NCERT Exercise Q & A

Q10.19: Exercise Problems

Q10.1: Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.

Answer: Wavelength of incident monochromatic light,
$\lambda=589 \mathrm{~nm}=589 \times 10^{-9} \mathrm{~m}$
Speed of light in air, $c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$
Refractive index of water, $\mu=1.33$
(a) The ray will reflect back in the same medium as that of incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be the same as that of the incident ray.
Frequency of light is given by the relation,
$\begin{aligned} \nu & =\frac{c}{\lambda} \\ & =\frac{3 \times 10^8}{589 \times 10^{-9}} \\ & =5.09 \times 10^{14} \mathrm{~Hz} \end{aligned}$
Hence, the speed, frequency, and wavelength of the reflected light are $3 \times 10^8 \mathrm{~m} / \mathrm{s}$, $5.09 \times 10^{14} \mathrm{~Hz}$, and 589 nm respectively.
(b) Frequency of light does not depend on the property of the medium in which it is travelling. Hence, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in air.
$\therefore$ Refracted frequency, $v=5.09 \times 10^{14} \mathrm{~Hz}$
Speed of light in water is related to the refractive index of water as:
$\begin{aligned} & v=\frac{c}{\mu} \\ & v=\frac{3 \times 10^8}{1.33}=2.26 \times 10^8 \mathrm{~m} / \mathrm{s} \end{aligned}$
Wavelength of light in water is given by the relation,
$\begin{aligned} \lambda & =\frac{\nu}{v} \\ & =\frac{2.26 \times 10^8}{5.09 \times 10^{14}} \\ & =444.007 \times 10^{-9} \mathrm{~m} \\ & =444.01 \mathrm{~nm} \end{aligned}$
Hence, the speed, frequency, and wavelength of refracted light are $2.26 \times 10^8 \mathrm{~m} / \mathrm{s}$, 444.01 nm , and $5.09 \times 10^{14} \mathrm{~Hz}$ respectively.

Q10.2: What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.

Answer: (a) The shape of the wavefront in case of a light diverging from a point source is spherical. The wavefront emanating from a point source is shown in the given figure.

(b) The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid. This is shown in the given figure.

(c) The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.

Q10.3: (a) The refractive index of glass is 1.5 . What is the speed of light in glass? (Speed of light in vacuum is $3.0 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$ )
(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?

Answer: (a) Refractive index of glass, $\mu=1.5$
Speed of light, $c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$
Speed of light in glass is given by the relation,
$\begin{aligned} v & =\frac{c}{\mu} \\ & =\frac{3 \times 10^8}{1.5}=2 \times 10^8 \mathrm{~m} / \mathrm{s} \end{aligned}$
Hence, the speed of light in glass is $2 \times 10^8 \mathrm{~m} / \mathrm{s}$.
(b) The speed of light in glass is not independent of the colour of light.
The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

Q10.4: In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm . Determine the wavelength of light used in the experiment.

Answer: Distance between the slits, $d=0.28 \mathrm{~mm}=0.28 \times 10^{-3} \mathrm{~m}$
Distance between the slits and the screen, $D=1.4 \mathrm{~m}$
Distance between the central fringe and the fourth $(n=4)$ fringe,
$u=1.2 \mathrm{~cm}=1.2 \times 10^{-2} \mathrm{~m}$
In case of a constructive interference, we have the relation for the distance between the two fringes as:
$u=n \lambda \frac{D}{d}$
Where,
$n=$ Order of fringes $=4$
$\lambda=$ Wavelength of light used
$\therefore=\frac{u d}{n D}$
$\begin{aligned} & =\frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4} \\ & =6 \times 10^{-7} \\ & =600 \mathrm{~nm} \end{aligned}$
Hence, the wavelength of the light is 600 nm.

Q10.5: In Young’s double-slit experiment using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda$, is $K$ units. What is the intensity of light at a point where path difference is $\lambda / 3$ ?

