• MCQ (Single correct option)

• MCQ (More than one correct option)

• Very short answer type questions

• Short answer type questions

• Long answer type questions

Ques. Consider a light beam incident from air to a glass slab at Brewster’s angle as shown in figure.
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.

Ques. to be continued …

Options :

(a) For a particular orientation, there shall be darkness as observed through the polaroid

(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

Thinking process :

When the light beam incident at Brewster’s angle, the transmitted beam is unpolarised and reflected beam is polarised.

Ans. (c)

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 \tos.

Sol. to be continued …

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

Ques. Consider sunlight incident on a slit of width $10^{4} \AA$. The image seen through the slit shall

Ans. (a)

Given, width of the slit $=10^{4} \AA$

$=10^{4} \times 10^{-10} m=10^{-6} m=1 \mu m$

Wavelength of (visible) sunlight varies from $4000 \AA$ to $8000 \AA$.

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

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

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

Ans. (a)

Consider the diagram, the ray $(P)$ is incident at an angle $\theta$ and gets reflected in the direction $P^{\prime}$ and refracted in the direction $P^{\prime \prime}$. Due to reflection from the glass medium, there is a phase change of $\pi$.

Time taken to travel along $O P^{\prime \prime}$

From Snell’s law, $\quad n=\frac{\sin \theta}{\sin r}$

$\Delta t=\frac{O P^{\prime \prime}}{v}=\frac{d / \cos r}{c / n}=\frac{n d}{c \cos r}$

Sol. to be continued …

$\Rightarrow \quad \sin r=\frac{\sin \theta}{n}$

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

Sol. to be continued …

$\Delta t=\frac{n d}{c 1-\frac{\sin ^{2} \theta}{n^{2}}}=\frac{n^{2} d}{c} 1-\frac{\sin ^{2} \theta}{n^{2}}$

Phase difference = $\Delta \varphi=\frac{2 \pi}{T} \times \Delta t=\frac{2 \pi n d}{\lambda} 1-\frac{\sin ^{2} \theta}{n^{2}}$

So, net phase difference $=\Delta \varphi+\pi$

$=\frac{4 \pi d}{\lambda} 1-\frac{1}{n^{2}} \sin ^{2} \theta^{-1 / 2}+\pi$

Ques. 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,

Ans. (c)

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.

Hence, in this case there shall be no interference fringes.

Ques. Figure 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 $P$ (figure).
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.

Ques. to be continued …

Options :

(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

Ans. (d)

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.

Ques. Two sources $S_1$ and $S_2$ of intensity $I_1$ and $I_2$ are placed in front of a screen [Fig. (a)]. The pattern of intensity distribution seen in the central portion is given by Fig. (b).

In this case, which of the following statements are true?

Ques. to be continued …

Options :

(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

(d) $S_1$ and $S_2$ have the same wavelength

Ans. (a, b, d)

Consider the pattern of the intensity shown in the 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.

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

Ans. (b, d)

Given, width of pinhole = 10$^{3}$ A = 1000 A

We know that wavelength of sunlight ranges from 4000 A to 8000 A.

Clearly, wavelength $\lambda<$ width of the slit.

Hence, light is diffracted from the hole. Due to diffraction from the slight the image formed on the screen will be different from the geometrical image.

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

Ans. (a, b)

(a) When a decreases $w$ increases.

So, size decreases.

(b) Now, light energy is distributed over a small area and intensity $\propto \frac{1}{\text{ area }}$ as area is decreasing so intensity increases.

Ques. For light diverging from a point source,

Ans. (a, b)

Consider the diagram in which light diverges from a point source $(0)$.

Sol. to be continued …

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.

Ques. Is Huygen’s principle valid for longitudinal sound waves?

Ans.

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.

Ques. 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?

Ans.

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.

Ques. hat is the shape of the wavefront on earth for sunlight?

Ans.

We know that the sun is at very large distance from the earth. Assuming sun as spherical, it can be considered as point source situated at infinity.

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

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

Ans.

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 $7000 \times 10^{-10} m$ 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.

Ques. The human eye has an approximate angular resolution of $\phi=5.8 \times 10^{-4}$ 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.

Ans.

