Question

In a double slit experiment, when light of wavelength \(400 nm\) was used, the angular width of the first minima formed on a screen placed \(1 m\) away, was found to be \(0.2°\). What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? \((\mu_{water}\)\(=\frac43)\) 

• A. 0.266°
• B. 0.15°
• C. 0.05°
• D. 0.1°

Answer 1

The Correct Option is B

To solve this problem, we need to understand how the angular width of the first minima in a double-slit experiment changes when the medium is changed from air to water.

The angular width of the first minima for a double-slit experiment is given by the formula:

\(\theta = \frac{\lambda}{d}\)

Where:

• \(\theta\) is the angular position of the minima.
• \(\lambda\) is the wavelength of the light used.
• d is the slit separation.

When the setup is immersed in water, the effective wavelength (\(\lambda_{\text{effective}}\)) of light in the medium changes according to the refractive index (\(\mu\)) of the medium:

\(\lambda_{\text{effective}} = \frac{\lambda}{\mu}\)

Given:

• Original wavelength, \(\lambda = 400 \, \text{nm}\)
• Refractive index of water, \(\mu_{\text{water}} = \frac{4}{3}\)
• Initial angular width of the first minima, \(\theta_{\text{air}} = 0.2^\circ\)

In water, the effective wavelength of light is:

\(\lambda_{\text{water}} = \frac{400 \, \text{nm}}{\frac{4}{3}} = 300 \, \text{nm}\)

Therefore, the new angular width of the first minima, \(\theta_{\text{water}}\), is:

\(\theta_{\text{water}} = \theta_{\text{air}} \times \frac{\lambda_{\text{water}}}{\lambda}\)

Plugging in the values:

\(\theta_{\text{water}} = 0.2^\circ \times \frac{300 \, \text{nm}}{400 \, \text{nm}} = 0.15^\circ\)

Therefore, the angular width of the first minima when the setup is immersed in water is 0.15^\circ.

Thus, the correct answer is 0.15°.