Answer: Let $I_1$ and $I_2$ be the intensity of the two light waves. Their resultant intensities can be obtained as:
$I^{\prime}=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi$
Where,
$\phi=\text { Phase difference between the two waves }$
For monochromatic light waves,
$\begin{aligned} & I_1=I_2 \\ & \begin{aligned} \therefore I^{\prime} & =I_1+I_1+2 \sqrt{I_1 I_1} \cos \phi \\ & =2 I_1+2 I_1 \cos \phi \end{aligned} \end{aligned}$
Phase difference $=\frac{2 \pi}{\lambda} \times$ Path difference
Since path difference $=\lambda$,
Phase difference, $\phi=2 \pi$
$\therefore I^{\prime}=2 I_1+2 I_1=4 I_1$
$\begin{aligned} &\text { Given, }\\ &\begin{aligned} & I^{\prime}=K \\ & \therefore I_1=\frac{K}{4} \dots(1) \end{aligned} \end{aligned}$
When path difference $=\frac{\lambda}{3}$,
Phase difference, $\quad \phi=\frac{2 \pi}{3}$
Hence, resultant intensity, $I_R^{\prime}=I_1+I_1+2 \sqrt{I_1 I_1} \cos \frac{2 \pi}{3}$
$=2 I_1+2 I_1\left(-\frac{1}{2}\right)=I_1$
Using equation (1), we can write:
$I_R=I_1=\frac{K}{4}$
Hence, the intensity of light at a point where the path difference is $\frac{\lambda}{3}$ is $\frac{K}{4}$ units.

Q10.6: A beam of light consisting of two wavelengths, 650 nm and 520 nm , is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm .
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Answer: Wavelength of the light beam, $\lambda_1=650 \mathrm{~nm}$
Wavelength of another light beam, $\lambda_2=520 \mathrm{~nm}$
Distance of the slits from the screen $=D$
Distance between the two slits $=d$
(a) Distance of the $n^{\text {th }}$ bright fringe on the screen from the central maximum is given by the relation,
$x=n \lambda_1\left(\frac{D}{d}\right)$
For third bright fringe, $n=3$
$\therefore x=3 \times 650 \frac{D}{d}=1950\left(\frac{D}{d}\right) \mathrm{nm}$
(b) Let the $n^{\text {th }}$ bright fringe due to wavelength $\lambda_2$ and $(n-1)^{\text {th }}$ bright fringe due to wavelength ${\lambda_1}$ coincide on the screen. We can equate the conditions for bright fringes as:
$\begin{aligned} & n \lambda_2=(n-1) \lambda_1 \\ & 520 n=650 n-650 \\ & 650=130 n \\ & \therefore n=5 \end{aligned}$
Hence, the least distance from the central maximum can be obtained by the relation:
$\begin{aligned} x & =n \lambda_2 \frac{D}{d} \\ & =5 \times 520 \frac{D}{d}=2600 \frac{D}{d} \mathrm{~nm} \end{aligned}$
Note: The value of $d$ and $D$ are not given in the question.

Exemplar Section

VSA

Q10.10: Is Huygen’s principle valid for longitudunal sound waves?

Answer: Yes it can. Huygen’s principle basically states that every point wave front can be considered as a secondary source of tiny wavelets that spread out in the forward direction of the wave itself. 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.

Q10.11: 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: Orientation of wave front is perpendicular to ray. The ray diagram of the situation is as shown in figure.

Parallel rays incident on lens $L_1$ forms the image $I_2$ at the focal point of the lens. This image acts as object for the lens $L_2$ Now, due to the converging lens $L_2$, let final image formed is I which is point image. Hence the wavefront for this image will be of spherical symmetry.

Q10.12: What is the shape of the wavefront on earth for sunlight?

Answer: The sun is at very large distance from the earth. Assuming sun as spherical, it can be considered as point source situated at infinity. We can treat it like a point object as seen from the surface of earth. Because of large distance, the radius of wavefront can be considered as Large (infinity) and hence, wavefront is almost plane.

Q10.13: Why is the diffraction of sound waves more evident in daily experience than that of light wave?

Answer: The frequencies of sound waves lie between 20 Hz to 20 kHz , their wavelength ranges between 15 m to 15 mm . The diffraction occurs if the wavelength of waves is nearly equal to slit width. The wavelength of light waves is $7000 \times 10^{-10} \mathrm{~m}$ to $4000 \times 10^{-10} \mathrm{~m}$. For observing diffraction of light we need very narrow slit width. In daily life experience we observe the slit width 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.