Given, angular resolution of human eye, $\phi=5.8 \times 10^{-4}$ rad and printer prints 300 dots per inch.

The linear distance between 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, $\phi=\frac{l}{z}$

$\therefore \quad z=\frac{l}{\phi}=\frac{0.84 \times 10^{-2} cm}{5.8 \times 10^{-4}}=14.5 cm$.

Ques. A polaroid (I) is placed infront of a monochromatic source. Another polariod (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.

Thinking process :
Natural light e.g., from the sun is unpolariser. This means the electric vector takes all possible direction in the transverse plane, rapidly.
Ans.
In the diagram shown, a monochromatic light is placed infront of polaroid (I) as shown below.

Sol. to be continued …

Monochromatic source

Polaroid I

Sol. to be continued …

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 $90^{\circ}$.

Sol. to be continued …

Monochromatic source

Sol. to be continued …

Polaroid I
Polaroid II
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 $90^{\circ}$ to the pass axis of (III).

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

Ans.

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 $\to$s.

Polarisation by reflection occurs when the angle of incidence is the Brewster’s angle

i.e.,

$\tan i_B={ }^{1} \mu_2=\frac{\mu_2}{\mu_1} \text{ where } \mu_2<\mu_1$

Sol. to be continued …

when the light rays travels in such a medium, the critical angle is

$\sin i_c=\frac{\mu_2}{\mu_1}$

Sol. to be continued …

where, $\mu_2<\mu_1$

As $|\tan i_B|>|\sin i_C|$ for large angles $i_B

Thus, the polarisation by reflection occurs definitely.

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

Thinking process :

Resolving power of a microscope is calculated by $\frac{2 \sin \beta}{1.22 \lambda}$, with $\mu$ as refractive index of the medium and $\beta$ is the angle subtented by the objective at the object.

Ans.

We know that

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

where, $\lambda$ is the wavelength of light and $\beta$ is the angle subtended by the objective at the object.

Sol. to be continued …

For the light of wavelength $5500 \AA$,

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

For electrons accelerated through $100 V$, the de-Broglie wavelength

$\lambda =\frac{12.27}{\sqrt{V}}=\frac{12.27}{\sqrt{100}}=0.12 \times 10^{-9} m$

$d _{\min } =\frac{1.22 \times 0.12 \times 10^{-9}}{2 \sin \beta}$

Ratio of the least separation

$\therefore \quad \frac{d _{\min }^{\prime}}{d _{\min }}=\frac{0.12 \times 10^{-9}}{5500 \times 10^{-10}}=0.2 \times 10^{-3}$

Ques. Consider a two slit interference arrangements (figure) 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 $O$.

Ans.

From the given figure of two slit interference arrangements, we can write

$T_2 P=T_2 O+O P=D+x$

and $\qquad T_1 P=T_1 O-O P=D-x$

$S_1 P=\sqrt{(S_1 T_1)^{2}+(P T_1)^{2}}=\sqrt{D^{2}+(D-x)^{2}}$

and $\qquad S_2 P=\sqrt{(S_2 T_2)^{2}+(T_2 P)^{2}}=\sqrt{D^{2}+(D+x)^{2}}$

Sol. to be continued …

The minima will occur when $S_2 P-S_1 P=(2 n-1) \frac{\lambda}{2}$

i.e., $\quad[D^{2}+(D+x)^{2}]^{1 / 2}-[D^{2}+(D-x)^{2}]^{1 / 2}=\frac{\lambda}{2} \quad$ [for first minima n = 1]

If $\qquad x=D$

we can write $\quad[D^{2}+4 D^{2}]^{1 / 2}-[D^{2}+0]^{1 / 2}=\frac{\lambda}{2}$

$\Rightarrow \quad [5 D^{2}]^{1 / 2}-[D^{2}]^{1 / 2}=\frac{\lambda}{2}$

$\Rightarrow \sqrt{5} D-D=\frac{\lambda}{2}$

Sol. to be continued …

$\Rightarrow D(\sqrt{5}-1)=\lambda / 2 \text{ or } D=\frac{\lambda}{2(\sqrt{5}-1)}$

Putting $\qquad \sqrt{5}=2.236$

$\Rightarrow \quad \sqrt{5}-1=2.236-1 =1.236$

$D=\frac{\lambda}{2(1.236)}= 0.404 \lambda$

Ques. Figure 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.

