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 the effect of immersing the experimental apparatus in water on the angular width of the first minima in a double slit experiment.

In a double slit experiment, the condition for the position of the first minima is given by:

\[\theta = \dfrac{\lambda}{d}\]

where \(\lambda\) is the wavelength of light, \(\theta\) is the angle of the minima, and \(d\) is the slit separation.

When the apparatus is immersed in water, the wavelength of light changes. The apparent wavelength \(\lambda_{water}\) in the medium is given by:

\[\lambda_{water} = \frac{\lambda_{vacuum}}{\mu_{water}}\]

Here, \(\mu_{water} = \frac{4}{3}\) is the refractive index of water, and \(\lambda_{vacuum} = 400\: nm\) is the wavelength of the light in a vacuum (or air).

Substituting the values, we get:

\[\lambda_{water} = \frac{400\: nm}{\frac{4}{3}} = 300\: nm\]

Now, the angular position of the first minima in water is calculated with this new wavelength:

\[\theta_{water} = \frac{\lambda_{water}}{d}\]

The ratio of the angular width in water to that in vacuum is given by:

\[\frac{\theta_{water}}{\theta_{vacuum}} = \frac{\lambda_{water}}{\lambda_{vacuum}}\]

Substituting the values:

\[\frac{\theta_{water}}{0.2^\circ} = \frac{300\: nm}{400\: nm} = \frac{3}{4}\]

This implies:

\[\theta_{water} = 0.2^\circ \times \frac{3}{4} = 0.15^\circ\]

Thus, the correct answer is 0.15°.