In two experiments, light is to be sent along the two paths shown in Fig. 35-35 by reflecting it from the various flat surfaces shown. In the first experiment, rays $\mathbf{1}$ and$\mathbf{2}$ are initially in phase and have a wavelength of $\mathbf{620}\mathbf{.}\mathbf{0}\mathbf{nm}$. In the second experiment, rays $\mathbf{1}$ and$\mathbf{2}$ are initially in phase and have a wavelength of $\mathbf{496}\mathbf{.}\mathbf{0}\mathbf{nm}$ . What least value of distance $\mathbf{L}$ is required such that the $\mathbf{620}\mathbf{.}\mathbf{0}\mathbf{nm}$waves emerge from the region exactly in phase but the $\mathbf{496}\mathbf{.}\mathbf{0}\mathit{n}\mathit{m}$waves emerge exactly out of phase?

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

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

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.

${\mathbf{L}}_{\mathbf{2}}\mathbf{-}{\mathbf{L}}_{\mathbf{1}}\mathbf{=}{\mathbf{m\lambda }}_{\mathbf{1}}$

The path difference in destructive interference is given as follows.

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

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 $m{\lambda }_1$.

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

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

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

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

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

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

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

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

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

$L=\frac{2{\lambda }_1}{4}$

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

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

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