Question

Two light waves having the same wavelength $\lambda$ in vacuum are in phase initially. Then the first wave travels a path $L_1$ through a medium of refractive index $n_1$ while the second wave travels a path of length $L_2$ through a medium of refractive index $n_2$. After this, the phase difference between the two waves is:

• A. $\frac{2\pi }{\lambda }(\frac{L_2}{n_1}-\frac{L_1}{n_2})$
• B. $\frac{2\pi }{\lambda }(\frac{L_1}{n_1}-\frac{L_2}{n_2})$
• C. $\frac{2\pi }{\lambda }(n_1{L}_1-n_2{L}_2)$
• D. $\frac{2\pi }{\lambda }(n_2{L}_1-n_1{L}_2)$

Answer 1

The correct option is C $\frac{2\pi }{\lambda }(n_1{L}_1-n_2{L}_2)$

The optical path of the light travelling a distance $d$ in a medium of refractive index $\mu$ is given by,

$x=\mu d$

Optical path for first ray, which travels a path $L_1$ through a medium of refractive index $n_1$

$x_1=n_1{L}_1$

Optical path for second ray, which travels a path $L_2$ through a medium of refractive index $n_2$

$x_2=n_2{L}_2$

Therefore, the Path difference,

$\Delta x=n_1{L}_1-n_2{L}_2$

Now, phase difference,

$\Delta \varphi =\frac{2\pi }{\lambda }\Delta x$

$\Rightarrow \Delta \varphi =\frac{2\pi }{\lambda }\times (n_1{L}_1-n_2{L}_2)$

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