Question

In the Young's double slit experiment, the distance between the slits varies in time as $d\left(t\right)=d_0+a_0\sin \omega t$; where $d_0,\omega$ and $a_0$ are constants. The difference between the largest fringe width and the smallest fringe width, obtained over time, is given as :

• A. $\frac{2\lambda D\left(d_0\right)}{(d_0^2-a_0^2)}$
• B. $\frac{\lambda D}{d_0^2}{a}_0$
• C. $\frac{\lambda D}{d_0+a_0}$
• D. $\frac{2\lambda D{a}_0}{(d_0^2-a_0^2)}$

Answer 1

The correct option is D $\frac{2\lambda D{a}_0}{(d_0^2-a_0^2)}$

Fring width, $\beta =\frac{\lambda D}{d}$

For $\beta ={\beta }_{max},d=d_{min}$

And for $\beta ={\beta }_{min},d=d_{max}$

Now, $d=d_0+a_0\sin \omega t$

$\Rightarrow d_{max}=d_0+a_0$ and $d_{min}=d_0-a_0$

$\therefore {\beta }_{min}=\frac{\lambda D}{d_0+a_0}$ and

$\therefore {\beta }_{max}=\frac{\lambda D}{d_0-a_0}$

${\beta }_{max}-{\beta }_{min}=\frac{\lambda D}{d_0-a_0}-\frac{\lambda D}{d_0+a_0}$

$=\frac{2\lambda D{a}_0}{d_0^2-a_0^2}$

Hence, $\left(B\right)$ is the correct answer.