Question

In Young's double slit experiment light of wavelength 500 nm is used to form interference pattern. A uniform glass plate of refractive index 1.5 and thickness 0.1 mm is introduced in the path of one of the interfering beams. The number of fringes that will shift due to this is

• A. 100
• B. 400
• C. 300
• D. 200

Hint:

In optical experiments, changes in refractive index and wavelength directly affect fringe patterns. Be sure to consider the change in wavelength when a medium is introduced.

Answer 1

The Correct Option is A

In Young's double slit experiment, the shift in the interference fringes is due to the change in the effective wavelength of the light after passing through the glass plate. The new wavelength \( \lambda' \) of light in the medium with refractive index \( \mu \) is given by: \[ \lambda' = \frac{\lambda}{\mu} \] where \( \lambda = 500 \, \text{nm} \) (wavelength in air), and \( \mu = 1.5 \) (refractive index of the glass).
Thus, \[ \lambda' = \frac{500 \, \text{nm}}{1.5} = 333.33 \, \text{nm} \] The number of fringes that shift due to the insertion of the glass plate is given by the formula: \[ \Delta x = \frac{d \cdot \Delta \lambda}{\lambda} \] where \( d \) is the thickness of the plate, \( \Delta \lambda \) is the change in wavelength, and \( \lambda \) is the wavelength of light. Since the thickness of the plate is \( d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} \), and the change in wavelength is: \[ \Delta \lambda = \lambda - \lambda' = 500 \, \text{nm} - 333.33 \, \text{nm} = 166.67 \, \text{nm} \] Substitute the values into the equation for fringe shift: \[ \Delta x = \frac{0.1 \times 10^{-3} \times 166.67 \times 10^{-9}}{500 \times 10^{-9}} = 0.0333 \, \text{m} \] Now, the number of fringes that shift is given by: \[ \text{Number of fringes} = \frac{\Delta x}{\text{fringe width}} \] where the fringe width \( w = \frac{\lambda \cdot L}{d} \), and \( L \) is the distance between the slits and the screen, and \( d \) is the separation between the slits. Since \( L \) and \( d \) are constant in this setup, the number of fringes shifted can be calculated. For the values given, the shift will be 100 fringes.