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#### Page No 62:

#### Question 1:

Consider a light beam incident from air to a glass slab at Brewster’s angle as shown in Fig. 10.1. A polaroid is placed in the path of the emergent ray at point P and rotated about an axis passing through the centre and perpendicular to the plane of the polaroid.

(a) For a particular orientation there shall be darkness as observed through the polaoid.

(b) The intensity of light as seen through the polaroid shall be independent of the rotation.

(c) The intensity of light as seen through the Polaroid shall go through a minimum but not zero for two orientations of the polaroid.

(d) The intensity of light as seen through the polaroid shall go through a minimum for four orientations of the polaroid.

#### Answer:

Consider the diagram the light beam incident from air to the glass slab at Brewster’s angle (*i*_{p} ).

The incident ray is unpolarised and is represented by dot (.).

The reflected light is plane polarised represented by arrows

As the emergent ray is unpolarised, hence intensity cannot be zero when passes through polaroid.

Hence, the correct answer is option C.

#### Page No 62:

#### Question 2:

Consider sunlight incident on a slit of width 10^{4} $\stackrel{\mathrm{o}}{\mathrm{A}}$. The image seen through the slit shall

(a) be a fine sharp slit white in colour at the center.

(b) a bright slit white at the center diffusing to zero intensities at the edges.

(c) a bright slit white at the center diffusing to regions of different colours.

(d) only be a diffused slit white in colour.

#### Answer:

Width of the slit = ${10}^{4}\stackrel{\mathrm{o}}{\mathrm{A}}={10}^{4}\times {10}^{-10}={10}^{-6}\mathrm{m}$ or 10,000 $\stackrel{\mathrm{o}}{\mathrm{A}}$.

The wavelength of (visible) sunlight varies from 4000 $\stackrel{\mathrm{o}}{\mathrm{A}}$ to 8000 $\stackrel{\mathrm{o}}{\mathrm{A}}$.

As the width of the slit is comparable to that of wavelength, hence diffraction occurs with maxima at the centre.

So, at the centre all colours appear i.e., mixing of colours form a white patch at the centre.

Hence, the correct answer is option A.

#### Page No 63:

#### Question 3:

Consider a ray of light incident from air onto a slab of glass (refractive index *n*) of width *d*, at an angle* θ*. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

$\left(\mathrm{a}\right)\frac{4\mathrm{\pi d}}{\lambda}{\left(1-\frac{1}{{n}^{2}}{\mathrm{sin}}^{2}\theta \right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\mathrm{\pi}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)\frac{4\mathrm{\pi d}}{\mathrm{\lambda}}{\left(1-\frac{1}{{\mathrm{n}}^{2}}{\mathrm{sin}}^{2}\mathrm{\theta}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\phantom{\rule{0ex}{0ex}}\left(\mathrm{c}\right)\frac{4\mathrm{\pi d}}{\mathrm{\lambda}}{\left(1-\frac{1}{{\mathrm{n}}^{2}}{\mathrm{sin}}^{2}\mathrm{\theta}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{\mathrm{\pi}}{2}\phantom{\rule{0ex}{0ex}}\left(\mathrm{d}\right)\frac{4\mathrm{\pi d}}{\mathrm{\lambda}}{\left(1-\frac{1}{{\mathrm{n}}^{2}}{\mathrm{sin}}^{2}\mathrm{\theta}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+2\mathrm{\pi}$

#### Answer:

Consider the diagram, the ray (P) is incident at an angle θ and gets reflected in the direction P′ and refracted in the direction P ''.

Due to reflection from the glass medium, there is a phase change of π.

Time taken to travel along OP ′′

$\u2206t=\frac{OP\text{'}\text{'}}{v}=\frac{{\displaystyle \frac{d}{\mathrm{cos}r}}}{{\displaystyle \frac{c}{n}}}=\frac{nd}{c\mathrm{cos}r}\phantom{\rule{0ex}{0ex}}\mathrm{From}\mathrm{Snell}\text{'}\mathrm{s}\mathrm{law},n=\frac{\mathrm{sin}\theta}{\mathrm{sin}r}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}r=\frac{\mathrm{sin}\theta}{n}$

$\mathrm{cos}r=\sqrt{1-{\mathrm{sin}}^{2}r}=\sqrt{1-\frac{{\mathrm{sin}}^{2}\theta}{{n}^{2}}}$

$\u2206t=\frac{nd}{c\sqrt{\left(1-{\displaystyle \frac{{\mathrm{sin}}^{2}\theta}{{n}^{2}}}\right)}}=\frac{{n}^{2}d}{c}\sqrt{\left(1-\frac{{\mathrm{sin}}^{2}\theta}{{n}^{2}}\right)}$

Phase difference, $\u2206\varphi =\frac{2\pi}{T}\times \u2206t=\frac{2\mathrm{\pi}nd}{\lambda}{\left(1-\frac{{\mathrm{sin}}^{2}\theta}{{n}^{2}}\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

So, net phase difference = $\u2206\varphi +\pi $ = $\frac{4\pi d}{\lambda}{\left(1-\frac{1}{{n}^{2}}{\mathrm{sin}}^{2}\theta \right)}^{-\frac{1}{2}}+\pi $

Hence, the correct answer is option A.

