91Ó°ÊÓ

In first experiment, the $1$$2$ and are in phase and the wavelength,${\mathrm{λ}}_1=620nm$ .

In second experiment, the $1$$2$ and are in phase and the wavelength,${\mathrm{λ}}_2=496nm$ .

The first ray travel total distance $L_1=2L$ and second ray travel total distance $L_2=6L$.

The path difference for the constructive interference is given as follows.

${\mathbf{L}}_{\mathbf{2}}\mathbf{-}{\mathbf{L}}_{\mathbf{1}}\mathbf{=}{\mathbf{³¾Î»}}_{\mathbf{1}}$

The path difference in destructive interference is given as follows.

${\mathbf{L}}_{\mathbf{2}}\mathbf{-}{\mathbf{L}}_{\mathbf{1}}\mathbf{=}\left(m\text{'}+1\right)\frac{{\mathbf{λ}}_{\mathbf{2}}}{\mathbf{2}}$

The extra distance travel by the first ray is given by.

$$\begin{gathered} L_2-L_1=6L-2L \\ =4L \end{gathered}$$

The path difference for the constructive interference is equal to $m{\mathrm{λ}}_1$.

$$\begin{gathered} L_2-L_1=m{\mathrm{λ}}_1 \\ 4L=m{\mathrm{λ}}_1 \end{gathered}$$

The path difference in destructive interference is equal to $\left(m\text{'}+1\right)\frac{{\mathrm{λ}}_2}{2}$.

$$\begin{gathered} L_2-L_1=\left(m\text{'}+1\right)\frac{{\mathrm{λ}}_2}{2} \\ 4L=\left(m\text{'}+1\right)\frac{{\mathrm{λ}}_2}{2} \end{gathered}$$

$m{\mathrm{λ}}_1=\left(m\text{'}+1\right)\frac{{\mathrm{λ}}_2}{2}$

Substitute $620nm$ for ${\mathrm{λ}}_1$and $496nm$ for ${\mathrm{λ}}_2$ into above equation.

$$\begin{gathered} m×620=\left(m\text{'}+1\right)\frac{496}{2} \\ =\left(m\text{'}+1\right)248 \end{gathered}$$

$$\begin{gathered} m=\left(m\text{'}+1\right)0.4 \\ =0.4m\text{'}+0.4 \end{gathered}$$

By substituting the difference values of $m\text{'}$, find the integer value of $m$

Therefore, the value of $L$ is given by,

$L=\frac{2{\mathrm{λ}}_1}{4}$

Substitute $620nm$ for ${\mathrm{λ}}_1$ into above equation.

$$\begin{gathered} L=\frac{2×620}{4} \\ =310\text{nm} \end{gathered}$$

Hence, the required value of $L$ is $310nm$.