Q10.14: The human eye has an approximate angular resolution of $\phi=5.8 \times 10^{-4} \mathrm{rad}$ and a typical photoprinter prints a minimum of 300 dpi (dots per inch, 1 inch $=2.54 \mathrm{~cm}$ ). 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 prints 300 dots per inch.
It is given, angular resolution of human eye $\phi=5.8 \times 10^{-4} \mathrm{rad}$ and printer prints 300 dots per inch.
The linear distance between two dots is $l=\frac{2.54}{300} \mathrm{~cm}=0.84 \times 10^{-2} \mathrm{~cm}$.
At a distance of $z \mathrm{~cm}$, this subtends an angle, $\phi=\frac{l}{z}$
$\therefore \quad z=\frac{1}{\phi}=\frac{0.84 \times 10^{-2} \mathrm{~cm}}{5.8 \times 10^{-4}}=14.5 \mathrm{~cm}$
If a printed page be held at a distance 14.5 cm , then one does not be able to see the individual dots.

Q10.15: 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: A polaroid $(I)$ is placed in front of a monochromatic source. Another polariod $(II)$ is placed in front of this polaroid $(I)$ when the pass axis of $(I I)$ is parallel to $(I)$, light passes through polaroid-II is unaffected.

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

SA

Q10.16: Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index?

Answer: If angle of incidence is equal to Brewster’s angle, the transmitted light is slightly polarised and reflected light is plane polarised.

Polarisation by reflection occurs when the angle of incidence is the Brewster’s angle.
i.e., $\quad \tan i_B={ }^1 \mu_2=\frac{\mu_2}{\mu_1}$ where $\mu_2<\mu_1$
When the light rays travels in such a medium, the critical angle is
$\sin i_c=\frac{\mu_2}{\mu_1}, \text { where } \mu_2<\mu_1$
As $\left|\tan i_B\right|>\left|\sin i_C\right|$ for large angles $i_B<i_C$.
Thus, the polarisation by reflection occurs definitely.

Important point: Brewster’s angle (also known as the polarization angle) is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. When unpolarized light is incident, at this angle, the light that is reflected from the surface is therefore perfectly polarized. This special angle of incidence is named after the Scottish physicist Sir David Brewster.

Q10.17: For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light of $5000 Å$ and electrons accelerated through 100 V used as the illuminating substance.

Answer: Key concept: Resolving power is the ability of an imaging device to separate (i.e., to see as distinct) points of an object that are located at a small angular distance or it is the power of an optical instrument to separate far away objects, that are close together, into individual images. The term resolution or minimum resolvable distance is the minimum distance between distinguishable objects in an image, although the term is loosely used by many users of microscopes and telescopes to describe resolving power. In scientific analysis, in general, the term “resolution” is used to describe the precision with which any instrument measures Ratio of the least separation,
For electrons accelerated through 100 V , the de-Broglie wavelength, 12.27 and records (in an image or spectrum) any variable in the specimen or sample under study.
It is defined as the reciprocal of the smallest distance $(\Delta x)$ between two point objects whose images are just resolved by the objective of the microscope. It is given by
$R=\frac{1}{\Delta x}=\frac{2 \sin \alpha}{1.22 \mu \lambda}$
where $\mu$ as refractive index of the medium, $\alpha$ is the angle subtended by the objective at the object and $\lambda$ is the wavelength of light.
de-Broglie wavelength:According to de-Broglie theory, the wavelength of de-Broglie wave is given by
$\lambda=\frac{h}{p}=\frac{h}{m v}=\frac{h}{\sqrt{2 m E}} \Rightarrow \lambda \propto \frac{1}{p} \propto \frac{1}{v} \propto \frac{1}{\sqrt{E}}$
where $h=$ Planick’s constant, $m=$ Mass of the particle, $v=$ Speed of the particle, $E=$ Energy of the particle.
The energy of a charged particle accelerated through potential difference $V$ is $E=\frac{1}{2} m v^2=q V$
Hence de-Broglie wavelength $\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m q V}}$
For electron, $\lambda_{\text {Electron }}=\frac{12.27}{\sqrt{V}} Å$

