Question

In $YDSE$, if a slab whose refractive index can be varied is placed in front of the upper slit. Then, the variation of resultant intensity $I$, at mid-point of screen with $\mu$ will be best represented by $(\mu \ge 1$ and $I_0$ is the maximum intensity)

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

After putting of glass slab, the path difference is given by,

$\Delta x=(\mu -1)t$

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

As, $I=I_0{\cos }^2\frac{\varphi }{2}$

$\Rightarrow I=I_0{\cos }^2\frac{\pi }{\lambda }(\mu -1)t$

As, $\mu =1\Rightarrow I=I_0$

And we know that, ${\cos }^2\theta$ graph is,


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