NCERT Exemplar Class 12 Physics Solutions Chapter 10 Wave Optics

Question:2

Consider sunlight incident on a slit of width $10^{4} $A. The image seen through the slit shall
A. be a fine sharp slit white in colour at the centre.
B. a bright slit white at the centre diffusing to zero intensities at the edges.
C. a bright slit white at the centre diffusing to regions of different colours.
D. only be a diffused slit white in colour.

Answer:

The answer is the option (a)
Diffraction is a phenomenon of bending of light rays around an obstacle or an aperture of a similar wavelength.

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
A. $\frac{4 \pi d}{\lambda}\left ( 1-\frac{1}{n^2}\sin^2 \theta \right )^{1/2}+\pi$
B.$\frac{4 \pi d}{\lambda}\left ( 1-\frac{1}{n^2}\sin^2 \theta \right )^{1/2}$
C. $\frac{4 \pi d}{\lambda}\left ( 1-\frac{1}{n^2}\sin^2 \theta \right )^{1/2}+\frac{\pi}{2}$
D.$\frac{4 \pi d}{\lambda}\left ( 1-\frac{1}{n^2}\sin^2 \theta \right )^{1/2}+2 \pi$

Answer:

The answer is the option (a)

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:

The answer is the option (c)
Here, in this case, due to the presence of red and blue filters. The waves of light will only be Red and Blue. In YDSE, the monochromatic light is used for the formation of fringe on the screen. Therefore, in this case, there will be no interference.

Question:5

Figure 10.2 shows a standard two-slit arrangement with slits S₁, S₂. P₁, P₂ are the two minima points on either side of P (Fig. 10.2). At P₂ on the screen, there is a hole and behind P₂ is a second 2- slit arrangement with slits S₃, S₄ 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:

The answer is the option (d)
Wavefront is the plane where every point on given wave front is the source of disturbance which are known as secondary wavelets.

The wavefront emitted by a narrow source is divided in two parts by reflection, refraction or diffraction. The coherent sources so obtained are imaginary.

Question:7

Consider sunlight incident on a pinhole of $10^{3}$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:

The correct answers are the options (b,d)
Diffraction of light can only be observed if the size of the aperture or obstacle is less than the wavelength of the light wave.
The given width of the pinhole is 103 angstrom. The wavelength of sunlight is 4000 angstrom to 8000 A. Therefore, the light is diffracted from the hole. And due to this, the image formed on the screen will be geometrically different.

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 correct answers are the options (a, b)
Key concept: The 'shadow' of the hole of dimeter d is spread out over an angle
$\Delta \theta =1.22\frac{\lambda}{D}\Rightarrow \Delta \theta\propto \frac{1}{D}$

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:

The correct answers are the options (a, b)

Due to the point source light propagates in all direction symmetrically and hence, wavefront will be spherical.

As the intensity of the source will be

$I\propto \frac{1}{r^{2}}$

where r is the radius of the wavefront at any time.
Hence the intensity decreases in proportion to the distance squared.

Question:11

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

Answer:

Here the orientation of the rays is perpendicular to L1. It forms an image at I1, i.e. the focal length. Again, the image is converged and goes through L2 after which the final image is formed at I. The nature of the wavefronts emerging from the final image is Spherical.

Question:12

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

Answer:

As the sun is at an exceptionally large distance from earth. Assuming it as a point source of light at infinity, as seen from earth, we can conclude that the radius of the wavefront from the sun to the earth is infinite. This would mean that the rays are perpendicular to earth and the wave front is almost a plane.

Question:13

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

Answer:

The wavelength of sound waves is 15 m to 15 mm for 20 Hz to 20,000 Hz respectively. Therefore, sound waves diffraction take place when the comparable size of the obstacle is confronted. While in the case of the light wave, the wavelength of visible light is 0.4 to 0.7 micron. So, the obstacles of this size are not easily present around us. Therefore, diffraction of light is not so evident in day to day life.

Question:14

The human eye has an approximate angular resolution of ? = 5.8×10^–4 rad and a typical photo printer 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:

It is given, angular resolution of human eye $\phi =5.8 \times 10^{-4}$ rad and printer print 300 dots per inch.
The linear distance between the two dots is
$l=\frac{2.54}{300} cm=0.84 \times 10^{-2}\; cm$
At a distance of z cm, this subtends an angle,
$\phi =\frac{l}{z}\\\\ z= \frac{l}{\phi }= \frac{0.84 \times 10^{-2}\;cm}{5.8 \times 10^{-4}}=14.5 \; cm$
If a printed page be held at a distance of 14.5 cm, then one does not be able to see the individual dots.

