Wave Optics

As intensity of emergent beam is reduced to half after passing through two polarisers. It means angle between A and B is 0$^\circ$.

Now, on placing polariser C between A and B.

Intensity after passing through A is ${I_A} = {I \over 2}$.

Let $\theta$ be the angle between A and C. Intensity of light after passing through C is given by

${I_C} = {I \over 2}{\cos ^2}\theta $

Intensity of light after passing through polariser B is ${I \over 3}$.

Angle between C and B is also $\theta$ as A is parallel to B.

So, ${I \over 3} = {I_C}{\cos ^2}\theta = {I \over 2}{\cos ^2}\theta \,.\,{\cos ^2}\theta $

${\cos ^4}\theta = {2 \over 3} \Rightarrow \cos \theta = {\left( {{2 \over 3}} \right)^{1/4}}$

According to Malus law, we have

$I = {I_0}{\cos ^2}\theta \Rightarrow {I \over {{I_0}}} = {\cos ^2}\theta $

Since the observed intensities are same for all values of $\theta$, therefore, for the angles given in options, the observed intensity should be same as well for all $\theta$, $\theta$ = 8$^\circ$, 38$^\circ$, 188$^\circ$ and 218$^\circ$, ${I \over {{I_0}}} \sim 0.9$; and for $\theta$ = 203$^\circ$, ${I \over I} \sim 0.9$.

Therefore, angle between direction of polarisation and x axis is 203$^\circ$.

The slit width is a = 0.1 mm = 10^$-$4 m.

The wavelength of the light is $\lambda$ = 6000 $\times$ 10^$-$10 = 6 $\times$ 10^$-$7.

The distance from the slit to diffraction bands is D = 0.5 m.

We calculate the third dark band from the central band as follows:

$A\sin \theta = 3\lambda \Rightarrow \sin \theta = {{3\lambda } \over a} = {x \over D}$

$ \Rightarrow x = {{3\lambda D} \over a} = {{3 \times 6 \times {{10}^{ - 7}} \times 0.5} \over {{{10}^{ - 4}}}} = 9$ mm

At any point P on circumference, the path difference, $\Delta$x = d sin$\theta$

At P₁, $\theta$ = 0$^\circ$ $\Rightarrow$ $\Delta$x = 0

At P₂, $\theta$ = 90$^\circ$ $\Rightarrow$ $\Delta$x = d

For constructive interference, $\Delta$x = n$\lambda$

$\therefore$ $n = {d \over \lambda } = {{1.8\,mm} \over {600\,nm}} = {{1.8 \times {{10}^{ - 3}}m} \over {600 \times {{10}^{ - 9}}m}} = 3000$

Since, n is integer, hence P₂ corresponds to bright spot and also it corresponds to maximum order of fringe.

Now, $\Delta$x = d sin$\theta$

d$\Delta$x = d cos$\theta$ d$\theta$ or R$\lambda$ = d cos$\theta$ d$\theta$

$\therefore$ $d\theta \propto {1 \over {\cos \theta }}$

Hence, as $\theta$ increases, cos $\theta$ decreases and consequently d$\theta$ increases.

At point O, the path difference is

${{2\pi } \over \lambda }({S_1}O - {S_2}O) = {{2\pi } \over \lambda }\left[ {\left( {D + {d \over 2}} \right) - \left( {D - {d \over 2}} \right)} \right]$

$ = {{2\pi d} \over \lambda } = {{2\pi (0.6003 \times {{10}^{ - 3}})} \over {600 \times {{10}^{ - 9}}}} = 2001\pi = (2n + 1)\pi $

Therefore, the dark fringe forms at O.

Hence, option (A) is correct.

The shape of wavefronts from both sources for D >> d is semi-circular, which are concentric with each other with the difference just in their phase resulting in concentric semicircular interference fringes. That is, the region nearby point O is dark.

Hence, option (B) is correct.

In a double slit experiment, the bright spots (maxima) from interference are found where:

  $d\,\sin \theta = m\lambda$, where $m$ can be 0, ±1, ±2, and so on.

The dark spots (minima) from single slit diffraction appear at:

  $a\,\sin \theta = \pm\lambda$.

For a bright interference fringe to be inside the central bright area of the single slit (central maximum), it must fall between these two first diffraction minima.

This means we need:

 $|m\lambda/d| \leq \lambda/a$, or $|m| \leq d/a$.

The problem says the distance between the slits ($d$) is 6.1 times the width of each slit ($a$), so $d/a = 6.1$.

