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Discussion Forum



1)

Two light rays having the same wavelength in vacuum are in phase initially. Then, the first ray travels a path L1 through a medium of refractive index  $\mu_{1}$ while the second ray travels a path L2 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) $\frac{2\pi}{\lambda}\left(\frac{L_{1}}{\mu_{1}}-\frac{L_{2}}{\mu_{2}}\right)$

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:

Option D

Explanation:

 First ray optical path=  $\mu_{1}L_{1}$

 Second ray optical path= $\mu_{2}L_{2}$

 So, phase difference 

       $\triangle\phi=\frac{2\pi}{\lambda}\times$ path difference=$\frac{2\pi}{\lambda}\times\triangle x$

$\triangle\phi=\frac{2\pi}{\lambda}(\mu_{1}L_{1}-\mu_{2}L_{2})$