye in normal adjustment,

*f*_{o} = 66.6 cm and *f*_{e} = 33.3 cm

Magnification $\left(m\right)$ in normal adjustment is given by

*m* = $\frac{{f}_{0}}{{f}_{e}}$

$\therefore m=\frac{66.6\mathrm{cm}}{33.3\mathrm{cm}}$= 2

So, with the relaxed eye, she can get the magnification of 2.

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