Thinking Process

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

Answer

(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 arrows.

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

Answer

(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 $4000 \AA$ ì 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.

Answer

(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, $\quad n=\frac{\sin \theta}{\sin r}$

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

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

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

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

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

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

Answer

(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.

Answer

(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.

Answer

$(a, b, d)$

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.

Answer

$(b, d)$

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

We know that wavelength of sunlight ranges from $4000 \AA$ to $8000 \AA$.

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.

Answer

$(a, b)$

(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.

Answer

$(a, b)$

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

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.

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.

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.

Answer

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.

Answer

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.

Answer

Given, angular resolution of human eye, $\varphi=5.8 \times 10^{-4} rad$.

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, $\varphi=\frac{l}{z}$

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

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.

Answer

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

Monochromatic source

Polaroid I

Polaroid II

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 $I I$ are at $90^{\circ}$.

Monochromatic source

Polaroid III

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).

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

i.e.,

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

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

$$ \sin i_{c}=\frac{\mu_2}{\mu_1} $$

where, $\mu_2<\mu_1$

As $|\tan i_{B}|>|\sin i_{C}|$ for large angles $i_{B}<i_{C}$.

Thus, the polarisation by reflection occurs definitely.

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.

Answer

We know that

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

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

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

$$ \begin{aligned} \lambda & =\frac{12.27}{\sqrt{V}}=\frac{12.27}{\sqrt{100}}=0.12 \times 10^{-9} m \\ d_{\min } & =\frac{1.22 \times 0.12 \times 10^{-9}}{2 \sin \beta} \end{aligned} $$

Ratio of the least separation

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

Answer

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

$$ \begin{aligned} & T_2 P=T_2 O+O P=D+x \\ \text{and}\qquad & T_1 P=T_1 O-O P=D-x \\ & S_1 P=\sqrt{(S_1 T_1)^{2}+(P T_1)^{2}}=\sqrt{D^{2}+(D-x)^{2}} \\ \text{and}\qquad & S_2 P=\sqrt{(S_2 T_2)^{2}+(T_2 P)^{2}}=\sqrt{D^{2}+(D+x)^{2}} \end{aligned} $$

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

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

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

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

$$ \begin{aligned} \Rightarrow \qquad \sqrt{5}-1=2.236-1 & =1.236 \\ D=\frac{\lambda}{2(1.236)}= & 0.404 \lambda \end{aligned} $$

Thinking Process

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

Answer

$A=$ Resultant amplitude

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

$$ \Rightarrow \quad A =A_{\perp}+A_{||} \\ \text{Without p} \qquad A =A_{\perp}+A_{||} \\ A_1=A^{1} \perp+A_{\perp}^{2} =A_{\perp}^{0} \sin (k x-\omega t)+A_{\perp}^{0} \sin (k x-\omega t+\phi) \\ A_{||} =A_{||}^{(1)}+A_{||}^{(2)} \\ A_{||} =A_{||}^{0}[\sin (k x-\omega t)+\sin (k x-\omega t+\phi)] $$

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

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

$$=\{|A_{\perp}^0|^2+|A_{|}^0|^2\} \frac{1}{2} \cdot 2(1+\cos \phi) $$

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

With $P$

Assume $A_{\perp}^{2}$ is blocked

$$ \begin{aligned} \text { Intensity } & =(A_{|}^{1}+A_{|}^{2})^{2}+(A_{\perp}^{1})^{2} \\ & =|A_{\perp}^{0}|^{2}(1+\cos \varphi)+|A_{\perp}^{0}|^{2} \cdot \frac{1}{2} \end{aligned} $$

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

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

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.

Answer

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

For the principal maxima, (path difference is zero)

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

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

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

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

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

$$ \begin{aligned} \therefore \quad 2 d \sin \theta_1+0.5 L & = \pm \frac{\lambda}{2} \\ \text { or } \quad \sin \theta_1 & =\frac{ \pm \lambda / 2-0.5 L}{2 d}=\frac{ \pm \lambda / 2-d / 8}{2 d} \\ & =\frac{ \pm \lambda / 2-\lambda / 8}{2 \lambda}= \pm \frac{1}{4}-\frac{1}{16} \end{aligned} $$

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

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

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

The first principal maxima on the positive side is at distance

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

The first principal minima on the negative side is at distance

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

Thinking Process

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

Answer

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 $\pi$.