Question:15

A polaroid (I) is placed in front of a monochromatic source. Another polaroid (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:

Monochromatic source of light is kept behind polaroid (I). Then some other polaroid (II) is placed in front of polaroid (I). So, the axes of both polaroid are parallel to each other; the light passes through (II) unaffected.

Now polaroid (II) is rotated till no light passes. In this situation the pass axis of polaroid (II) is perpendicular to polaroid (I), then (I) and (II) are set in a crossed position. No light passes through a polaroid- (II)

Now third polaroid (III) is now placed in between (I) and (II). Only in the special cases when the pass axis of (III) is parallel to (I) or (II) there shall be no light emerging. In all other cases, there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III).


Now, the polaroid (II) is rotated such that (I) and (II) are in a crossed position. Therefore, no light will pass through (II).

Question:17

For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for the light of $5000 Å$ and electrons accelerated through 100V used as the illuminating substance.

Answer:

$5000 Å = 5000 \times 10^{-10} m\\ \\ \frac{1}{d} = \frac{2 \sin \beta}{1.22 \lambda } \\\\ d\;min = \frac{1.22 \lambda}{2 \sin \beta}\\d_{min}=\frac{1.22\times5000\times10^{-10}}{2sin\beta}$
When we use 100V light,
$\lambda_d =\frac{1.27}{\sqrt{V}}nm=\frac{1.27}{\sqrt{100}}nm$
$d' min =\frac{1.22\lambda_d }{2 \sin \beta}$
$d'min = \frac{1.22 \times 1.27 \times 10 ^{-10} }{2 \sin \beta}$
The required ratio $=\frac{d \;min}{d'min} = \frac{1.22 }{5000} =0.244 \times 10^{-3}$

Question:18

Consider a two-slit interference arrangement (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 such that the first minima on the screen fall at a distance D from the centre O.

Answer:

$\begin{aligned} & \quad S_1 P=\sqrt{D^2}+(D-x)^2 \\ & S_2 P=\sqrt{D^2}+(D-x)^2 \\ & T_2 P=D+x \\ & T_2 P=D-x \\ & {\left[D^2+(D+x)^2\right]^{1 / 2}-\left[D^2+(D-x)^2\right]^{-1 / 2}=\frac{\lambda}{2}} \\ & D=0.404 \lambda\end{aligned}$

Question:20

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:

For the principal maxima,(path difference is zero)

$
\begin{aligned}
& \Delta x=2 d \sin \theta+[(\mu-1) L]=0 \\
& \sin \theta_0=-\frac{L(\mu-1)}{2 d}=-\frac{-L(0.5)}{2 d}[\therefore L=d / 4] \\
& \text { or } \quad \Rightarrow \sin \theta_0=\frac{-1}{16}
\end{aligned}
$

$\theta_0$ is the angular position corresponding to the principal maxima.

$
\Rightarrow \quad O P=D \tan \theta_0 \approx D \sin \theta_0=\frac{-D}{16}
$

For the first minima, the path difference is

$
\begin{aligned}
& \pm \frac{\lambda}{2} \\
& \Delta x=2 d \sin \theta_1+0.5 L= \pm \frac{\lambda}{2} \\
& \sin \theta_1=\frac{ \pm \lambda / 2-0.5 L}{2 d}=\frac{ \pm \lambda / 2-d / 8}{2 d} \\
& \Rightarrow \sin \theta_1=\frac{ \pm \lambda / 2-\lambda / 8}{2 \lambda}= \pm \frac{1}{4}-\frac{1}{16}
\end{aligned}
$

Question:21

Four identical monochromatic sources A, B, C, D, as shown in the (Fig.10.7) produce waves of the same wavelength and are coherent. Two receiver R1 and R2 are at great but equal distances 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 us consider the disturbances at the receiver R₁, which is at a distance d from B.
Let the equation of wave at R₁, because of A be
$y_A=a \cos \omega t$ $......(i)$
The path difference of the signal from A with that from B is $\lambda/2$ and hence, the phase difference
$\Delta \phi = \frac{2\pi }{\lambda }\times \left ( path\;dif\! \! ference \right )=\frac{2\pi }{\lambda }\times \frac{\lambda }{2}$
Thus, the wave equation at R1, because of B is
$y_B=a \cos \left ( \omega t-\pi \right )= -a \cos \omega t$ $......(ii)$
The path difference of the signal from C with that from A is $\lambda$ and hence the phase difference
$\Delta \phi = \frac{2\pi }{\lambda }\times \left ( path\;dif\! \! ference \right )=\frac{2\pi }{\lambda }\times \lambda=2 \pi$
Thus, the wave equation at R1 because of C is
$y_C=a \cos \omega t =a \cos\left ( \omega t-2\pi \right )= a \cos \omega t$ $......(iii)$
The path difference between the signal from D with that of A is