We know that
$\text { Resolving power }=\frac{1}{d}=\frac{2 \sin \alpha}{1.22 \lambda} \Rightarrow d_{\min }=\frac{122 \lambda}{2 \sin \alpha}$
where, $\lambda$ is the wavelength of light and $\beta$ is the angle subtended by the objective at the object.
For the light of wavelength $5500 Å$,
$d_{\min }=\frac{1.22 \times 5500 \times 10^{-10}}{2 \sin \alpha} \dots(i)$
For electrons accelerated through 100 V , the de-Broglie wavelength,
$\begin{array}{r} \lambda=\frac{12.27}{\sqrt{V}}=\frac{12.27}{\sqrt{100}}=0.12 \times 10^{-9} \mathrm{~m} \\ d_{\min }=\frac{122 \times 0.12 \times 10^{-9}}{2 \sin \alpha} \end{array}$
Ratio of the least separation, $\frac{d_{\min }^{\prime}}{d_{\min }}=\frac{0.12 \times 10^{-9}}{5500 \times 10^{-10}}=0.2 \times 10^{-3}$

Q10.18: 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 $\lambda$ such that the first minima on the screen falls at a distance $D$ from the centre O.

Answer: Key Point: The condition for destructive interference is
$\Delta x=S_2 P-S_1 P= \pm\left(\frac{2 n-1}{2}\right) \lambda \text { where } n=1,2, \ldots$
For nth minima to be formed on the screen path difference ( $\Delta x$ ) between the rays coming from $S_1$ and $S_2$ must be $\left(\frac{2 n-1}{2}\right) \lambda$.

The minima will occur when $\Delta x=S_2 P-S_1 P=\left(\frac{2 n-1}{2}\right) \lambda \dots(i)$

From the given figure,
$S_1 P=\sqrt{\left(S_1 T_1\right)^2+\left(P T_1\right)^2}=\sqrt{D^2+(D-x)^2}$
and
$\begin{aligned} & S_2 P=\sqrt{\left(S_2 T_2\right)^2+\left(T_2 P\right)^2}=\sqrt{D^2+(D+x)^2} \\ & T_2 P=T_2 O+O P=D+x \end{aligned}$
And $\quad T_1 P=T_1 O-O P=D-x$
Hence, $\left[D^2+(D+x)^2\right]^{-1 / 2}-\left[D^2+(D-x)^2\right]^{1 / 2}=\frac{\lambda}{2} \quad[$ for first minima $n=1]$ If $x=D$
we can write, $\left[D^2+4 D^2\right]^{1 / 2}-\left[D^2+0\right]^{1 / 2}=\frac{\lambda}{2}$
$\begin{array}{ll} \Rightarrow & {\left[5 D^2\right]^{1 / 2}-\left[D^2+0\right]^{1 / 2}=\frac{\lambda}{2}} \\ \Rightarrow & \sqrt{5} D-D=\frac{\lambda}{2} \text { or } D=\frac{\lambda}{2(\sqrt{5}-1)}=0.404 \lambda \end{array}$

LA

Q10.19: 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 $I_o$ 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: The resultant amplitude of wave reaching on screen will be the sum of amplitude of either wave in perpendicular and parallel polarisation.
Amplitude of the wave in perpendicular polarisation
$\begin{aligned} & A_{\perp}=A_{\perp}^1+A_{\perp}^2=A_{\perp}^0 \sin (k x-\omega t)+A_{\perp}^0 \sin (k x-\omega t+\phi) \\ & \Rightarrow A_{\perp}=A_{\perp}^0(\sin (k x-\omega t)+\sin (k x-\omega t+\phi)) \end{aligned}$
Amplitude of the wave in parallel polarisation
$A_{\|}=A_{\|}^{(1)}+A_{\|}^{(2)}$
$\Rightarrow \quad A_{\|}=A_{\|}^0[\sin (k x-\omega t)+\sin (k x-\omega t+\phi)]$
$\therefore$ Average Intensity on the screen
$I=\left\{\left|A_{\perp}^0\right|^2+\left|A_{\|}^0\right|^2\right\}\left[\sin ^2(k x-\omega t)\left(1+\cos ^2 \phi+2 \sin \phi\right)+\sin ^2(k x-\omega t) \sin ^2 \phi\right]_{\text {average }}$
$=\left\{\left|A_{\perp}^0\right|^2+\left|A_{\|}^0\right|^2\right\}\left(\frac{1}{2}\right) \cdot 2(1+\cos \phi)$
With polariser $P$,
Assume $A_{\perp}^2$ is blocked
$\begin{aligned} \text { Intensity } & =\left(A_{\|}^1+A_{\|}^2\right)^2+\left(A_{\perp}^1\right)^2 \\ & =\left|A_{\perp}^0\right|^2(1+\cos \phi)+\left|A_{\perp}^0\right|^2 \cdot \frac{1}{2} \end{aligned}$
Given, $I_0=4\left|A_{\perp}^0\right|^2=$ Intensity without polariser at principal maxima.
Intensity at first minima with polariser
$=\left|A_{\perp}^0\right|^2(1-1)+\frac{\left|A_{\perp}^0\right|^2}{2}=\frac{I_0}{8} .$

