91Ó°ÊÓ

The wavelength, $\mathrm{λ}=589\text{nm}$.

Angular separation, ${\mathrm{θ}}_1=3.5×{10}^{-3}\text{rad}$.

Increment for the wavelength in angular separation is $10\%$.

The condition for the maxima in Young’s experiment is given as follows.

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\mathbf{=}\mathit{m}\mathit{λ}$ …… (1)

Here, d is the distance between the slits, $\mathit{λ}$is the wavelength, mis the order, and$\mathit{θ}$is the angular separation.

Calculate the distance between the slits.

Substitute 1 for m, $3.50×{10}^{-3}\text{rad}$for ${\mathrm{θ}}_1$and $589\text{nm}$for $\mathrm{λ}$into equation (1).

$d\sin \left(3.5×{10}^{-3}\right)=1×589×{10}^{-9}$

$$\begin{gathered} d=\frac{589×{10}^{-9}}{\sin \left(3.5×{10}^{-3}\right)} \\ =1.683×{10}^{-4} \end{gathered}$$

The angular separation increased by 10%. Therefore, the new value of the angular separation.

$$\begin{gathered} {\mathrm{θ}}_2=3.5×{10}^{-3}+3.5×{10}^{-3}×\frac{10}{100} \\ =3.85×{10}^{-3}\text{rad} \end{gathered}$$

Calculate the value of the wavelength at which angular separation increased by 10%.

Substitute $1.683×{10}^{-4}\text{m}$ for d and $3.85×{10}^{-3}$for ${\mathrm{θ}}_2$ and 1 for m into equation (1).

$$\begin{gathered} 1.683×{10}^{-4}×\sin \left(3.85×{10}^{-3}\right)=1×{\mathrm{λ}}_2 \\ 1.683×{10}^{-4}×0.00383={\mathrm{λ}}_2 \\ 6.48×{10}^{-7}\text{m}={\mathrm{λ}}_2 \\ {\mathrm{λ}}_2=648\text{nm} \end{gathered}$$

Hence, the wavelength at which the angular separation increased by 10% is $648\text{nm}$.