A double-slit arrangement produces interference fringes for sodium light $\left(\lambda =589\text{nm}\right)$that have an angular separation of $\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathit{r}\mathit{a}\mathit{d}$. For what wavelength would the angular separation be 10% greater?

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

Angular separation, ${\theta }_1=3.5\times {10}^{-3}\text{rad}$.

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

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

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

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

Calculate the distance between the slits.

Substitute 1 for m, $3.50\times {10}^{-3}\text{rad}$for ${\theta }_1$and $589\text{nm}$for $\lambda$into equation (1).

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

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

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

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

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

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

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

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