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: \( 0.33 \, \text{mm} \).