Thinking process :

The resultant amplitude will be the sum of amplitude of either beam in perpendicular and parallel polarisation.

Ans.

$A=$ Resultant amplitude

$\quad =A$ parallel $(A _{||})+A$ perpendicular $(A _{\perp})$

$\Rightarrow \quad A =A _{\perp}+A _{||}$

Sol. to be continued …

Without p $\qquad A =A _{\perp}+A _{||}$

$A_1=A^{1} \perp+A _{\perp}^{2} =A _{\perp}^{0} \sin (k x-\omega t)+A _{\perp}^{0} \sin (k x-\omega t+\phi)$

$A _{||} =A _{||}^{(1)}+A _{||}^{(2)}$

$A _{||} =A _{||}^{0}[\sin (k x-\omega t)+\sin (k x-\omega t+\phi)]$

where $A _{\perp}^{0}, A _{||}^{0}$ are the amplitudes of either of the beam in perpendicular and parallel polarisations.

$\therefore$ Intensity $=\{|A _{\perp}^{0}|^{2}+|A _{||}^{0}|^{2}\} [\sin ^{2}(k x-\omega t)(1+\cos ^{2} \varphi+2 \sin \varphi)+\sin ^{2}(k x-\omega t) \sin ^{2} \varphi]$

Sol. to be continued …

$=\{|A _{\perp}^0|^2+|A _{|}^0|^2\} \frac{1}{2} . 2(1+\cos \phi)$

$=2|A _ {\perp} ^ 0| ^ 2(1+\cos \phi), \text{ since, }|A _ {\perp} ^ 0 | _ {av} =|A _ {||} ^ 0| _ {av}$

With $P$

Assume $A _{\perp}^{2}$ is blocked

Intensity $=(A _{|}^{1}+A _{|}^{2})^{2}+(A _{\perp}^{1})^{2}$

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

Sol. to be continued …

Given, $I_0=4|A _{\perp}^{0}|^{2}=$ Intensity without polariser at principal maxima.

Intensity at principal maxima with polariser

$=|A _{\perp}^{0}|^{2} \quad 2+\frac{1}{2}=\frac{5}{8} I_0$

Intensity at first minima with polariser

$=|A _{\perp}^{0}|^{2}(1-1)+\frac{|A _{\perp}^{0}|^{2}}{2}=\frac{I_0}{8}$

Ques.

$AC = CO = D, S_1C = S_2C = d « D$

A small transparent slab containing material of $\mu=1.5$ is placed along $A S_2$ (figure). What will be the distance from 0 of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?

Thinking process :

Whenever a transparent slab of refractive index $\mu$ and thickness $t$ is inserted in the path of the ray the fringes on the screen shifts by $(\mu-1)$ t towards the slab.

Ans.

In case of transparent glass slab of refractive index $\mu$, the path difference will be calculated as $\Delta x=2 d \sin \theta+(\mu-1) L$.

In case of transparent glass slab of refractive index $\mu$,

the path difference $=2 d \sin \theta+(\mu-1) L$.

Sol. to be continued …

For the principal maxima, (path difference is zero)

$\text{i.e.,}\qquad 2 d \sin \theta_0+(\mu-1) L=0$

$or\qquad \sin \theta_0=-\frac{L(\mu-1)}{2 d}=\frac{-L(0.5)}{2 d} \quad[\because L=d / 4]$

$or\qquad \sin \theta_0=\frac{-1}{16}$

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

For the first minima, the path difference is $\pm \frac{\lambda}{2}$

Sol. to be continued …

$\therefore \quad 2 d \sin \theta_1+0.5 L = \pm \frac{\lambda}{2}$

$\text{ or } \quad \sin \theta_1 =\frac{ \pm \lambda / 2-0.5 L}{2 d}=\frac{ \pm \lambda / 2-d / 8}{2 d}$

