Question

The width of the fringes obtained in the Young's double slit experiment is 2.6 mm when light of wavelength 6000$^\circ$A is used. If the whole apparatus is immersed in a liquid of refractive index 1.3 the new fringe width will be:

• A. 2.6 mm
• B. 5.2 mm
• C. 2 mm
• D. 4 mm

Hint:

When the apparatus is immersed in a medium with refractive index \( n \), the fringe width decreases by a factor of \( n \).

Answer 1

The Correct Option is C

The fringe width in the Young's double slit experiment is given by the formula: \[ \beta = \frac{\lambda D}{d} \] Where: - \( \beta \) is the fringe width, - \( \lambda \) is the wavelength of the light, - \( D \) is the distance between the screen and the slits, - \( d \) is the separation between the slits. 
When the apparatus is immersed in a liquid of refractive index \( n \), the wavelength of the light in the liquid becomes \( \lambda' = \frac{\lambda}{n} \). 
Thus, the new fringe width \( \beta' \) will be: \[ \beta' = \frac{\lambda' D}{d} = \frac{\frac{\lambda}{n} D}{d} = \frac{\lambda D}{n d} = \frac{\beta}{n} \] Given: 
- Initial fringe width \( \beta = 2.6 \, \text{mm} \), 
- Refractive index \( n = 1.3 \). 
Now, calculating the new fringe width: \[ \beta' = \frac{2.6}{1.3} = 2 \, \text{mm} \] Thus, the new fringe width is \( 2 \, \text{mm} \).