Q10.20: A small transparent slab containing material of $\mu=1.5$ is placed along $\mathrm{AS}_2$ (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: Here the separation between the slits $=S_1 S_2=2 d$. Hence for calculating path difference, equation (i) becomes $\Delta x=2 d \sin \theta+[\mu(t-1)]$
For the principal maxima, (path difference is zero)
$\begin{aligned} & \Delta x=2 d \sin \theta_0+[\mu(t-1)]=0 \\ & \sin \theta_0=-\frac{L(\mu-1)}{2 d}=\frac{-L(0.5)}{2 d} \quad[\because L=d / 4] \\ & \Rightarrow \sin \theta_0=\frac{-1}{16} \end{aligned}$
$\theta_0$ 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 $\pm \frac{\lambda}{2}$.
$\begin{aligned} & \Delta x=2 d \sin \theta_1+0.5 L= \pm \frac{\lambda}{2} \\ & \sin \theta_1=\frac{ \pm \lambda / 2-0.5 L}{2 d}=\frac{ \pm \lambda / 2-d / 8}{2 d} \\ \Rightarrow \quad & \sin \theta_1=\frac{ \pm \lambda / 2-\lambda / 8}{2 \lambda}= \pm \frac{1}{4}-\frac{1}{16} \end{aligned}$
$[\because$ The diffraction occurs if the wavelength of waves is nearly equal to the side width (d).]
On the positive side $\sin \theta_1^{+}=+\frac{1^{+}}{4}-\frac{1}{16}=\frac{3}{16}$
On the negative side $\sin \theta_1^{-}=\frac{1}{4}-\frac{1}{16}=-\frac{5}{16}$
The first principal maxima on the positive side is at distance
$D \tan \theta_1^{\prime+}=D \frac{\sin \theta_1^{\prime+}}{\sqrt{1-\sin ^2 \theta_1^{\prime}}}=D \frac{3}{\sqrt{16^2-3^2}}=\frac{3 D}{\sqrt{247}}$ above point $O$
The first principal minima on the negative side is at distance
$D \tan \theta_1^{+}=\frac{5 D}{\sqrt{16^2-5^2}}=\frac{5 D}{\sqrt{231}}$ below point $O$.