The biggest whole number for $m$ is 6, so there are maxima for $m = \pm1, \pm2, ..., \pm6$.

If we count only the bright spots on the sides (not the middle one), that's 6 on one side and 6 on the other, making 12 bright maxima.

If the central bright spot ($m=0$) is counted too, there would be 13, but usually “within the central maximum” means only the fringes between the two minima, not the central line.

So, you will see 12 bright interference fringes inside the central maximum of the single slit envelope.

Answer = 12. (Option C)

The smallest detail a microscope can see (its resolution limit) is given by the formula:

$ d_{\text{min}} \approx \frac{\lambda}{2 \text{N.A.}} $

Here, N.A. stands for numerical aperture. The formula for numerical aperture is:

$ \text{N.A.} = n \sin \alpha $

where n is the refractive index of the oil (the medium), and α is half the angle made by the cone of light rays coming from the specimen.

The resolving power (RP) of the microscope tells us how well it can see small details. It is the inverse of the resolution limit:

$ \text{RP} = \frac{1}{d_{\text{min}}} \propto n \sin \alpha $

Now, the angle subtended by the lens diameter at the focal point is β. From geometry, we see that α (half the cone angle) is related to β by:

$ \sin \alpha = \sin 2\beta $

This means the resolving power becomes:

$ \text{RP} \propto n \sin 2\beta $

So, the microscope will see finer details as the value of $ n \sin 2\beta $ increases. Answer: C.

Wavelength of light is $\lambda$.

Now, using the relation for intensity, we have

$I = {I_{\max }}{\cos ^2}\left( {{\phi \over 2}} \right)$

${1 \over 2} = {\cos ^2}\left( {{\phi \over 2}} \right) \Rightarrow \cos {\phi \over 2} = {1 \over {\sqrt 2 }}$

$ \Rightarrow {\phi \over 2} = {\pi \over 4} \Rightarrow \phi = {\pi \over 2}$

$ \Rightarrow \phi = {\pi \over 2},{{3\pi } \over 2},{{5\pi } \over 2},{{7\pi } \over 2}$

$\phi = {{2\pi } \over \lambda }\Delta x \Rightarrow {\pi \over 2} = {{2\pi } \over 2}\Delta x \Rightarrow \Delta x = {\lambda \over 4},{{3\lambda } \over 4},{{5\lambda } \over 4}$

$ \Rightarrow \Delta x = (2n + 1){\lambda \over 4}$

Fringe width, $\beta = {{\lambda D} \over d}$

$\therefore$ $\beta \propto \lambda $

As, ${\lambda _R} > {\lambda _G} > {\lambda _B}$ $\therefore$ ${\beta _R} > {\beta _G} > {\beta _B}$

Statement - 1 : When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of π.

This statement is true. According to the physics of wave optics, when light reflects from a denser medium (glass in this case) to a rarer medium (air), there is a phase change of π (or 180 degrees).

Statement - 2 : The center of the interference pattern is dark.

This statement is also true. When a thin air film is formed between a convex lens and a plane surface, the path difference at the center of the pattern is zero. However, due to the phase change of π at the lower surface (glass surface), the two interfering waves are out of phase at the center, resulting in destructive interference and creating a dark spot at the center.

So both statements are true. However, statement - 2 is not a direct explanation of statement - 1. Statement - 2 arises because of the combination of the path difference and the phase change mentioned in statement - 1, but it doesn't directly explain why the phase change occurs.

Hence, the answer is :

Option B : Statement - 1 is true, Statement - 2 is true, Statement - 2 is not the correct explanation of Statement - 1.

If I is the intensity of the incident light then R + T = I. Since the incident ray is travelling in a denser medium (glass), it will be totally reflected at a certain critical angle $\theta$_c. For $\theta$ < $\theta$_c a part of the incident intensity is reflected and the remaining part is transmitted. But for $\theta$ > $\theta$_c, there is no refracted ray. Hence for $\theta$ > $\theta$_c, the value of R is 100%. Hence the only correct option is (c).

(A)$\to$(P), (S)

Since the path difference is

$S_1P_0=S_2P_0=0$

the phase difference becomes

$\delta(P_0)=0$

The intensity at any point is

$I = {I_{\max }}{\cos ^2}\left( {{8 \over 2}} \right)$

Now, $I({P_0}) = {I_{\max }}$ [$\because$ $\delta ({P_0}) = 0$]

$I({P_1}) = {I_{\max }}{\cos ^2}\left( {{\pi \over 4}} \right) = {{{I_{\max }}} \over 2}$

Therefore, $I({P_0}) > I({P_1})$

(B)$\to$(Q)

In this case, the path difference at P$_1$ is

${P_1} = {\lambda \over 4} - (\mu - 1)t = {\lambda \over 4} - {\lambda \over 4} = 0$

Therefore, $\delta ({P_1}) = 0$.