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 $Y_{C}=a \cos (\omega t-2 \pi)=a \cos \omega t$

The path difference between the signal from $D$ with that of $A$ is

$$ \begin{aligned} & \sqrt{d^{2}+\frac{\lambda}2^{2}}-(d-\lambda / 2)=d \quad 1+{\frac{\lambda}{4 d^{2}}}^{1 / 2}-d+\frac{\lambda}{2} \\ & =d \quad 1+{\frac{\lambda^{2}}{8 d^{2}}}^{1 / 2}-d+\frac{\lambda}{2} \approx \frac{\lambda}{2} \quad(\because d \gg \lambda) \end{aligned} $$

Therefore, phase difference is $\pi$.

$$ \therefore \quad Y_{D}=a \cos (\omega t-\pi)=-a \cos \omega t $$

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

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

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

and

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

Hence,

Similarly,

$$ \begin{aligned} & y_{A}=a_1 \cos (\omega t-\varphi) \\ & y_{C}=a_1 \cos (\omega t-\varphi) \end{aligned} $$

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

$$ y_{A}+y_{B}+y_{C}+y_{D} =y=2 a_1 \cos (\omega t-\phi) $$ $$\therefore |y|^2 =4 a_1^2 \cos^2(\omega t-\phi) $$ S Thus, $R_1$ picks up the larger signal.

(ii) If $B$ is switched off,

$$ \begin{aligned} R_1 \text { picks up } y =a \cos \omega t \\ \therefore \langle I_{R_1}\rangle =\frac{1}{2} a^{2} \\ R_2 \text { picks up } y =a \cos \omega t \\ \therefore \langle I_{R_2}\rangle =a^{2}<\cos ^{2} \omega t>=\frac{a^{2}}{2} \end{aligned} $$

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

If $D$ is switched off.

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

$$ \begin{aligned} \therefore \langle I_{R_1}\rangle =\frac{1}{2} a^{2} \\ R_2 \text { picks up } y =3 a \cos \omega t \\ \therefore \langle I_{R_2}\rangle =9 a^{2}<\cos ^{2} \omega t>=\frac{9 a^{2}}{2} \end{aligned} $$

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.

Answer

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

(i)

(ii)

(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.

$$ \begin{aligned} \text{Thus} \qquad -\sqrt{\varepsilon_{r} \mu_{r}} A E & =B C-\sqrt{\varepsilon_{r} \mu_{r}} C D \\ \text{or} \qquad B C & =\sqrt{\varepsilon_{r} \mu_{r}}(C D-A E) \\ B C & >0, C D>A E \end{aligned} $$

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)

$$ \begin{aligned} \text{Then,} \qquad -\sqrt{\varepsilon_{r} \mu_{r}} A E & =B C-\sqrt{\varepsilon_{r} \mu_{r}} C D \\ \text{or} \qquad B C & =\sqrt{\varepsilon_{r} \mu_{r}}(C D-A E) \end{aligned} $$

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 2nd quadrant to 4th quadrant) as shown in Fig. (i)., then proceeding as above,

$$ \begin{aligned} -\sqrt{\varepsilon_{r} \mu_{r}} A E & =B C-\sqrt{\varepsilon_{r} \mu_{r}} C D \\ \text{or} \qquad B C & =\sqrt{\varepsilon_{r} \mu_{r}}(C D-A E) \end{aligned} $$

As $A E>C D$, therefore $B C<0$ which is not possible. Hence, the given postulate is correct.

(ii) From Fig. (i)

$$ \begin{aligned} B C & =A C \sin \theta_{i} \\ \text{and}\qquad C D-A E & =A C \sin \theta_{r} \\ \text{As} \qquad B C & =\sqrt{\mu_{r} \varepsilon_{r}} \\ \therefore \qquad A C \sin \theta_{i} & =\sqrt{\varepsilon_{r} \mu_{r}} A C \sin \theta_{r} \\ \text{or} \qquad \frac{\sin \theta_{i}}{\sin \theta_{r}} & =\sqrt{\varepsilon_{r} \mu_{r}}=n \end{aligned} $$

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

Which proves Snell’s law.

Answer

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

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.

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

$$ \begin{aligned} A D & =C D=\frac{d}{\cos r} \\ A B & =A C \sin i \\ \frac{A C}{2} & =d \tan r \\ A C & =2 d \tan r \end{aligned} $$

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

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

$$ \begin{aligned} & =2 \cdot \frac{\sin i d}{\sin r \cos r}-2 d \frac{\sin r}{\cos r} \sin i \\ & =2 d \sin \frac{1-\sin ^{2} r}{\sin r \cos r} \\ & =2 n d \cos r \end{aligned} $$

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

$$ \begin{matrix} \Rightarrow & 2 n d \cos r=\frac{\lambda}{2} \\ \Rightarrow & n d \cos r=\frac{\lambda}{4} \end{matrix} $$

For photographic lenses, the sources are normally in vertical plane $\therefore$

From Eq. (i),

$$ \begin{aligned} i & =r=0^{\circ} \\ n d \cos 0^{\circ} & =\frac{\lambda}{4} \\ d & =\frac{\lambda}{4 n} \\ & =\frac{5500 \AA}{4 \times 1.38} \approx 1000 \AA \end{aligned} $$