Question

An electromagnetic wave of wavelengths ${\lambda }_0$ (in vacuum) based from $\text{P}$ towards $\text{Q}$ crossing three different media of refractive indices $\mu ,2\mu \text{and}3\mu$ respectively as shown in figure. ${\varphi }_P$ and ${\varphi }_Q$ be the phase of the wave at points $\text{P}$ and $\text{Q}$. The phase difference ${\varphi }_Q-{\varphi }_P$ is
[Take : $\mu =1$]

• A. $0$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

The correct option is C $\frac{\pi }{2}$


As light is travelling in a medium other than air, we will have to consider the optical path, i.e. $x=\mu d$.

For given diagram, optical path difference will be

$\Delta x=(2.25{\lambda }_0\times \mu )+(3.5{\lambda }_0\times 2\mu )+(3{\lambda }_0\times 3\mu )$

$=18.25\mu {\lambda }_0$

$\therefore \Delta x=18.25{\lambda }_0[\because \mu =1]$

Further, phase difference,
$\varphi =\frac{2\pi }{{\lambda }_0}\times \Delta x$

$=\frac{2\pi }{{\lambda }_0}\times 18.25{\lambda }_0$

$=36.5\pi =18\left(2\pi \right)+0.5\pi$

$\therefore {\varphi }_Q-{\varphi }_P=0.5\pi =\frac{\pi }{2}$

Hence, option $\left(\text{C}\right)$is correct.

Why this question ? In dealing with interference of two waves in a medium, we need the differences between the optical paths travelled by the waves. The geometrical path and the optical path are equal only when light travels in vacuum or in air.