Question: Q. 3. (a) There are two sets of apparatus of Young’s double slit experiment. In set $A$, the phase difference between the two waves emanating from the slits does notchange with time, whereas in set $B$, the phase difference between the two waves from the slits changes rapidly with time. What difference will be observed in the pattern obtained on the screen in the two set ups?
(b) Deduce the expression for the resultant intensity in both the above mentioned set ups ( $A$ and $B$ ), assuming that the waves emanating from the two slits have the same amplitude $a$ and same wavelength $\lambda$.
U] [SQP 2018-19]
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Solution:
Ans. (a) Set $A$ : Stable interference pattern, the positions of maxima and minima do not change with time.
Set $B$ : Positions of maxima and minima will change rapidly with time and an average uniform intensity distribution will be observed on the screen.
(b) Expression for intensity of stable interference pattern in set- $A$ If the displacement produced by slit $S_{1}$ is
$$ y_{1}=a \cos \omega t $$
then, the displacement produced by $S_{2}$ would be
$$ y_{2}=a \cos (\omega t+\phi) $$
and the resultant displacement will be given by
$$ \begin{aligned} y & =y_{1}+y_{2} \ & =a[\cos \omega t+\cos (\omega t+\phi)] \ & =2 a \cos (\phi / 2) \cos (\omega t+\phi / 2) \end{aligned} $$
The amplitude of the resultant displacement is $2 a \cos (\phi / 2)$ and therefore the intensity at that point will be
$$ \begin{aligned} I & =4 I_{0} \cos ^{2}(\phi / 2) \ \phi & =0 \ I & =4 I_{0} \end{aligned} $$
$\therefore \quad I=4 I_{0}$ In set $B$, the intensity will be given by the average intensity
$$ \begin{aligned} & =4 I_{0}<\cos ^{2}(\phi / 2)> \ I & =2 I \end{aligned} $$
[CBSE Marking Scheme 2018-19]