(C)$\to$(T)

In this case, the path difference at P$_1$ is

${P_1} = {\lambda \over 4} - (\mu - 1)t = {\lambda \over 4} - {\lambda \over 2} = {{ - \lambda } \over 4}$

Therefore, the phase difference at P$_1$ is

${P_1} = {{2\pi } \over \lambda } \times \left( {{{ - \lambda } \over 4}} \right) = - {\pi \over 2}$

Therefore, $I({P_1}) = {I_{\max }}{\cos ^2}\left( {{\pi \over 4}} \right) = {{{I_{\max }}} \over 2}$

The path difference at P$_2$ is

${P_2} = {\lambda \over 3} - (\mu - 1)t = {\lambda \over 3} - {\lambda \over 2} = {{ - \lambda } \over 6}$

Therefore, the phase difference at P$_1$ is

${P_2} = {{2\pi } \over \lambda } \times \left( {{{ - \lambda } \over 6}} \right) = - {\pi \over 3}$

Therefore, $I({P_2}) = {I_{\max }}{\cos ^2}\left( {{\pi \over 6}} \right) = {3 \over 4}{I_{\max }}$

Therefore, $I({P_2}) > I({P_1})$.

(D)$\to$(R), (S), (T)

In this case, the path difference at P$_1$ is

${P_1} = {\lambda \over 4} - (\mu - 1)t = {\lambda \over 4} - {{3\lambda } \over 4} = {{ - \lambda } \over 2}$

Therefore, the phase difference at P$_1$ is

${P_1} = {{2\pi } \over \lambda } \times \left( { - {\lambda \over 2}} \right) = - \pi $

Therefore, $I({P_1}) = {I_{\max }}{\cos ^2}\left( {{\pi \over 2}} \right) = 0$

The path difference at P$_0$ is

${P_0} = 0 - (\mu - 1)t = {{ - 3\lambda } \over 4}$

Therefore, the phase difference at P$_0$ is

${P_0} = {{2\pi } \over \lambda } \times \left( {{{ - 3\lambda } \over 4}} \right) = {{ - 3\lambda } \over 2}$

Therefore, $I({P_0}) = {I_{\max }}{\cos ^2}{{3\pi } \over 4} = {{{I_{\max }}} \over 2}$

Now, the path difference at P$_1$ is

${P_1} = {\lambda \over 4} - (\mu - 1)t = {\lambda \over 4} - {{3\lambda } \over 4} = - {\lambda \over 2}$

The phase difference at P$_1$ is

${P_1} = {{2\pi } \over \lambda } \times \left( {{{ - \lambda } \over 2}} \right) = - \pi $

Therefore, $I({P_1}) = {I_{\max }}\cos \left( {{\pi \over 2}} \right) = 0$

If $d=\lambda$, the maximum path difference (path difference is given by $d \sin\theta$) will be less than $\lambda$. So there will be only central maximum on the screen, because in the equation $d\sin\theta=n\lambda,n$ can take only one value.

If $\lambda < d < 2\lambda$, then the maximum path difference will be less than 2$\lambda$. So there will be two more maximum on the screen in addition to the central maximum.

Intensity of dark fringes becomes zero when intensities at the two slits are equal. Initial intensity at both the slits is unequal so there will some brightness at dark fringes. Hence, when intensity of both slits is made equal, the intensity at dark fringes on screen will reduces to zero.

The gap between consecutive wave fronts in medium 2 is less than that in medium 1. Therefore, wavelength of light in medium 2 in less than that in medium 1. Therefore, speed of light is more in medium 1 and less in medium 2.

The interference fringes formed in a Young's double slit experiment are straight lines. These fringes are a result of the constructive and destructive interference of the light waves coming from the two slits, and they appear as alternating bright and dark straight bands or lines.

Hence, the answer is :

Option D : straight line.

Remember for double hole experiment a hyperbola is generated.

To observe the phenomenon of interference, you need two sources that emit radiation of the same frequency and have a definite phase relationship. This is because interference is a result of the superposition of waves, which requires the waves to be coherent. Coherence is achieved when the waves have a constant phase difference, which implies that they are of the same frequency and have a definite phase relationship (a phase relationship that does not change with time).

So, the correct answer is :

Option D : of the same frequency and having a definite phase relationship.