Question

Two light waves having the same wavelength $\lambda$ in vacuum are in phase initially. Then the first ray travels a path of length $L_1$ through a median 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, is

• 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 $\frac{2\pi }{\lambda }({{\mu }_{1}L}_1-{\mu }_2{L}_2)$

Optical path length is the product of geometric path length multiplied by refractive index.
Hence, path length for the first wave is: ${\mu }_1{L}_1$
path length for the first wave is: ${\mu }_2{L}_2$
path difference: $\Delta ={\mu }_1{L}_1-{\mu }_2{L}_2$
Phase difference between the two waves is: $\delta =\frac{2\pi }{\lambda }\left(\Delta \right)=\frac{2\pi }{\lambda }({\mu }_1{L}_1-{\mu }_2{L}_2)$