Question

Two light ways 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. $\frac{2 \pi }{\lambda }\left(\frac{L_{1}}{\left(\mu \right)_{1}} - \frac{L_{2}}{\left(\mu \right)_{2}}\right)$
• B. $\frac{2 \pi }{\lambda }\left(L_{2} - L_{1}\right)$
• C. $\frac{2 \pi }{\lambda }\left(\left(\mu \right)_{2} L_{1} - \left(\mu \right)_{1} L_{2}\right)$
• D. $\frac{2 \pi }{\lambda }\left(\left(\mu \right)_{1} L_{1} - \left(\mu \right)_{2} L_{2}\right)$

Answer 1

Answer: (D)

Optical path for first ray $=\mu _{1}L_{1}$
Optical path for second ray $=\mu _{2}L_{2}$
So, phase difference is given by
$\Delta \phi=\frac{2 \pi }{\lambda }\times $ path difference
$=\frac{2 \pi }{\lambda \, }\times \Delta x$
$\Rightarrow \, \, \Delta \phi=\frac{2 \pi }{\lambda }\left(\left(\mu \right)_{1} L_{1} - \left(\mu \right)_{2} L_{2}\right)$