Q10.21: Four identical monochromatic sources A,B,C,D as shown in the (Fig.10.7) produce waves of the same wavelength $\lambda$ 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 us consider the disturbances at the receiver $R_1$, which is at a distance $d$ from $B$.
Let the equation of wave at $R_1$, because of $A$ be
$y_A=a \cos \omega t \dots(i)$
The path difference of the signal from $A$ with that from $B$ is $\lambda / 2$ and hence, the phase difference
$\Delta \phi=\frac{2 \pi}{\lambda} \times(\text { path difference })=\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}=\pi$
Thus, the wave equation at $R_1$, because of $B$ is
$y_B=a \cos (\omega t-\pi)=-a \cos \omega t \dots(ii)$
The path difference of the signal from $C$ with that from $A$ is $\lambda$ and hence the phase difference
$\Delta \phi=\frac{2 \pi}{\lambda} \times(\text { path difference })=\frac{2 \pi}{\lambda} \times \lambda=2 \pi$
Thus, the wave equation at $R_1$, because of $C$ is
$y_C=a \cos \omega t=a \cos (\omega t-2 \pi)=a \cos \omega t \dots(iii)$
The path difference between the signal from $D$ with that of $A$ is
$\begin{aligned} & \Delta x_{R_1}=\sqrt{d^2+\left(\frac{\lambda}{2}\right)^2}-\left(d-\frac{\lambda}{2}\right)=d\left(1+\frac{\lambda^2}{4 d^2}\right)^{1 / 2}-d+\frac{\lambda}{2} \\ & =d\left(1+\frac{\lambda^2}{8 d^2}\right)-d \cdot \frac{\lambda}{2} \approx \frac{\lambda}{2}(\because d \gg \lambda) \end{aligned}$
Therefore, phase difference is $\pi$.
$\therefore \quad y_D=a \cos (\omega t-\pi)=-a \cos \omega t \dots(iv)$
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 $R_1$ is zero.
Now let us consider the resultant signal received at $R_2$. Let the equation of wave at $R_2$, because of $B$ be
$y_B=a_1 \cos \omega t \dots(v)$
The path difference of the signal from $D$ with that from $B$ is $\lambda / 2$ and hence, the phase difference
$\Delta \phi=\frac{2 \pi}{\lambda} \times(\text { path difference })=\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}=\pi$
Thus, the wave equation at $R_2$, because of $D$ is
$y_B=a_1 \cos \left(\omega_t-\pi\right)=-a_1 \cos \omega t \dots(vi)$
The path difference between signal at $A$ and that at $B$ is
$\Delta x_{R_2}=\sqrt{(d)^2+\left(\frac{\lambda}{2}\right)^2}-d=d\left(1+\frac{\lambda^2}{4 d^2}\right)^{1 / 2}-d \simeq \frac{\lambda^2}{8 d^2}$
As $d \gg \lambda$, therefore this path differences $\Delta x_{R_2} \rightarrow 0$
and phase difference $\Delta \phi=\frac{2 \pi}{\lambda} \times$ (path difference)
$=\frac{2 \pi}{\lambda} \times 0 \rightarrow 0(\text { very small })=\phi(\text { say })$
Hence, $y_B=a_1 \cos \left(\omega_t-\phi\right)$
Similarly, $y_B=a_1 \cos \left(\omega_1-\phi\right)$
The resultant signal picked up at $R_2$, from all the four sources is the summation of all four waves, $y_{R_2}=y_A+y_B+y_C+y_D$
$\begin{aligned} y_{R_2} & =a_1 \cos \omega t-a_1 \cos \omega t+a_1 \cos (\omega t-\phi)+a_1 \cos (\omega t-\phi) \\ & =2 a_1 \cos (\omega t-\phi) \end{aligned}$
$\therefore$ The signal picked up $R_2$ is $y_{R_2}=2 a_1 \cos (\omega t-\phi)$
$\therefore \quad\left[V_{R_2}\right]^2=4 a_1^2 \cos ^2(\omega t-\phi) \Rightarrow\left\langle I_{R_2}\right\rangle=2 a_1^2$
Thus, $R_2$ picks up the larger signal.

(ii) (ii) If $B$ is switched off,
$R_1$, picks up $y=a \cos \omega t$
$\therefore \quad\left\langle I_{R_1}\right\rangle=a^2<\cos ^2 \omega t>=\frac{a^2}{2}$
$R_2$ 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 $y=a \cos \omega t$
$\therefore \quad\left\langle I_{R_1}\right\rangle=\frac{1}{2} a^2$
$R_2$ picks up $y=3 a \cos \omega t$

$\left.\therefore \quad\left\langle I_{R_2}\right\rangle=9 a^2<\cos ^2 \omega t\right\rangle=\frac{9 a^2}{2}$
Thus, $R_2$ picks up a 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.

Q10.22: The optical properties of a medium are governed by the relative permitivity $\left(\varepsilon_r\right)$ and relative permeability $\left(\mu_r\right)$. The refractive index is defined as $\sqrt{\mu_r \varepsilon_r}=n$. For ordinary material $\varepsilon_r>0$ and $\mu_{\mathrm{r}}>0$ and the positive sign is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with $\varepsilon_r$ $<0$ and $\mu_{\mathrm{r}}<0$. Since then such ‘metamaterials’ have been produced in the laboratories and their optical properties studied.
For such materials $n=-\sqrt{\mu_r \varepsilon_r}$. 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 $\theta$ in $2^{\text {nd }}$ quadrant, them the refracted beam is in the $3^{\text {rd }}$ quadrant.
(ii) Prove that Snell’s law holds for such a medium.

