Question

Light of wave length $\lambda$ air enters into two medium of refractive indices $\mu$ and $\mu /2$. Two points $P_1$ and $P_2$ lying along the path of this light as shown in the figure. The phase difference between these two point is $k\frac{\left(\pi \mu X_0\right)}{\lambda }$. Then find the value of k.

• A. 3
• B. 4
• C. 5
• D. 6
• E. None of these

Answer 1

The correct option is B 4

Let $t$ be the distance between points S and P${}_1$

Optical path length traveled by light via

$S{P}_2=\mu t+\mu \left(X_0\right)+\frac{\mu }{2}\left(2X_0\right)=\mu (t+2X_0)$

Optical path traveled by light via

$S{P}_1=\mu t$

Path difference between points P${}_1$ and P${}_2$,

$\Delta x=S{P}_2-S{P}_1=2\mu X_0$

Now, phase difference,

$\varphi =\Delta x\times \frac{2\pi }{\lambda }=2\mu X_0\times \frac{2\pi }{\lambda }=4\frac{\pi \mu X_0}{\lambda }$

But given the phase difference is

$k\frac{\pi \mu X_0}{\lambda }$.

Hence, K=4