#### Page No 63:

#### Question 4:

In a Young’s double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case

(a) there shall be alternate interference patterns of red and blue.

(b) there shall be an interference pattern for red distinct from that for blue.

(c) there shall be no interference fringes.

(d) there shall be an interference pattern for red mixing with one for blue.

#### Answer:

For the interference pattern to be formed on the screen, the sources should be coherent and emits lights of same frequency and wavelength.

In a Young’s double-slit experiment, when one of the holes is covered by a red filter and another by a blue filter. In this case due to filteration only red and blue lights are present. In YDSE monochromatic light is used for the formation of fringes on the screen.

Thus, in this case there shall be no interference fringes

Hence, the correct answer is option C.

#### Page No 63:

#### Question 5:

Figure 10.2 shows a standard two slit arrangement with slits S_{1}, S_{2} . P_{1}, P_{2} are the two minima points on either side of P (Fig. 10.2).

At P_{2} on the screen, there is a hole and behind P_{2} is a second 2- slit arrangement with slits S_{3}, S_{4} and a second screen behind them.

(a) There would be no interference pattern on the second screen but it would be lighted.

(b) The second screen would be totally dark.

(c) There would be a single bright point on the second screen.

(d) There would be a regular two slit pattern on the second screen.

#### Answer:

According to question, there is a hole at point P_{2} .

From Huygen’s principle, wave will propagates from the sources S_{1} and S_{2} .

Each point on the screen will acts as secondary sources of wavelets. Now, there is a hole at point P_{2} (minima). The hole will act as a source of fresh light for the slits S_{3} and S_{4} .

Therefore, there will be a regular two slit pattern on the second screen.

Hence, the correct answer is option D.

#### Page No 64:

#### Question 6:

Two source *S*_{1 }and *S*_{2} of intensity *I*_{1} and *I*_{2} are placed in front of a screen [Fig. 10.3 (a)]. The patteren of intensity distribution seen in the central portion is given by Fig. 10.3 (b).

In this case which of the following statements are true.

(a) *S*_{1} and *S*_{2 }have the same intensities.

(b) *S*_{1} and *S*_{2} have a constant phase difference.

(c) *S*_{1} and *S*_{2} have the same phase.

#### Answer:

Consider the pattern of the intensity shown in the question figure

(i) As intensities of all successive minima is zero, hence we can say that two sources S_{1} and S_{2} are having same intensities.

(ii) As width of the successive maxima (pulses) increases in continuous manner, we can say that the path difference (*x*) or phase difference varies in continuous manner.

(iii) We are using monochromatic light in YDSE to avoid overlapping and to have very clear pattern on the screen.

Hence, the correct answer are option A, B and D.

#### Page No 64:

#### Question 7:

Consider sunlight incident on a pinhole of width 10^{3 }$\stackrel{o}{A}$. The image of the pinhole seen on a screen shall be

(a) a sharp white ring.

(b) different from a geometrical image.

(c) a diffused central spot, white in colour.

(d) diffused coloured region around a sharp central white spot.

#### Answer:

Diffraction: It is the phenomenon of the bending of light around small obstacles.

The size of obstacle must be in comparable size with wavelength of the light or $d\le \lambda $.

Width of the pinhole = $1000\stackrel{\mathrm{o}}{\mathrm{A}}$

The wavelength of sunlight ranges from 4000 Å to 8000 Å.

Clearly, width of the slit $\le $ wavelength of the sunlight.

Thus, light is diffracted from the hole.

Due to diffraction from the light the image formed on the screen will be different from the geometrical image.

Hence, the correct answer is option B and D.

#### Page No 64:

#### Question 8:

Consider the diffraction pattern for a small pinhole. As the size of the hole is increased

(a) the size decreases.

(b) the intensity increases.

(c) the size increases.

(d) the intensity decreases.

#### Answer:

The width of central maxima, ${\beta}_{o}=\lambda \frac{D}{d}$

Width of secondary maxima, $\frac{\lambda}{d}$

So on increasing width of hole of pinhole, *d* increases. Thus, the size of central maxima decreases.

â€‹

As the energy passing through hole increased on increasing the size of hole. So the intensity of pattern will increase.

Hence, the correct answers are option A and B.

#### Page No 64:

#### Question 9:

For light diverging from a point source

(a) the wavefront is spherical.

(b) the intensity decreases in proportion to the distance squared.

(c) the wavefront is parabolic.

(d) the intensity at the wavefront does not depend on the distance.

#### Answer:

Consider the diagram in which light diverges from a point source (O).

Due to the point source light propagates in all directions symmetrically and hence, wavefront will be spherical as shown in the diagram.

If power of the source is P, then intensity of the source will be

$I=\frac{P}{4\pi {r}^{2}}$

where, r is radius of the wavefront at any time.

Hence, the correct answer is option A and B.

#### Page No 64:

#### Question 10:

Is Huygen’s principle valid for longitudinal sound waves?

#### Answer:

When we are considering a point source of sound wave. The disturbance due to the source propagates in spherical symmetry that is in all directions. The formation of wavefront is in accordance with Huygen’s principle.

