Question

In a $\text{YDSE}$ experiment, if a slab whose refractive index can be varied is placed in front of one of the slits then the variation of resultant intensity at the mid-point of the screen with $\mu$ will be best represented by ($\mu \ge 1)$. $[$Assume slits of equal width and there is no absorption by slab$]$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C


The path difference due to the slab is,

$(\Delta x{)}_0=(\mu -1\left)t\right(\text{t =thickness})$

And its corresponding phase difference is,

$\Delta \varphi =\frac{2\pi }{\lambda }(\mu -1)t$

$I_{res}=4I_0{\cos }^2\frac{\Delta \varphi }{2}$

For $I_{max};{\cos }^2\left(\frac{\Delta \varphi }{2}\right)=1$

$\Rightarrow \Delta \varphi =0$

$\Rightarrow \Delta \varphi =\frac{2\pi }{\lambda }(\mu -1)t=0$

$\Rightarrow \mu =1\left(\text{and function will be cosine}\right)$



<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.