$=\frac{ \pm \lambda / 2-\lambda / 8}{2 \lambda}= \pm \frac{1}{4}-\frac{1}{16}$

$[\because$ The diffraction occurs if the wavelength of waves is nearly equal to the side width $(d)]$

On the positive side $\sin \theta_1^{\prime}+=+\frac{1}{4}-\frac{1}{16}=\frac{3}{16}$

On the negative side $\sin \theta^{\prime \prime} _1=-\frac{1}{4}-\frac{1}{16}=-\frac{5}{16}$

Sol. to be continued …

The first principal maxima on the positive side is at distance

$D \tan \theta_1^{+}=D \frac{\sin \theta_1^{+}}{\sqrt{1-\sin ^{2} \theta_1^{\prime}}}=D \frac{3}{\sqrt{16^{2}-3^{2}}}=\frac{3 D}{\sqrt{247}}$ above point

The first principal minima on the negative side is at distance

$D \tan \theta^{\prime} _1=\frac{5 D}{\sqrt{16^{2}-5^{2}}}=\frac{5 D}{\sqrt{231}}$ below point

Ques. Four identical monochromatic sources $A, B, C, D$ as shown in the (figure) produce waves of the same wavelength $\lambda$ and are coherent. Two receiver $R_1$ and $R_2$ are at great but equal distances from $B$.

Thinking process :

The resultant disturbance at a point will be calculated by some of disturbances due to individual sources.

Ans.

Consider the disturbances at the receiver $R_1$ which is at a distanced from $B$.

Let the wave at $R_1$ because of $A$ be $Y_A=a \cos w$.

The path difference of the signal from $A$ with that from $B$ is $\lambda / 2$ and hence, the phase difference is $\pi$.

Sol. to be continued …

Thus, the wave at $R_1$ because of $B$ is

$y_B=a \cos (\omega t-\pi)=-a \cos \omega t .$

The path difference of the signal from $C$ with that from $A$ is $\lambda$ and hence the phase difference is $2 \pi$.

Thus, the wave at $R_1$ because of $C$ is $Y_C=a \cos (\omega t-2 \pi)=a \cos \omega t$

Sol. to be continued …

Sol. to be continued …

Thus, the signal picked up at $R_1$ from all the four sources is $Y _{R_1}=y_A+y_B+y_C+y_D$

$=a \cos \omega t-a \cos \omega t+a \cos \omega t-a \cos \omega t=0$

(i) Let the signal picked up at $R_2$ from $B$ be $y_B=a_1 \cos \omega t$.

The path difference between signal at $D$ and that at $B$ is $\lambda / 2$.

$\therefore \quad y_D=-a_1 \cos \omega t$

The path difference between signal at $A$ and that at $B$ is

Sol. to be continued …

$\sqrt{(d)^{2}+\frac{\lambda}2^{2}}-d=d 1+{\frac{\lambda^{2}}{4 d^{2}}}^{1 / 2}-d=\frac{1 \lambda^{2}}{8 d^{2}}$

As $d \gg \lambda$, therefore this path difference $\rarr 0$

and

phase difference = $\frac{2 \pi}{\lambda} \frac{1}{8} \frac{\lambda^{2}}{d^{2}} \to 0$

Hence,

Similarly,

Sol. to be continued …

$y_A=a_1 \cos (\omega t-\varphi)$

$y_C=a_1 \cos (\omega t-\varphi)$

$\therefore$ Signal picked up by $R_2$ is

$y_A+y_B+y_C+y_D =y=2 a_1 \cos (\omega t-\phi)$

$\therefore |y|^2 =4 a_1^2 \cos^2(\omega t-\phi)$

Thus, $R_1$ picks up the larger signal.

Sol. to be continued …

(ii) If $B$ is switched off,

$R_1 \text{ picks up } y =a \cos \omega t$

$\therefore \langle I _{R_1}\rangle =\frac{1}{2} a^{2}$

$R_2 \text{ picks up } y =a \cos \omega t$

$\therefore \langle I _{R_2}\rangle =a^{2}<\cos ^{2} \omega t>=\frac{a^{2}}{2}$

(iii) Thus, $R_1$ and $R_2$ pick up the same signal.

Sol. to be continued …

If $D$ is switched off.

