Question

Two light waves having the same wavelength $\left(\lambda \right)$ in vacuum are in phase initially. Then, the first ray travels path length $L_1$ through a medium of refractive index ${\mu }_1$. The second ray travels a path of length $L_2$ through a medium of refractive index ${\mu }_2$. The two waves are then combined to observe interference effects. The phase difference between the two, when they interfere-

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

Answer 1

The correct option is B $\frac{2\pi }{\lambda }({\mu }_1{L}_1-{\mu }_2{L}_2)$

The phase difference is given by,

$\Delta \varphi =\frac{2\pi }{\lambda }\Delta x.......\left(1\right)$

Since, the ray $1$ and ray $2$ travels through medium of refractive index ${\mu }_1$ and ${\mu }_2$ respectively. Hence, their path lengths as follows,

Path length of ray $1$ is, $l_1={\mu }_1{L}_1$

Path length of ray $2$ is, $l_2$=${\mu }_2{L}_2$

Therefore, path difference between the two rays is,

$\Delta x=l_1-l_2={\mu }_1{L}_1-{\mu }_2{L}_2$

Putting this in (1) we get,

$\Delta \varphi =\frac{2\pi }{\lambda }({\mu }_1{L}_1-{\mu }_2{L}_2)$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.

Why this question ? Key concept : The relation between the path difference and phase difference is given by, $\text{Phase difference}=\frac{2\pi }{\lambda }\times \text{path difference}$ $\Delta \varphi =\frac{2\pi }{\lambda }\Delta x$