Answer: (i) If given postulate is true, then two parallel rays would proceed as shown in the figure (1).

Again consider figure (i), let $A B$ represent the incident wavefront and $D E$ represent the refracted wavefront. All points on a wavefront must be in same phase and in turn, must have the same optical path length.
Thus
$\begin{aligned} & -\sqrt{\varepsilon_r \mu_r} A E=B C-\sqrt{\varepsilon_r \mu_r} C D \\ & B C=\sqrt{\varepsilon_r \mu_r}(C D-A E) \\ & B C>0, C D>A E \end{aligned}$
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, $\quad-\sqrt{\varepsilon_r \mu_r} A E=B C-\sqrt{\varepsilon_r \mu_r} C D$
or $\quad B C=\sqrt{\varepsilon_r \mu_r}(C D-A E)$
If . $\quad B C>0$, then $C D>A E$
which is obvious from Fig. (1). Hence, the postulate is reasonable.
However, if the light proceed in the sense it does for ordinary material, (going from 2nd quadrant to 4th quadrant) as shown in Fig. (1). then proceeding as above,
$\begin{array}{ll} & -\sqrt{\varepsilon_r \mu_r} A E=B C-\sqrt{\varepsilon_r \mu_r} C D \\ \text { or } & B C=\sqrt{\varepsilon_r \mu_r}(C D-A E) \end{array}$
As $A E>C D$, therefore $B C<0$ which is not possible. Hence, the given postulate is correct.

(ii) From Fig. (1),
$\begin{array}{ll} & B C=A C \sin \theta_i \\ \text { and } & C D-A E=A C \sin \theta_r \\ \text { As } & B C=-\sqrt{\mu_r \varepsilon_r}(A E-C D) \\ \therefore & A C \sin \theta_i=\sqrt{\varepsilon_r \mu_r} A C \sin \theta_r \\ \text { or } & \frac{\sin \theta_i}{\sin \theta_r}=\sqrt{\varepsilon_r \mu_r}=n \end{array}$
which proves Snell’s law.

Q10.23: 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 $\mathrm{MgF}_2(n=1.38)$. What should the thickness of the film be so that at the center of the visible speetrum ( $5500 Å$ ) there is maximum transmission.

Answer:

Consider a ray incident at angle $i$. A part of this ray is reflected from the air-film interface and a part refracted inside. This is partly reflected at the film-glass interface and a part transmitted. A part of the reflected ray is reflected at the film-air interface and a part transmitted as $r_2$ parallel to $r_1$ Of course successive reflections and transmissions will keep on decreasing the amplitude of the wave. Hence rays $r_1$ and $r_2$ shall dominate the behavior. If incident light is to be transmitted through the lens, $r_1$ and $r_2$ should interfere destructively. Both the reflections at $A$ and $D$ are from lower to higher refractive index and hence there is no phase change on reflection. The optical path difference between $r_2$ and $r_1$ is
$n(A D+C D)-A B$
If $d$ is the thickness of the film, then
$\begin{aligned} & A D=C D=\frac{d}{\cos r} \\ & A B=A C \sin i \\ & \frac{A C}{2}=d \tan r \\ & \therefore A C=2 d \tan r \end{aligned}$
Hence, $A B=2 d \tan r \sin i$
Thus the optical path difference is
$\begin{aligned} & 2 n \frac{d}{\cos r}-2 d \tan r \sin i \\ & =2 \cdot \frac{\sin i}{\sin r} \frac{d}{\cos r}-2 d \frac{\sin r}{\cos r} \sin i \\ & =2 d \sin i\left[\frac{1-\sin ^2 r}{\sin r \cos r}\right] \\ & =2 n d \cos r \end{aligned}$
For these waves to interfere destructively this must be $\lambda / 2$.
$\begin{aligned} & 2 n d \cos r=\frac{\lambda}{2} \\ & n d \cos r=\frac{\lambda}{4} \end{aligned}$
For a camera lens, the sources are in the vertical plane and hence
$\begin{aligned} & i \simeq r \simeq 0 \\ & \therefore n d=\frac{\lambda}{4} \\ & \Rightarrow d=\frac{5500 Å}{1.38 \times 4} \simeq 1000 Å \end{aligned}$