$R_1$ picks up $y=a \cos \omega t$

$\therefore \langle I _{R_1}\rangle =\frac{1}{2} a^{2}$

$R_2 \text{ picks up } y =3 a \cos \omega t$

$\therefore \langle I _{R_2}\rangle =9 a^{2}<\cos ^{2} \omega t>=\frac{9 a^{2}}{2}$

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.

Ques. The optical properties of a medium are governed by the relative permittivity $(\varepsilon_r)$ and relative permeability $(\mu_r)$. The refractive index is defined as $\sqrt{\mu_r \varepsilon_r}=n$. For ordinary material, $\varepsilon_r>0$ and $\mu_r>0$ 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 $\mu_r<0$. 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, then the refracted beam is in the 3$^\text{rd}$ quadrant.

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

Ans.

Let us assume that the given postulate is true, then two parallel rays would proceed as shown in the figure below

Sol. to be continued …

(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 $\qquad -\sqrt{\varepsilon_r \mu_r} A E =B C-\sqrt{\varepsilon_r \mu_r} C D$

or $\qquad B C =\sqrt{\varepsilon_r \mu_r}(C D-A E)$

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

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)

Sol. to be continued …

or $\qquad B C =\sqrt{\varepsilon_r \mu_r}(C D-A E)$

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

which is obvious from Fig (i).

Hence, the postulate reasonable.

However, if the light proceeded in the sense it does for ordinary material, (going from 2$^\text{nd}$ quadrant to 4$^\text{th}$ quadrant) as shown in Fig. (i)., then proceeding as above,

Sol. to be continued …

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

or $\qquad B C =\sqrt{\varepsilon_r \mu_r}(C D-A E)$

As $A E>C D$, therefore $B C<0$ which is not possible.

Hence, the given postulate is correct.

(ii) From Fig. (i)

$B C =A C \sin \theta_i$

Sol. to be continued …

and $\qquad C D-A E =A C \sin \theta_r$

As $\qquad B C =\sqrt{\mu_r \varepsilon_r}$

$\therefore \qquad A C \sin \theta_i =\sqrt{\varepsilon_r \mu_r} A C \sin \theta_r$

or $\qquad \frac{\sin \theta_i}{\sin \theta_r} =\sqrt{\varepsilon_r \mu_r}=n$

$[C D-A E=B C]$

Which proves Snell’s law.

Ques. To ensure almost $100 \%$ 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 $\text{MgF}_2(n=1.38)$. What should the thickness of the film be so that at the centre of the visible spectrum ($5500 \AA$) there is maximum transmission.

Ans.

In this figure, we have shown a dielectric film of thickness $d$ deposited on a glass lens.

Sol. to be continued …

Refractive index of film $=1.38$ and refractive index of glass $=1.5$.

Given,

$\lambda=5500 \AA .$

Consider a ray incident at an 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.

Sol. to be continued …

Hence, rays $r_1$ and $r_2$ shall dominate the behaviour. 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

$A D =C D=\frac{d}{\cos r}$

Sol. to be continued …

$A B =A C \sin i$

$\frac{A C}{2} =d \tan r$

$A C =2 d \tan r$

Hence, $A B=2 d \tan r \sin i$.

Thus, the optical path difference $=\frac{2 n d}{\cos r}-2 d \tan r \sin i$

$=2 \cdot \frac{\sin i d}{\sin r \cos r}-2 d \frac{\sin r}{\cos r} \sin i$

Sol. to be continued …

$=2 d \sin \frac{1-\sin ^{2} r}{\sin r \cos r}$

$=2 n d \cos r$

For these waves to interfere destructively path difference $=\frac{\lambda}{2}$.

$\Rightarrow 2 n d \cos r=\frac{\lambda}{2}$

$\Rightarrow n d \cos r=\frac{\lambda}{4}$

For photographic lenses, the sources are normally in vertical plane

Sol. to be continued …

$\therefore$ From Eq. (i),

$i =r=0^{\circ}$

$n d \cos 0^{\circ} =\frac{\lambda}{4}$

$d =\frac{\lambda}{4 n}$

$=\frac{5500 \AA}{4 \times 1.38} \approx 1000 \AA$

Thank You