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  • Two light waves having the same wavelength  in vacuum are in phase initially. Then the first wave travels a path 

Two light waves having the same wavelength \lambda in vacuum are in phase initially. Then the first wave travels a path L_{1} through a medium of refractive index n_{1} while the second wave travels a path of length L_{2} through a medium of refractive index n_{2}. After this the phase difference between the two waves is :
Option: 1 \frac{2\pi }{\lambda }\left ( \frac{L_{2}}{n_{1}}-\frac{L_{1}}{n_{2}} \right )  
Option: 2 \frac{2\pi }{\lambda }\left ( \frac{L_{1}}{n_{1}}-\frac{L_{2}}{n_{2}} \right )
Option: 3 \frac{2\pi }{\lambda }(n_{1}L_{1}-n_{2}L_{2})
Option: 4 \frac{2\pi }{\lambda }(n_{2}L_{1}-n_{1}L_{2})

Answers (1)

best_answer

As we know

Phase difference =  \frac{2 \pi}{\lambda}*( path difference)

S0

\begin{array}{l} \lambda_{n 1}=\frac{\lambda}{n_{1}} \\ \\ \lambda_{n 2}=\frac{\lambda}{n_{2}} \\ \\ (\Delta \phi)_{1}=\frac{2 \pi}{\lambda_{n 1}} *L_{1} \\ \\ (\Delta \phi)_{2}=\frac{2 \pi}{\lambda_{n 2}} *L_{2} \\ \\ \Delta \phi=\Delta \phi_{1}-\Delta \phi_{2}=\frac{2 \pi}{\lambda}\left(n_{1} L_{1}-n_{2} L_{2}\right) \end{array} 

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avinash.dongre

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