Question

Two light rays having the same wavelength λ in vacuum are in phase initially then the first ray travels a path L_1 through a medium of refractive index n_1, while the second ray travel the path of length L_2 through the medium of refractive index n_2. the two ways are then combined to produce interference. the phase difference between the two waves at the point of interference is

Answer 1

SOLUTION:-

$\tt{First \: ray -L_{1} ,μ_{1}}$

$\tt{ Second \: ray -L_{2} ,μ_{2}}$

$\tt{ if \: they \: travel \: L_{1} and \: L_{2}in \: vaccum, Then}$

$\tt{ \triangle x = L_{1} - \: L_{2}}$

optical path Length = n * L

$\tt{ = > o_{1} = μ_{1}L_{1} } \\ \tt{o_{2} = μ_{2}L_{2}}$

$\tt{ \triangle x= μ_{1}L_{1} - μ_{2}L_{2}}$

$\tt{ \triangle Φ= \frac{2 \pi}{ \lambda} \triangle x}$

$\tt{ \triangle Φ= \frac{2 \pi}{ \lambda} (μ_{1}L_{1} - μ_{2}L_{2})}$

Hence, the face difference is

$\tt{ \frac{2 \pi}{ \lambda} (μ_{1}L_{1} - μ_{2}L_{2})}$

Answer 2

AnswEr:-

first ray−L1,μ1

Second ray -$L_{2} ,μ_{2}$

Let they travel $L_{1} and \:L_{2}$ in vaccum,

∆ x =$L_{1} - \: L_{2}$

Length of optical path  = n * L

$\begin{gathered}\tt{ \implies o_{1} = μ_{1}L_{1} } \\ \tt{o_{2} = μ_{2}L_{2}}\end{gathered}\\</p><p>\tt{ \triangle x = μ_{1}L_{1} - μ_{2}L_{2}}\\ \\ </p><p>\tt{ \triangle \phi= \frac{2 \pi}{ \lambda} \triangle x}\\ \\ </p><p>\tt{ \triangle \phi= \frac{2 \pi}{ \lambda} (μ_{1}L_{1} - μ_{2}L_{2})}$

Hence, the face difference is

$\boxed{\tt{ \frac{2 \pi}{ \lambda} (μ_{1}L_{1} - μ_{2}L_{2})}}$

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HOPE IT HELPS :)