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Two light waves having the same wavelength $\lambda$ in vacuum are in phase initially. Then first ray travels a path of length L₁ through a medium of refractive index $\mu_{1}$ . The second ray travels a path of length L₂through refractive index $\mu_{2}$ . The two waves are combined to observe interference effect the phase difference between the two when they interfere is:
Option: 1
$\frac{2\pi}{\lambda}\left ( L_{1} -L_{2}\right )$

Option: 2
$\frac{2\pi}{\lambda}\left ( \mu_{1}L_{1}-\mu_2 L_{2} \right )$

Option: 3
$\frac{2\pi}{\lambda}\left ( \mu_{2} L_{1}-\mu_{1}L_{2}\right )$

Option: 4
$\frac{2\pi}{\lambda}\left ( \frac{L_{1}}{\mu_1}-\frac{L_{2}}{\mu_{2}} \right )$

Answers (1)

best_answer

As we learnt

Optical Path -

$x{}'=\mu \cdot x$

- wherein

$x{}'=$ Distance travelled in vacuum

$x=$ Distance travelled in a medium of refractive index $\mu$

Optical path length = $\mu L$

Path difference $\Delta x=\mu_{1}L_{1}-\mu_{2}L_{2}$

Phase difference $\Delta \phi=\frac{2\pi}{\lambda} \left ( \mu_{1}L_{1}-\mu_{2}L_{2} \right )$

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manish

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