Question

In the Young's double slit experiment shown in the figure, the medium between slit plane and the screen has refractive index varying with time as $n=n_0+kt$, then

• A. The y co-ordinates of central maximum is $\frac{D\sin \varphi }{n_0+kt}$.
• B. Velocity of central maximum as a function of time is $\frac{-kD\cos \varphi }{(n_0+kt{)}^2}$.
• C. If a glass plate of small thickness $p$ is placed in front of $S_1$, then to form central maximum at O, refractive index of glass should vary as $n_0+kt+\frac{d\sin \varphi }{p}$.
• D. If a glass plate of small thickness $p$ is placed in front of $S_1$, then to form central maxima at O, refractive index of glass should vary as $n_0+kt+\frac{d\cos \varphi }{p}$

Answer 1

Answer: (A), (C)

The correct options are
A The y co-ordinates of central maximum is $\frac{D\sin \varphi }{n_0+kt}$.
C If a glass plate of small thickness $p$ is placed in front of $S_1$, then to form central maximum at O, refractive index of glass should vary as $n_0+kt+\frac{d\sin \varphi }{p}$.


(a) $S_{1}P-S_{2}P=\frac{dy}{D}$
Then, path difference
$\Delta x=(n_0+kt)\frac{dy}{D}-d\sin \varphi$
For central maxima,
$\Delta x=0$
$\Rightarrow (n_0+kt)\frac{dy}{D}-d\sin \varphi =0$
$\therefore y=\frac{D\sin \varphi }{n_0+kt}$

(y-coordinates of central maximum)

(b) Velocity of central maximum
$\frac{dy}{dt}=\frac{-kD\sin \varphi }{(n_0+kt{)}^2}$

(c) Path difference produced due to glass plate of thickness $p$
$=n^{\prime }(\frac{n}{n^{{}^{\prime }}}-1)p$ where $n$ is the refractive index of the plate.


For central maximum to be formed at O, path difference produced by glass slab should cancel the path difference between light entering $S_1$ and $S_2$.
i.e $n^{{}^{\prime }}(\frac{n}{n^{{}^{\prime }}}-1)p=d\sin \varphi$.
where $n^{{}^{\prime }}=n_0+kt$
$\Rightarrow n-n^{\prime }=\frac{dsin\varphi }{p}$
$\Rightarrow n=n_0+kt+\frac{d\sin \varphi }{p}$