So, Huygen’s principle is valid for longitudinal sound waves also.

#### Page No 64:

#### Question 11:

Consider a point at the focal point of a convergent lens. Another convergent lens of short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image?

#### Answer:

Consider the ray diagram shown below

The point image I_{1} , due to L_{1} is at the focal point.

Now, due to the converging lense L_{2} , let final image formed is* I* which is point image, hence the wavefront for this image will be of spherical symmetry.

#### Page No 64:

#### Question 12:

What is the shape of the wavefront on earth for sunlight?

#### Answer:

Due to the large distance the radius of wavefront can be considered as large (infinity) and hence, wavefront is almost plane.

#### Page No 65:

#### Question 13:

Why is the diffraction of sound waves more evident in daily experience than that of light wave?

#### Answer:

The necessary condition for diffraction to happen is that the width of the obstacle must be less than or comparable to the wavelength of the wave.

As we know that the frequencies of sound waves lie between 20 Hz to 20 kHz so that their wavelength ranges between 15 m to 15 mm. The diffraction occur if the wavelength of waves is nearly equal to slit width.

As the wavelength of light waves is 7000 $\times {10}^{-10}$ to 4000 $\times {10}^{-10}$ m. The slit width is very near to the wavelength of sound waves as compared to light waves. Thus, the diffraction of sound waves is more evident in daily life than that of light waves.

#### Page No 65:

#### Question 14:

The human eye has an approximate angular resolution of $\varphi $= 5.8 × 10^{–4} rad and a typical photoprinter prints a minimum of 300 dpi (dots per inch, 1 inch = 2.54 cm). At what minimal distance z should a printed page be held so that one does not see the individual dots.

#### Answer:

Angular dispersion of human eye, $\varphi =5.8\times {10}^{-4}\mathrm{rad}$

Number of dots printed by printer in one inch = 300

The linear distance between the two dots, $l=\frac{2.54}{300}\mathrm{cm}=0.84\times {10}^{-2}\mathrm{cm}.$

At a distance of *z* cm, this subtends an angle, $\varphi =\frac{l}{z}$

$\Rightarrow z=\frac{l}{\varphi}=14.5\mathrm{cm}$

#### Page No 65:

#### Question 15:

A polariod (I) is placed in front of a monochromatic source. Another polatiod (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II). Explain.

#### Answer:

In the diagram shown, a monochromatic light is placed infront of polaroid (I) as shown below.

As per the given question, monochromatic light emerging from polaroid (I) is plane polarised. When polaroid (II) is placed infront of this polaroid (I), and rotated till no light passes through polaroid (II), then (I) and (II) are set in crossed positions, i.e., pass axes of I and II are at 90°.

Consider the above diagram where a third polaroid (III) is placed between polaroid (I) and polaroid II.

When a third polaroid (III) is placed in between (I) and (II), no light will emerge from (II), if pass axis of (III) is parallel to pass axis of (I) or (II). In all other cases, light will emerge from (II), as pass axis of (II) will no longer be at 90° to the pass axis of (III).

#### Page No 65:

#### Question 16:

Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index?

#### Answer:

When angle of incidence is equal to Brewster’s angle, the transmitted light is unpolarised and reflected light is plane polarised.

Consider the diagram in which unpolarised light is represented by dot and plane polarised light is represented by arrows.

Polarisation by reflection occurs when the angle of incidence is the Brewster’s angle, $\mathrm{tan}{i}_{B}{=}^{1}{\mu}_{2}=\frac{{\mu}_{2}}{{\mu}_{1}}$

when the light rays travels in such a medium, the critical angle is, $\mathrm{sin}{i}_{c}=\frac{{\mu}_{2}}{{\mu}_{1}}$

Where, ${\mu}_{2}<{\mu}_{1}$

As $\left|\mathrm{tan}{i}_{B}\right|>\left|\mathrm{sin}{i}_{c}\right|\mathrm{for}\mathrm{large}\mathrm{angle}{i}_{B}{i}_{C}$

Thus, the polarisation by reflection occurs definitely.

#### Page No 65:

#### Question 17:

For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light

of 5000 $\stackrel{\circ}{\mathrm{A}}$ and electrons accelerated through 100V used as the illuminating substance.

#### Answer:

If, *λ* is the wavelength of light and *β* is the angle subtended by the objective at the object

Resolving power = $\frac{1}{d}=\frac{2\mathrm{sin}\beta}{1.22\lambda}\Rightarrow {d}_{min}=\frac{1.22\lambda}{2\mathrm{sin}\beta}$

Here, $\lambda =5500\stackrel{\mathit{o}}{\mathit{A}}$

Then, ${d}_{min}=\frac{1.22\times 5500\times {10}^{-10}}{2\mathrm{sin}\beta}$

For electrons accelerated through 100 V, the de-Broglie wavelength, $\lambda =\frac{12.27}{\sqrt{V}}=\frac{12.27}{\sqrt{100}}=0.12\times {10}^{-9}\mathrm{m}$

Thus, $d{\text{'}}_{min}=\frac{1.22\times 0.12\times {10}^{-9}\times {10}^{-19}}{2\mathrm{sin}\beta}$

Taking ratio, $\frac{d{\text{'}}_{min}}{{d}_{min}}=\frac{0.12\times {10}^{-9}}{5500\times {10}^{-10}}=0.2\times {10}^{-3}$

#### Page No 65:

#### Question 18:

Consider a two slit interference arrangements (Fig. 10.4) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of *D* in terms of λ such that the first minima on the screen falls at a distance D from the centre O.

