Question

In a YDSE fringes are observed by using light of wavelength 4800 Å, if a glass plate (μ = 1.5) is introduced in the path of one of the wave and another plates is introduced in the path of the (μ = 1.8) other wave. The central fringe takes the position of fifth bright fringe. The thickness of plate will be

• A. 8 micron
• B. 80 micron
• C. 0.8 micron
• D. None of these

Answer 1

Answer: (A)

8 micron

Answer 2

Shift due to the first plate $x_{1} = \frac{\beta}{\lambda}(\mu_{1} - 1)t$ (Upward)

and shift due to the second $x_{2} = \frac{\beta}{\lambda}(\mu_{2} - 1)t$ (Downward)

Hence net shift = x₂ – x₁$= \frac{\beta}{\lambda}(\mu_{2} - \mu_{1})t$

⇒$5p = \frac{\beta}{\lambda}(1.8 - 1.5)t$

⇒ $t = \frac{5\lambda}{0.3} = \frac{5 \times 4800 \times 10^{- 10}}{0.3} = 8 \times 10^{- 6}m = 8micron$.