Question

An electromagnetic wave of wavelength ${\lambda }_0$ (in vacuum) passes from $P$ towards $Q$ crossing three different media of refractive index $\mu ,2\mu$ and $3\mu$ respectively as shown in figure. ${\varphi }_P$ and ${\varphi }_Q$ are the phase of the wave at point $P$ and $Q$. Find the phase difference ${\varphi }_Q-{\varphi }_P$ [Take: $\mu =1$]

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

Answer 1

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

The relation between phase difference ($\Delta \varphi$) and path difference ($\Delta x$) is
$\Delta \varphi =\frac{2\pi }{\lambda }\times \Delta x$

In a medium different from from vaccum,
$\Delta \varphi =\frac{2\pi }{{\lambda }^{\prime }}\times \Delta x$
If ${\lambda }^{\prime }=\frac{\lambda }{\mu },\Delta \varphi =\frac{2\pi }{\lambda }\times \left(\mu \Delta x\right)$
Here, $\mu \Delta x$ is called the path length of light in the medium.


In this question,
Total optical path length $={\mu }_1\times 2.25{\lambda }_0+{\mu }_2\times 3.5{\lambda }_0+{\mu }_3\times 3{\lambda }_0$ (given $\mu =1$)
$=1\times 2.25{\lambda }_0+2\times 3.5{\lambda }_0+3\times 3{\lambda }_0$
$=18.25{\lambda }_0$
Therefore phase difference between P and Q
$=\frac{2\pi }{{\lambda }_0}\times 18.25{\lambda }_0$
$=18\times 2\pi +0.25\times 2\pi$
$=0.5\pi =\frac{\pi }{2}$
(because $n\pi +\varphi =\varphi$)