$\Delta _{x_{R_1}}= \sqrt{d^2+\left ( \frac{\lambda }{2} \right )^{2}}-\left (d- \frac{\lambda }{2} \right )=d\left ( 1+\frac{\lambda ^{2}}{4d^{2}} \right )^{1/2}-d+\frac{\lambda }{2}$
$=d\left ( 1+\frac{\lambda ^{2}}{8d^{2}} \right )-d.\frac{\lambda }{2}=\frac{\lambda }{2}\left ( \because d>>\lambda \right )$
Therefore, phase difference is $\pi$
$y_D=a \cos\left ( \omega t-\pi \right )= -a \cos \omega t$ $......(iii)$
The resultant signal picked up at R_1, from all the four sources is the summation of all four waves,
$y_{R_1}=y_A+y_B+y_C+y_D$
$y_{R_1}=a \cos \omega t-a \cos \omega t+a \cos \omega t-a \cos \omega t=0$
Thus, the signal picked up at R₁ is zero.
Now let us consider the resultant signal received at R_2. Let the equation of wave at R₂ . Let the equation
of wave at R₂, because of B be
$y_{B}=a_1 \cos \omega t$
The path difference of the signal from from D with that from B is $\frac{\lambda }{2}$ and
hence, the phase difference
$\Delta \phi =\frac{2\pi }{\pi }\times \left ( \text {path difference} \right )=\frac{2\pi}{\lambda }\times \frac{\lambda }{2}=\pi$
Thus, the wave equation at $R_{2},$ because of D is
$y_{B}=a_{1}\cos \left ( \omega _{t}-\pi \right )=-a_{1}\cos\; \omega t\; \; \; \; \; \; \; \; \; \; \; \; \; ....(ii)$
The path difference between signal at A and that at B is
$\Delta x_{R_{2}}=\sqrt{\left ( d \right )^{2}+\left ( \frac{\lambda }{2} \right )^{2}}-d=d\left ( 1+\frac{\lambda ^{2}}{4d^{2}} \right )^{\frac{1}{2}}-d\simeq \frac{\lambda ^{2}}{8d^{2}}$
As $d>>\lambda ,$ therefore this path differences $\Delta x_{R_{2}}\rightarrow 0$
and phase difference $\Delta \phi =\frac{2\pi}{\lambda }\times \left ( \text {path difference} \right )$
$=\frac{2\pi}{\lambda }\times 0\rightarrow 0\left ( \text {very small} \right )=\phi \left ( \text {say} \right )$
Hence, $y_{B}=a_{1}\cos \left ( \omega _{t}-\phi \right )$
Similarly, $y_{B}=a_{1}\cos \left ( \omega _{t}-\phi \right )$
The resultant signal picked up at $R_{2},$ from all the four sources is the summation of all four waves, $y_{R_{2}}=y_{A}+y_{B}+y_{C}+y_{D}$
$y_{R_{2}}=a_{1}\cos \; \omega t-a_{1}\cos \omega t+a_{1}\cos\left ( \omega t-\phi \right )+a_{1}\cos \left ( \omega t-\phi \right )$
$= 2a_{1}\cos \left ( \omega t-\phi \right )$
$\therefore$ Signal picked up by $R_{2}$ is $y_{R_{2}}=2a_{1}\cos \left ( \omega t-\phi \right )$
$\therefore$ $\left [ V_{R_{2}} \right ]^{2}=4a_{1}^{2}\cos^{2}\left ( \omega t-\phi \right )\Rightarrow \left \langle I_{R_{2}} \right \rangle=2a_{1}^{2}$
Thus, $R_{2}$ picks up the larger signal.
(ii) If B is switched off,
$R_{1}$, picks up $y=a\cos \; \omega t$
$\therefore$ $\left \langle I_{R_{1}} \right \rangle=a^{2}<\cos^{2}\omega t>=\frac{a^{2}}{2}$
$R_{2}$ picks up $y=a\; \cos\; \omega t$
$\left \langle I_{R_{2}} \right \rangle=a^{2}<\cos^{2}\omega t>=\frac{a^{2}}{2}$
Thus , $R_{1}$ and $R_{2}$ pick up the same signal
(iii) If D is switched off.
$R_{1}$ picks up $y=a\; \cos\; \omega t$
$\therefore$ $\left \langle I_{R_{1}} \right \rangle=\frac{1}{2}a^{2}$
$R_{2}$ picks up $y=3a\; \cos\; \omega t$
$\therefore$ $\left \langle I_{R_{2}} \right \rangle=9a^{2}<\cos^{2}\omega t>=\frac{9a^{2}}{2}$
Thus, $R_{2}$ picks up larger signal compared to $R_{1}$.
(iv) Thus, a signal at $R_{1}$ indicates B has been switched off and an enhanced signal at $R_{2}$ indicates D has been switched off.