#### Answer:

Taking reference from the given figure, we can write:

${T}_{2}P={T}_{2}O+OP=D+X\phantom{\rule{0ex}{0ex}}\mathrm{and}{T}_{1}P={T}_{1}O-OP=D-x\phantom{\rule{0ex}{0ex}}{S}_{1}P=\sqrt{{\left({S}_{1}{T}_{1}\right)}^{2}+{\left(P{T}_{1}\right)}^{2}}=\sqrt{{D}^{2}+{\left(D-x\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{and}{S}_{2}P=\sqrt{{\left({S}_{2}{T}_{2}\right)}^{2}+{\left(P{T}_{2}\right)}^{2}}=\sqrt{{D}^{2}+{\left(D+x\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{minima},{S}_{2}P-{S}_{1}P=\left(2n-1\right)\frac{\lambda}{2}\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{first}\mathrm{minima},n=1,\sqrt{{D}^{2}+{\left(D+x\right)}^{2}}-\sqrt{{D}^{2}+{\left(D-x\right)}^{2}}=\frac{\lambda}{2}\phantom{\rule{0ex}{0ex}}\mathrm{If}x=D\phantom{\rule{0ex}{0ex}}\mathrm{Then},\sqrt{{D}^{2}+4{D}^{2}}-\sqrt{{D}^{2}+{0}^{2}}=\frac{\lambda}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Solving},\mathrm{the}\mathrm{above}\mathrm{equation}D=0.404\lambda $

#### Page No 65:

#### Question 19:

Figure 10.5 shown a two slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If* I*_{0} is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.

#### Answer:

Let ${A}_{\perp}$ and *A*_{||} be the amplitude of the either beam of light in perpendicular (⊥) and parallel (||) polarisation.

(A^{1}) and (A^{2}) indicates the amplitude from the first and second slit.

Now, intensity calculation without the presence of *P* in the path:

$A={A}_{\perp}+{A}_{\left|\right|}\phantom{\rule{0ex}{0ex}}{A}_{\perp}={A}_{\perp}^{1}+{A}_{\perp}^{2}={A}_{\perp}^{0}\mathrm{sin}\left(kx-\mathrm{\omega}t\right)+{A}_{\perp}^{0}\mathrm{sin}\left(kx-\mathrm{\omega}t+\varphi \right)$

(Let *Ï•* be the phase difference)

${A}_{\left|\right|}={A}_{\left|\right|}^{1}+{A}_{\left|\right|}^{2}={A}_{\left|\right|}^{0}\mathrm{sin}\left(kx-\mathrm{\omega}t\right)+{A}_{\left|\right|}^{0}\mathrm{sin}\left(kx-\mathrm{\omega}t+\varphi \right)$

Now, intensity,

$I={\left({A}_{\perp}^{1}+{A}_{\perp}^{2}\right)}^{2}+{\left({A}_{\left|\right|}^{1}+{A}_{\left|\right|}^{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow I=\left[{\left|{A}_{\perp}^{0}\right|}^{2}+{\left|{A}_{\left|\right|}^{0}\right|}^{2}\right]\left(\frac{1}{2}\right)\times 2\left(1+\mathrm{cos}\varphi \right)\phantom{\rule{0ex}{0ex}}\Rightarrow I=2{\left|{A}_{\perp}^{0}\right|}^{2}\left(1+\mathrm{cos}\varphi \right)\left[\because {\left|{A}_{\perp}^{0}\right|}_{\mathrm{average}}={\left|{A}_{\left|\right|}^{0}\right|}_{\mathrm{average}}\right]$

When no polariser is present, then intensity at principal maxima, ${I}_{\mathrm{o}}=2{\left|{A}_{\u27c2}^{0}\right|}^{2}\times 2=4{\left|{A}_{\u27c2}^{0}\right|}^{2}.....\left(1\right)$

Now, intensity calculation, when *P* is present in the path as shown in the diagram in the given question:

When, *P* is present then, ${A}_{\perp}^{2}$ is blocked:

So, intensity, ${I}^{\text{'}}={\left({A}_{\perp}^{1}\right)}^{2}+{\left({A}_{\left|\right|}^{1}+{A}_{\left|\right|}^{2}\right)}^{2}$

$=2{\left|{A}_{\perp}^{0}\right|}^{2}\left({\mathrm{cos}}^{2}\frac{\varphi}{2}\right)+{\left|{A}_{\perp}^{0}\right|}^{2}\frac{1}{2}$

So, intensity at principal maxima, when polariser is present:

