91Ó°ÊÓ

The wavelength in air,$\mathrm{λ}=620\text{nm}$

The phase difference,$\mathrm{Δ}\mathrm{ϕ}=\mathrm{π}$

The refractive index of medium 1, $n_1=1.45$

The refractive index of medium 2, $n_2=1.65$

The expression to calculate the phase difference is given as follows.

$\mathit{Δ}\mathit{ϕ}\mathbf{=}\mathbf{2}\mathit{π}\mathit{L}\left(\frac{n_2}{\mathrm{λ}}-\frac{n_1}{\mathrm{λ}}\right)$

$\mathit{L}\mathbf{=}\frac{\mathbf{λ}\mathbf{Δ}\mathbf{ϕ}}{\mathbf{2}\mathbf{π}\left(n_2-n_1\right)}$ ......(1)

The expression to calculate the second smallest value of Lis given as follows.

Calculate the smallest value of the L .

Substitute the given values into equation (1).

$\begin{array}{rcl}L& =& \frac{\mathrm{π}×620×{10}^{-9}}{2\mathrm{π}\left(1.65-1.45\right)}\\ & =& \frac{310×{10}^{-9}}{0.20}\\ & =& 1550×{10}^{-9}\\ & =& 1.55×{10}^{-6}\text{m}\end{array}$

Hence, the smallest value of $L$ is $1.55×{10}^{-6}\text{m}$

Calculate the smallest value of the$L$.

Substitute the given values into equation (1).

$\begin{array}{rcl}L& =& \frac{3\mathrm{π}×620×{10}^{-9}}{2\mathrm{π}\left(1.65-1.45\right)}\\ & =& \frac{930×{10}^{-9}}{0.20}\\ & =& 4650×{10}^{-9}\\ & =& 4.65×{10}^{-6}\text{m}\end{array}$

Hence the second smallest value of $L$ is $4.65×{10}^{-6}\text{m}$.