Question

Two light rays having the same wavelength in vacuum are in phase initially. Then, the first ray travels a path $L_1$ through a medium of refractive index ${\mu }_1$ while the second ray travels a path $L_2$ through a medium of refractive index ${\mu }_2$. The two waves are then combined to observe interference. The Phase difference between the two waves is

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

Answer 1

Answer: (D)

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

First ray optical path $={\mu }_1{L}_1$
second ray optical path $={\mu }_2{L}_2$
So, phase difference
$\Delta \varphi =\frac{2\pi }{\lambda }\times$ path difference $=\frac{2\pi }{\lambda }\times \Delta x$
$\therefore \Delta \varphi =\frac{2\pi }{\lambda }({\mu }_1{L}_1-{\mu }_2{L}_2)$