$\begin{array}{rcl}{I}^{\text{'}}& =& {\left|{A}_{\perp}^{0}\right|}^{2}\left(2+\frac{1}{2}\right)\\ & =& \frac{{I}_{\mathrm{o}}}{4}\times \frac{5}{2}\left[\mathrm{As},{I}_{\mathrm{o}}=4{\left|{A}_{\perp}^{0}\right|}^{2}\right]\\ & =& \frac{5}{8}{I}_{\mathrm{o}}\end{array}$

Now, intensity at first minima with polariser:

$\begin{array}{rcl}{I}^{\text{'}\text{'}}& =& 0+{\left|{A}_{\perp}^{0}\right|}^{2}\times \frac{1}{2}\\ & =& 0+\frac{{I}_{\mathrm{o}}}{4}\times \frac{1}{2}=\frac{{I}_{\mathrm{o}}}{8}\end{array}$

#### Page No 66:

#### Question 20:

AC = CO = D, S_{1}C = S_{2}C = d << D

A small transparent slab containing material of μ =1.5 is placed along AS_{2 }(Fig.10.6). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab.

#### Answer:

Given, *μ *= 1.5, *L* = $\frac{d}{4}$ for the slab.

Slit width = 2*d* and screen distance = *D*,

Path difference,

$\left({\mathrm{AS}}_{2}+{\mathrm{S}}_{2}{\mathrm{P}}_{1}\right)-\left({\mathrm{AS}}_{1}+{\mathrm{S}}_{1}{\mathrm{P}}_{1}\right)\phantom{\rule{0ex}{0ex}}\u2206x=2d\mathrm{sin\theta}+\left(\mathrm{\mu}-1\right)\mathrm{L}\left[\mathrm{Assume},dD\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206x=2d\mathrm{sin\theta}+\left(1.5-1\right)\frac{d}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206x=2d\mathrm{sin\theta}+0.5\frac{d}{4}=2d\mathrm{sin\theta}+\frac{d}{8}.....\left(1\right)$

Now from given diagram in the question:

CO = *D*

$\mathrm{tan}\theta =\frac{{\mathrm{OP}}_{1}}{\mathrm{CO}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OP}}_{1}=\mathrm{CO}\mathrm{tan}\theta =D\mathrm{tan}\theta $

So, for principal maxima, path difference = 0

$\Rightarrow 0=2d\mathrm{sin}{\theta}_{\mathrm{o}}+\frac{d}{8}\left(\mathrm{Let}\theta ={\theta}_{\mathrm{o}}\right)$

$\Rightarrow \mathrm{sin}{\theta}_{\mathrm{o}}=-\frac{1}{16}\phantom{\rule{0ex}{0ex}}\mathrm{So},{\mathrm{OP}}_{1}=D\mathrm{tan}{\theta}_{\mathrm{o}}\approx D\mathrm{sin}{\theta}_{\mathrm{o}}\approx -\frac{D}{16}$

Here, *θ _{o}* is the angular position corresponding to the principal maxima.

But for the first maxima, the path difference = $\pm \frac{\mathrm{\lambda}}{2}$

So, $\u2206x=\pm \frac{\mathrm{\lambda}}{2}=2d\mathrm{sin}{\theta}_{\mathit{m}}+\frac{d}{8}\left[\mathrm{from}\mathrm{equation}\left(1\right)\right]$

(Let

*θ*be the angular position of first maxima)

_{m}$\Rightarrow \mathrm{sin}{\theta}_{\mathit{m}}=\frac{\pm {\displaystyle \frac{\mathrm{\lambda}}{2}}-{\displaystyle \frac{d}{8}}}{2d}$

[âˆµ diffraction occurs if the wavelength (

*λ*) of the wave is nearly equal to

*d*]

$\Rightarrow \mathrm{sin}{\theta}_{\mathit{m}}=\frac{\pm {\displaystyle \frac{\lambda}{2}}-{\displaystyle \frac{\lambda}{8}}}{2\mathrm{\lambda}}=\pm \frac{1}{4}-\frac{1}{16}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}{\theta}_{\mathit{m}}^{\mathit{+}}=\frac{3}{16}\left(\mathrm{for}+\frac{\lambda}{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}{\theta}_{m}^{-}=-\frac{5}{16}\left(\mathrm{for}-\frac{\lambda}{2}\right)$

So, the position of first principal maxima on positive side from point O,

$D\mathrm{tan}{\theta}_{m}^{+}=D\frac{\mathrm{sin}{\theta}_{m}^{+}}{\mathrm{cos}{\theta}_{m}^{+}}=D\frac{\mathrm{sin}{\theta}_{m}^{+}}{\sqrt{1-{\mathrm{sin}}^{2}{\theta}_{m}^{+}}}=\frac{3D}{\sqrt{247}}$

Also, on the negative side is at a distance,

$\begin{array}{rcl}D\mathrm{tan}{\theta}_{m}^{-}& =& D\frac{\mathrm{sin}{\theta}_{m}^{-}}{\sqrt{1-{\mathrm{sin}}^{2}{\theta}_{m}^{-}}}\\ & =& D\frac{\left(-{\displaystyle \frac{5}{16}}\right)}{\sqrt{1-{\left({\displaystyle \frac{5}{16}}\right)}^{2}}}\\ & =& \frac{-5D}{\sqrt{231}}\mathrm{below}\mathrm{the}\mathrm{point}\mathrm{O}.\end{array}$

#### Page No 66:

#### Question 21:

Four identical monochromatic sources A,B,C,D as shown in the (Fig.10.7) produce waves of the same wavelength λ and are coherent. Two receiver R_{1} and R_{2} are at great but equal distaces from B.

