Question

In a double slit experiment, the screen is placed at a distance of $1.25 \,m$ from the slits. When the apparatus is immersed in water $ ({{\mu }_{w}}=4/3) $ , the angular width of a fringe is found to be $ {{0.2}^{o}} $ . When the experiment is performed in air with same set up, the angular width of the fringe is

• A. $ {{0.4}^{o}} $
• B. $ {{0.27}^{o}} $
• C. $ {{0.35}^{o}} $
• D. $ {{0.15}^{o}} $

Answer 1

The Correct Option is D

When the apparatus is immersed in water the angular width of a fringe
$ \theta =\frac{\lambda }{d} $
and $ \theta ={{0.2}^{o}} $
and the angular width of a fringe in air
$ \theta =\frac{\lambda }{d} $
$ \frac{1}{{{\mu }_{w}}}=\frac{\lambda }{\lambda } $
$ \frac{\lambda }{\lambda }=\frac{3}{4} $
Now, $ \frac{\theta }{\theta }=\frac{\lambda }{\lambda } $
$ \theta =\frac{\lambda }{\lambda }\times \theta $
$ \theta =\frac{3}{4}\times {{0.2}^{o}} $
$ \theta =0.15{}^\circ $

Answer 2

The formula for the angular width of a fringe:
\(\theta = \frac{\lambda}{d}\)

Where:
- \(\theta\) is the angular width of the fringe.
- \(\lambda\) is the wavelength of the light used.
- d is the distance between the slits.

In Air
Given the angular width in the air:
\(\theta_{\text{air}} = 0.2^\circ\)

In Water
\(\lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{\mu_{\text{medium}}}\)

The refractive index of water is approximately \(\mu_w = \frac{4}{3}\).

So, the wavelength of light in water becomes:
\(\lambda_w = \frac{\lambda_{\text{air}}}{\mu_w} = \frac{\lambda_{\text{air}}}{4/3} = \lambda_{\text{air}} \times \frac{3}{4}\)

Angular Width in Water
\(\theta_{\text{water}} = \frac{\lambda_{\text{water}}}{d}\)

Since \(\lambda_{\text{water}} = \frac{3}{4} \lambda_{\text{air}}\), we substitute this into the formula:
\(\theta_{\text{water}} = \frac{\frac{3}{4} \lambda_{\text{air}}}{d}\)

This shows that:
\(\theta_{\text{water}} = \frac{3}{4} \left( \frac{\lambda_{\text{air}}}{d} \right) = \frac{3}{4} \theta_{\text{air}}\)
Given: \(\theta_{\text{air}} = 0.2^\circ\)
Therefore:
\(\theta_{\text{water}} = \frac{3}{4} \times 0.2^\circ = 0.15^\circ\)

So, the correct option is (D): \(0.15\degree\)