Question

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

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

Answer 1

The correct option is A

$\frac{2\mathrm{\pi }}{\lambda }(n_1{L}_1-n_2{L}_2)$

Step 1: Given data:

The optical path length is the product of the geometric path length multiplied by the refractive index.

Step 2: Calculation:

Since, the length of the path for the first wave is: $n_1{L}_1$

The length of the path for the second wave is: $n_2{L}_2$

Difference between two paths:

$∆=n_1{L}_1-n_2{L}_2$

The phase difference between the two waves is:

$=\frac{2\mathrm{\pi }}{\lambda }(∆)$

$=\frac{2\mathrm{\pi }}{\lambda }(n_1{L}_1-n_2{L}_2)$

Thus, the phase difference between the two waves is $\frac{2\mathrm{\pi }}{\lambda }(n_1{L}_1-n_2{L}_2)$.