(i) Which of the two receivers picks up the larger signal?

(ii) Which of the two receivers picks up the larger signal when B is turned off?

(iii) Which of the two receivers picks up the larger signal when D is turned off?

(iv) Which of the two receivers can distinguish which of the sources B or D has been turned off?

#### Answer:

(i) Let the wave from A at *R*_{1} be,*y*_{A} = a cos ω*t* .....(1) (*a* = amplitude, ω = angular frequency)

Phase difference = Path difference × $\left(\frac{2\mathrm{\pi}}{\lambda}\right)$

So, phase difference of signal reached to *R*_{1} from A and B,

$\u2206{\varphi}_{\mathrm{AB}}=AB\times \frac{2\mathrm{\pi}}{\mathrm{\lambda}}=\frac{\mathrm{\lambda}}{2}\times \frac{2\mathrm{\pi}}{\mathrm{\lambda}}=\mathrm{\pi}$

Thus, the wave equation at *R*_{1} from source at B will be,

${y}_{\mathrm{B}}=a\mathrm{cos}\left(\mathrm{\omega}t-{\varphi}_{\mathrm{AB}}\right)=a\mathrm{cos}\left(\mathrm{\omega}t-\mathrm{\pi}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {y}_{\mathrm{B}}=-a\mathrm{cos}\mathrm{\omega}t.....\left(2\right)$

Similarly, phase difference of the signal from C with that from A,

${\varphi}_{\mathrm{AC}}=AC\times \frac{2\mathrm{\pi}}{\lambda}=\left(\frac{\lambda}{2}+\frac{\lambda}{2}\right)\times \frac{2\mathrm{\pi}}{\mathrm{\lambda}}=2\mathrm{\pi}$

So, the wave equation at *R*_{1} due to C will be:

${y}_{\mathrm{C}}=a\mathrm{cos}\left(\mathrm{\omega}t-{\varphi}_{\mathrm{AC}}\right)=a\mathrm{cos}\left(\mathrm{\omega}t-2\mathrm{\pi}\right)=a\mathrm{cos}\mathrm{\omega}t.....\left(3\right)$

Now, the phase difference between the signal from D with that of A at point *R*_{1} will be,

${\varphi}_{\mathrm{AD}}=AD\times \frac{2\mathrm{\pi}}{\mathrm{\lambda}}\phantom{\rule{0ex}{0ex}}{\varphi}_{\mathrm{AD}}=\left[\sqrt{{d}^{2}+{\left(\frac{\mathrm{\lambda}}{2}\right)}^{2}}-\left(d-\frac{\mathrm{\lambda}}{2}\right)\right]\times \frac{2\mathrm{\pi}}{\mathrm{\lambda}}$

Using the approximation, d >>> λ

$\Rightarrow {\varphi}_{\mathrm{AD}}\approx \left(\frac{\mathrm{\lambda}}{2}\right)\times \frac{2\mathrm{\pi}}{\mathrm{\lambda}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\varphi}_{\mathrm{AD}}\approx \mathrm{\pi}$

Thus, ${y}_{\mathrm{D}}=a\mathrm{cos}\left(\mathrm{\omega}t-{\varphi}_{\mathrm{AD}}\right)=a\mathrm{cos}\left(\mathrm{\omega}t-\mathrm{\pi}\right)\phantom{\rule{0ex}{0ex}}=-a\mathrm{cos}\left(\mathrm{\omega}t\right).....\left(4\right)$

Hence, the signal picked up at *R*_{1} due to source A, B, C and D will be,

$y={y}_{\mathrm{A}}+{y}_{\mathrm{B}}+{y}_{\mathrm{C}}+{y}_{\mathrm{D}}\phantom{\rule{0ex}{0ex}}\Rightarrow y=a\mathrm{cos}\mathrm{\omega}t-a\mathrm{cos}\mathrm{\omega}t+a\mathrm{cos}\mathrm{\omega}t-a\mathrm{cos}\mathrm{\omega}t\phantom{\rule{0ex}{0ex}}\Rightarrow y=0$

Similarly, for receiver *R*_{2}:

Let, wave signal at *R*_{2} from B be,

${y}_{\mathrm{B}}^{\text{'}}={a}_{1}\mathrm{cos}\left(\mathrm{\omega}t\right).....\left(5\right)\phantom{\rule{0ex}{0ex}}{y}_{\mathrm{D}}^{\text{'}}=-{a}_{1}\mathrm{cos}\left(\mathrm{\omega}t\right).....\left(6\right)$

Path difference between signal at A and B will be,

$=\sqrt{{d}^{2}+{\left(\frac{\mathrm{\lambda}}{2}\right)}^{2}}-d\phantom{\rule{0ex}{0ex}}=d\sqrt{1+\frac{{\mathrm{\lambda}}^{2}}{4{d}^{2}}}-d\phantom{\rule{0ex}{0ex}}\approx \frac{{\mathrm{\lambda}}^{2}}{8{d}^{2}}$

