Question

A light wave of wavelength ${\lambda }_0$ propagates from point $\text{A}$ to point $\text{B}$. We introduce in its path a glass plate of refractive index $n$ and thickness $l.$ The introduction of the plate alters the phase of the light at $B$ by an angle $\varphi .$ If $\lambda$ is the wavelength of light on emerging from the plate , then

• A. $\Delta \varphi =0$
• B. $\Delta \varphi =\frac{2\pi l}{{\lambda }_0}$
• C. $\Delta \varphi =2\pi l(\frac{1}{{\lambda }_0}-\frac{1}{\lambda })$
• D. $\Delta \varphi =\frac{2\pi l}{{\lambda }_0}(n-1)$

Answer 1

The correct option is D $\Delta \varphi =\frac{2\pi l}{{\lambda }_0}(n-1)$

When a light wave propagates from points $\text{A}$ to $\text{B}$, the Phase difference between those two points is zero.

After adding the glass plate:

The phase difference in incident light is,

${\varphi }_i=\frac{2\pi l}{{\lambda }_0}$

The phase difference in emerging light is,

${\varphi }_f=\frac{2\pi l}{\lambda }$

$\Delta \varphi =2\pi l(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})$

Further by snell's law,

$n\lambda ={\lambda }_0\Rightarrow \lambda =\frac{{\lambda }_0}{n}$

Substituting this we get, $\Delta \varphi =\frac{2\pi l}{{\lambda }_0}(n-1)$