Question:22

The optical properties of a medium are governed by the relative permittivity (εr) and relative permeability (μr ). The refractive index is defined as $\sqrt{\mu, \epsilon }=n$ For ordinary material εr > 0 and μr > 0, and the positive sign is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with εr < 0 and μr < 0. Since then, such ‘metamaterials’ have been produced in the laboratories and their optical properties studied. For such materials $n=\sqrt{\mu, \epsilon }$
As light enters a medium of such refractive index, the phases travel away from the direction of propagation.

(i) According to the description above shows that if rays of light enter such a medium from the air (refractive index = 1) at an angle in 2^nd quadrant, them the refracted beam is in the 3^rd quadrant.
(ii) Prove that Snell’s law holds for such a medium.

Answer:

Again consider figure (i), let AB represent the incident wavefront and DE represent the refracted wavefront. All point on a wavefront must be in same phase and in turn, must have the same optical path length.

Thus $\quad-\sqrt{\epsilon_r \mu_r} A E=B C-\sqrt{\epsilon_r \mu_r} C D$
Or $\quad B C=\sqrt{\epsilon_r \mu_r}(C D-A E)$

$
B C>0, C D>A E
$

As showing that the postulate is reasonable. If however, the light proceeded in the sense it does for ordinary material (viz. in the fourth quadrant, Fig.2)

Then $\quad-\sqrt{\epsilon_r \mu_r} A E=B C-\sqrt{\epsilon_r \mu_r} C D$
Or $\quad B C=\sqrt{\epsilon_r \mu_r}(C D-A E)$
if $B C>0$, then $C D>A E$
Which is obvious from Fig. (i). Hence, the postulate is reasonable.
However, if the light proceeds in the sense it does for ordinary material, (going from the second quadrant to 4th quadrant)as shown in Fig. (i). then proceeding as above,

$
-\sqrt{\epsilon_r \mu_r} A E=B C-\sqrt{\epsilon_r \mu_r} C D
$

Or $\quad B C=\sqrt{\epsilon_r \mu_r}(C D-A E)$
As $A E>C D$ Therfore $B C<0$
Which is not possible. Hence, the postulate is correct

(ii) From Fig (i),
$\; \; \; \; \; \; BC=AC \sin \theta,\\ and\: \; \; \; \;CD-AE=AC \sin \theta \\As\; \; \; \; \; \; BC=-\sqrt{\mu_r \epsilon_r }(AE-CD)\\\\ \therefore \; \; \; \; \; \; AC \sin \theta_i =-\sqrt{\epsilon_r\mu_r }AC \sin \theta\\\\or\; \; \; \; \; \; \; \; \frac{\sin \theta_i}{\sin \theta_r}=\sqrt{\epsilon_r\mu_r }\;n$
Which proves Snell's law

Question:23

To ensure almost 100 per cent transitivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediated between that of (which makes the optical element of the lens). A typically used dielectric film is MgF₂ (n = 1.38). What should the thickness of the film be so that at the center of the visible spectrum (5500 Å) there is a maximum transmission

Answer:

IA is an incident ray at point A is such that the incident angle I is formed from air to the film surface.
AR₁ and AD are the reflected and refracted rays, respectively. D is the point on which the partial reflection of glass and film interface. CR₂ and AR₁ are parallel.
$\begin{aligned} & \mu(A D+C D)-A B \\ & A D=A C=\frac{d}{\cos r} \\ & d \tan r=\frac{A C}{2} \\ & d=1000 A\end{aligned}$