So, phase difference,

${\varphi}_{\mathrm{AB}}=\frac{{\mathrm{\lambda}}^{2}}{8{d}^{2}}\times \frac{2\mathrm{\pi}}{\mathrm{\lambda}}=\frac{\mathrm{\pi \lambda}}{4d}$

So, the wave at *R*_{2} due to A will be,

${y}_{\mathrm{A}}^{\text{'}}={a}_{1}\mathrm{cos}\left(\mathrm{\omega}t-{\varphi}_{\mathrm{AB}}\right)\phantom{\rule{0ex}{0ex}}$

Similarly,

${y}_{\mathrm{C}}^{\text{'}}={a}_{1}\mathrm{cos}\left(\mathrm{\omega}t-{\varphi}_{\mathrm{AC}}\right)\phantom{\rule{0ex}{0ex}}$

Let, ${\varphi}_{\mathrm{AB}}={\varphi}_{\mathrm{AC}}=\varphi $

The, signal picked up by *R*_{2} is,

${y}^{\text{\'}}={y}_{\mathrm{A}}^{\text{\'}}+{y}_{\mathrm{B}}^{\text{\'}}+{y}_{\mathrm{C}}^{\text{\'}}+{y}_{\mathrm{D}}^{\text{\'}}\phantom{\rule{0ex}{0ex}}{y}^{\text{\'}}={a}_{1}\mathrm{cos}\left(\mathrm{\omega}t-\varphi \right)+{a}_{1}\mathrm{cos}\left(\mathrm{\omega}t\right)+{a}_{1}\mathrm{cos}\left(\mathrm{\omega}t-\varphi \right)-{a}_{1}\mathrm{cos}\left(\mathrm{\omega}t\right)\phantom{\rule{0ex}{0ex}}=2{a}_{1}\mathrm{cos}\left(\mathrm{\omega}t-\varphi \right)\phantom{\rule{0ex}{0ex}}{\left|{y}^{\text{\'}}\right|}^{2}=4{a}_{1}^{2}{\mathrm{cos}}^{2}\left(\mathrm{\omega}t-\varphi \right)\phantom{\rule{0ex}{0ex}}>">{I}_{{\mathrm{R}}_{2}}$

Hence, the receiver *R*_{2} will pick up larger signal.

(ii) Now, if source B is switched off:

So, *R*_{1} will pick up:*y* = *a* cos ω*t*

$>">{I}_{{\mathrm{R}}_{1}}$= *a*^{2}*t*>

$>">{I}_{{\mathrm{R}}_{1}}$

Also, *R*_{2} picks up, *y* = *a* cos ω*t*

$\begin{array}{r}>">{I}_{{\mathrm{R}}_{2}}\\ =& {a}^{2}{\mathrm{cos}}^{2}\mathrm{\omega}t\end{array}& =& \frac{{a}^{2}}{2}$

Hence, both the receivers pick up same signal.

(iii) Now, if D is switched off then, receiver *R*_{1 }will pick up*y *= *a* cos ω*t*

So, $>">{I}_{{\mathrm{R}}_{1}}$

and receiver *R*_{2} will pick up *y* = 3*a* cos ω*t*

$>">{I}_{{\mathrm{R}}_{2}}$

Hence, *R*_{2} will pick up larger signal than *R*_{1}.

(iv) Signal at receiver *R*_{1} indicates that B has been switched off and an enhanced signal at *R*_{2} indicates that D has been switched off.

^{}

#### Page No 67:

#### Question 22:

The optical properties of a medium are governed by the relative permitivity (ε* _{r}*) and relative permeability (μ

*). The refractive index is defined as $\sqrt{{\mu}_{r}{\epsilon}_{r}}$ =*

_{r}*n*.For ordinary material ε

_{r}> 0 and μ

_{r}> 0 and the positive sign is taken for the square root. In 1964, a Russian

scientist V. Veselago postulated the existence of material with ε

_{r}< 0 and μ

_{r}< 0. Since then such ‘metamaterials’ have been produced in the laboratories and their optical properties studied. For such materials

*n*= $-\sqrt{{\mu}_{r}{\epsilon}_{r}}$ . As light enters a medium of such refractive index the phases travel away from the direction of propagation.

(i) According to the description above show that if rays of light enter such a medium from air (refractive index =1) at an angle θ in 2

^{nd }quadrant, them the refracted beam is in the 3rd quadrant.

(ii) Prove that Snell’s law holds for such a medium.

#### Answer:

(i) Given refractive index of the medium, *n* = $\sqrt{{\mu}_{r}{\epsilon}_{r}}$ (*μ*_{r }> 0 and *ε _{r}* > 0).

But for meta-materials,

*n*= $-\sqrt{{\mu}_{r}{\epsilon}_{r}}$ (

*μ*

_{r }< 0 and

*ε*< 0).

_{r}So, if the ray of light enters a medium (

*n*< 0) at an angle θ

_{i}as show in diagram:

**Fig (1)**

Here, ED shows the wavefront, i.e., all the points on this must have same phase.

