Question

Young's double slit interference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm. The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm. The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m, will be:

• A. \( 0.33 \, \text{mm} \)
• B. \( 0.23 \, \text{mm} \)
• C. \( 0.46 \, \text{mm} \)
• D. \( 0.63 \, \text{mm} \)

Hint:

In a Young's double slit experiment, the fringe width depends on the wavelength of light, the distance between the slits, and the distance between the screen and the slits. Remember to adjust the wavelength according to the refractive index when in a medium other than air.

Answer 1

The Correct Option is A

The fringe width \( \beta \) in a Young's double slit experiment is given by: \[ \beta = \frac{\lambda D}{d}, \] where: - \( \lambda \) is the wavelength of light in the medium, - \( D \) is the distance between the screen and the slits, - \( d \) is the separation between the slits. The wavelength in the liquid is: \[ \lambda' = \frac{\lambda}{n}, \] where \( n = 1.44 \) is the refractive index of the liquid. Substituting the values: \[ \lambda' = \frac{690 \times 10^{-9}}{1.44} = 479.17 \, \text{nm}. \] Now, the fringe width is: \[ \beta = \frac{479.17 \times 10^{-9} \times 0.72}{1.5 \times 10^{-3}} = 0.33 \, \text{mm}. \] Final Answer: \( 0.33 \, \text{mm} \).

Answer 2

Step 1: Understand the problem setup.
We are given a Young's double slit interference apparatus immersed in a liquid of refractive index 1.44. The slit separation is \( d = 1.5 \, \text{mm} \), and the wavelength of light in air is \( \lambda_{\text{air}} = 690 \, \text{nm} \). The screen is placed at a distance of \( L = 0.72 \, \text{m} \) behind the slits. We need to find the fringe width on the screen.

Step 2: Determine the effective wavelength of light in the liquid.
The wavelength of light in a medium is given by: \[ \lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{n}, \] where \( n \) is the refractive index of the medium. In this case, the refractive index of the liquid is \( n = 1.44 \), so the wavelength of light in the liquid is: \[ \lambda_{\text{liquid}} = \frac{690 \, \text{nm}}{1.44} = 479.17 \, \text{nm}. \]

Step 3: Use the formula for fringe width.
The fringe width \( \beta \) is given by the formula: \[ \beta = \frac{\lambda_{\text{liquid}} L}{d}, \] where: - \( \lambda_{\text{liquid}} \) is the wavelength of light in the liquid, - \( L \) is the distance between the screen and the slits, - \( d \) is the slit separation.
Substituting the known values: \[ \beta = \frac{479.17 \times 10^{-9} \, \text{m} \times 0.72 \, \text{m}}{1.5 \times 10^{-3} \, \text{m}} = 0.33 \, \text{mm}. \]

Final Answer:
The fringe width is \( 0.33 \, \text{mm} \).