Question 64: Waves Optics

Question

Column I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits S$$_1$$ and S$$_2$$. In each of these cases, S$$_1$$P$$_0$$ = S$$_2$$P$$_0$$, S$$_1$$P$$_1$$ $$-$$ S$$_2$$P$$_1$$ = $$\lambda/4$$ and S$$_1$$P$$_2$$ $$-$$ S$$_2$$P$$_2$$ = $$\lambda/3$$, where $$\lambda$$ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index $$\mu$$ and thickness t is pasted on slit S$$_2$$. The thickness of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by $$\delta$$(P) and the intensity by I(P). Match each situation given in Column I with the statement(s) in Column II valid for that situation:

Column I

Column II

(A)

(P)
$$\delta ({P_0}) = 0$$

(B)
$$(\mu-1)t=\lambda/4$$
(Q)
$$\delta ({P_1}) = 0$$

(C)
$$(\mu-1)t=\lambda/2$$
(R)
$$I({P_1}) = 0$$

(D)
$$(\mu-1)t=3\lambda/4$$
(S)
$$I({P_0}) > I({P_1})$$

(T)
$$I({P_2}) > I({P_1})$$

Accepted Answer

(A)$$	o$$(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
$$I = {I_{\max }}{\cos ^2}\left( {{8 \over 2}} ight)$$
Now, $$I({P_0}) = {I_{\max }}$$ [$$\because$$ $$\delta ({P_0}) = 0$$]
$$I({P_1}) = {I_{\max }}{\cos ^2}\left( {{\pi  \over 4}} ight) = {{{I_{\max }}} \over 2}$$
Therefore, $$I({P_0}) > I({P_1})$$
(B)$$	o$$(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)$$	o$$(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
$${P_1} = {{2\pi } \over \lambda } 	imes \left( {{{ - \lambda } \over 4}} ight) =  - {\pi  \over 2}$$
Therefore, $$I({P_1}) = {I_{\max }}{\cos ^2}\left( {{\pi  \over 4}} ight) = {{{I_{\max }}} \over 2}$$
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
$${P_2} = {{2\pi } \over \lambda } 	imes \left( {{{ - \lambda } \over 6}} ight) =  - {\pi  \over 3}$$
Therefore, $$I({P_2}) = {I_{\max }}{\cos ^2}\left( {{\pi  \over 6}} ight) = {3 \over 4}{I_{\max }}$$
Therefore, $$I({P_2}) > I({P_1})$$.
(D)$$	o$$(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
$${P_1} = {{2\pi } \over \lambda } 	imes \left( { - {\lambda  \over 2}} ight) =  - \pi $$
Therefore, $$I({P_1}) = {I_{\max }}{\cos ^2}\left( {{\pi  \over 2}} ight) = 0$$
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
$${P_0} = {{2\pi } \over \lambda } 	imes \left( {{{ - 3\lambda } \over 4}} ight) = {{ - 3\lambda } \over 2}$$
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
$${P_1} = {{2\pi } \over \lambda } 	imes \left( {{{ - \lambda } \over 2}} ight) =  - \pi $$
Therefore, $$I({P_1}) = {I_{\max }}\cos \left( {{\pi  \over 2}} ight) = 0$$

Answer

(A)$$	o$$(P), (S); (B)$$	o$$(Q); (C)$$	o$$(T); (D)$$	o$$(R), (S), (T)

Answer

(A)$$	o$$(P), (S); (B)$$	o$$(R); (C)$$	o$$(T); (D)$$	o$$(R), (S), (T)

Answer

(A)$$	o$$(P), (S); (B)$$	o$$(Q); (C)$$	o$$(S); (D)$$	o$$(R), (S), (T)

Answer

(A)$$	o$$(P), (R); (B)$$	o$$(Q); (C)$$	o$$(T); (D)$$	o$$(R), (S), (T)

Waves Optics

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