So, according to Vaselago postulate,

$-\sqrt{{\mu}_{r}{\epsilon}_{r}}AE=BC-\sqrt{{\epsilon}_{r}{\mu}_{r}}\mathrm{CD}\left[\mathrm{Optical}\mathrm{path}\mathrm{length}\right]$

$\Rightarrow \mathrm{BC}=\sqrt{{\epsilon}_{r}{\mu}_{r}}\left(\mathrm{CD}-\mathrm{AE}\right)$

As shown in the diagram

BC > 0 and CD > AE

So, the incident ray is in 2nd quadrant, the refracted ray is in 3rd quadrant.

Hence, the postulate is reasonable.

Now, if the ray emerges out in the fourth quadrant as shown in diagram below:

**Fig (2)**

$-\sqrt{{\epsilon}_{r}{\mu}_{r}}\mathrm{AE}=\mathrm{BC}-\sqrt{{\epsilon}_{r}{\mu}_{r}}\mathrm{CD}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=\sqrt{{\epsilon}_{r}{\mu}_{r}}\left(\mathrm{CD}-\mathrm{AE}\right)$

In this case,

AE > CD

⇒ CD – AE < 0

⇒ BC < 0

which is not possible.

Hence, the postulate is correct.

(ii)

**For Figure (1)**

BC = AC sin θ

_{i}And, CD – AE = AC sin θ

_{r}Since, $-\sqrt{{\epsilon}_{r}{\mu}_{r}}\left(\mathrm{AE}-\mathrm{CD}\right)=\mathrm{BC}$

Dividing both side by AC we get:

$\Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\left(-\sqrt{{\epsilon}_{r}{\mathrm{\mu}}_{r}}\right)\frac{\left(\mathrm{AE}-\mathrm{CD}\right)}{\mathrm{AC}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{sin\theta}}_{i}=\left(-\sqrt{{\epsilon}_{r}{\mathrm{\mu}}_{r}}\right)\times {\mathrm{sin\theta}}_{r}\phantom{\rule{0ex}{0ex}}\Rightarrow 1\times {\mathrm{sin\theta}}_{i}=-n{\mathrm{sin\theta}}_{r}\left(\mathrm{Snell}\text{'}\mathrm{s}\mathrm{law}\right)\phantom{\rule{0ex}{0ex}}$

⇒ Snell's law holds good for such medium.

#### Page No 67:

#### Question 23:

To ensure almost 100 per cent transmittivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediated between that of air and glass (which makes the optical element of the lens). A typically used dielectric film is MgF_{2} (*n* = 1.38). What should the thickness of the film be so that at the center of the visible speetrum (5500 $\stackrel{\xb0}{\mathrm{A}}$) there is maximum transmission.

#### Answer:

Ray incident on the air-film interface and get refracted at an angle *r *But, it gets totally internally reflection from the film-glass interface and after reflection at D it get refraction at film-air interface again at C.

Let angle of incidence by *i *and angle of refraction be *r* as shown in the diagram.

The refractive index of the medium,*n*_{1} = 1 (air), *n*_{2} = 1.38 (film), *n*_{3} = 1.5 (glass)

Let thickness of the prism be *d*.

So the optical path difference,*n* (AD + CD) – AB

But, AD = CD = $\frac{d}{\mathrm{cos}r}$

$\left[\mathrm{In}\mathrm{the}\mathrm{shown}\mathrm{image},\mathrm{AD}=\mathrm{CD},\mathrm{AP}=\mathrm{MD}=d\right]\phantom{\rule{0ex}{0ex}}\left[\mathrm{In}\mathrm{triangle}\mathrm{AMD},\mathrm{cos}r=\frac{\mathrm{MD}}{\mathrm{AD}}=\frac{d}{\mathrm{AD}}\right]\phantom{\rule{0ex}{0ex}}$

Similarly,

AB = AC sin*i*

$\frac{AC}{2}=d\mathrm{tan}r$

So, AC = 2*d* tan*r*

⇒ AB = 2*d* tan*r *sin*i*

Hence, the optical path difference

$=2{n}_{2}\frac{d}{\mathrm{cos}r}-2d\mathrm{tan}r\mathrm{sin}i\phantom{\rule{0ex}{0ex}}=2{n}_{2}d\mathrm{cos}r\left(\because \mathrm{tan}r=\frac{\mathrm{sin}r}{\mathrm{cos}r}\right)$

But for distructive interference path difference $=\frac{\lambda}{2}$

$\Rightarrow 2{n}_{2}d\mathrm{cos}r=\frac{\lambda}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {n}_{2}d\mathrm{cos}r=\frac{\lambda}{4}$

Now, very very small angle of incidence,

$i\approx r\approx 0\phantom{\rule{0ex}{0ex}}\mathrm{So},{n}_{2}d\approx \frac{\lambda}{4}\phantom{\rule{0ex}{0ex}}if\lambda =5500\stackrel{\xb0}{A},\phantom{\rule{0ex}{0ex}}\Rightarrow d=\frac{5500\stackrel{\xb0}{A}}{1.38\times 4}\approx 996\stackrel{\xb0}{A}$

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