Question

In Young's double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in a liquid of refractive index \( \frac{4}{3} \) without changing the geometrical arrangement, the new fringe width will be:

• A. 0.45 mm
• B. 0.4 mm
• C. 0.53 mm
• D. 0.30 mm

Hint:

The fringe width decreases when the refractive index of the medium increases.

Answer 1

The Correct Option is D

Step 1: Formula for fringe width in Young's double slit experiment

In Young's double slit experiment, the fringe width \( \beta \) is given by the formula: \[ \beta = \frac{\lambda D}{d} \] where: - \( \lambda \) is the wavelength of light, - \( D \) is the distance between the screen and the slits, - \( d \) is the distance between the slits. The fringe width \( \beta \) is directly proportional to the wavelength \( \lambda \).

Step 2: Effect of immersion in a liquid

When the apparatus is immersed in a liquid of refractive index \( n = \frac{4}{3} \), the wavelength of light in the liquid is reduced. The new wavelength \( \lambda_{\text{new}} \) in the liquid is given by: \[ \lambda_{\text{new}} = \frac{\lambda}{n} \] where \( n = \frac{4}{3} \). Therefore, the wavelength in the liquid will be: \[ \lambda_{\text{new}} = \frac{3}{4} \lambda \]

Step 3: New fringe width

Since the fringe width \( \beta \) is proportional to the wavelength \( \lambda \), the new fringe width \( \beta_{\text{new}} \) in the liquid will be: \[ \beta_{\text{new}} = \frac{\lambda_{\text{new}} D}{d} = \frac{\frac{3}{4} \lambda D}{d} = \frac{3}{4} \beta \] Given that the initial fringe width \( \beta = 0.4 \, \text{mm} \), the new fringe width \( \beta_{\text{new}} \) is: \[ \beta_{\text{new}} = \frac{3}{4} \times 0.4 \, \text{mm} = 0.3 \, \text{mm} \]

Final Answer:

The new fringe width after immersing the apparatus in the liquid is \( \boxed{0.3 \, \text{mm}} \).