Question

In Young’s double slit experiment, light of wavelength $ 5000\, \text{Å} $ is used. The slit separation is $ 3\, \text{mm} $, and the screen is $ 20\, \text{cm} $ away. If a $ 1\, \text{mm} $ transparent plate is introduced in front of one slit, and the fringe pattern shifts by $ 6\, \text{mm} $, then the refractive index of the plate is:

• A. 1.08
• B. 1.09
• C. 1.1
• D. 1.2

Hint:

Use formula \( x = \frac{t(\mu - 1)D}{d} \) to calculate fringe shift due to transparent material.

Answer 1

The Correct Option is B

Fringe shift due to introduction of transparent plate: \[ x = \frac{t(\mu - 1)D}{d} \] Where: - \( x = 6\, \text{mm} \) - \( t = 1\, \text{mm} \) - \( D = 20\, \text{cm} = 200\, \text{mm} \) - \( d = 3\, \text{mm} \) Substitute: \[ 6 = \frac{1(\mu - 1) \cdot 200}{3} \Rightarrow 6 \cdot 3 = 200(\mu - 1) \Rightarrow 18 = 200(\mu - 1) \Rightarrow \mu - 1 = \frac{18}{200} = 0.09 \Rightarrow \mu = 1.09 \]