Question

A double-slit arrangement produces interference fringes for sodium light$\left(\lambda =589\text{nm}\right)$that are $\mathbf{0}\mathbf{.}{\mathbf{20}}^{\mathbf{0}}\mathit{C}$apart. What is the angular separation if the arrangement is immersed in water $\left(n=1.33\right)$?

Short answer

The angular separation when the arrangement immerged in water $0.{15}^{0}C$.

Step-by-step solution

Write the given data from the question

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

The angular separation, ${\theta }_1=0.20°$.

The refractive index of water, $n=1.33$.

Determine the formulas to calculate the angular separation when the arrangement in immerged in water

Young's double-slit experiment. When monochromatic light passing through two narrow slits illuminates a distant screen, a characteristic pattern of bright and dark stripes is observed. This interference pattern is caused by the superposition of overlapping light waves originating from the two slits.

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

$d\sin \theta =m\lambda$ …… (1)

Here, d is the distance between the slits, $\lambda$ is the wavelength, m is the order and $\theta$is the angular separation.

The expression to calculate the wavelength in the vacuum is given as follows.

${\mathit{\lambda }}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{n}}$ …… (2)

Calculate the angular separation when the arrangement is immersed in water:

Calculate the distance between the slits.

Substitute 1 for m, $0.{20}^{0}C$for ${\theta }_1$and $589\text{nm}$for $\lambda$into equation (1).

$d\sin \left(0.20\right)=1\times 589\times {10}^{-9}$

$$\begin{gathered} d=\frac{589\times {10}^{-9}}{\sin \left(0.20\right)} \\ =1.687\times {10}^{-4}\text{m} \end{gathered}$$

Calculate the wavelength in the water.

Substitute $589\text{nm}$for $\lambda$into equation (2).

$$\begin{gathered} {\lambda }_n=\frac{589\times {10}^{-9}}{1.33} \\ =442.85\times {10}^{-9}\text{m} \\ =442.85\text{nm} \end{gathered}$$

Calculate the angular separation when the arrangement immerged in water.

Substitute $1.687\times {10}^{-4}\text{m}$ for d, $442.85\text{nm}$for ${\lambda }_n$and 1 for m into equation (1).

$$\begin{gathered} 1.687\times {10}^{-4}\times \sin \left({\theta }_2\right)=1\times 442.85\times {10}^{-9} \\ \sin {\theta }_2=\frac{442.85\times {10}^{-9}}{1.687\times {10}^{-4}} \end{gathered}$$

$$\begin{gathered} {\theta }_2={\sin }^{-1}\left(2.62\times {10}^{-3}\right) \\ =0.15° \end{gathered}$$

Hence, the angular separation when the arrangement immerged in water $0.{15}^{0}C$.