YDSE: Film, Medium & Two Wavelengths question from JEE Main 2014 (9 Apr, Shift 1)

Answer: (b)

1. Setup 2 (sheet of thickness $t = 1.8\ \mu\text{m}$, $\mu = 1.6$ as read from the figure) shifts the central fringe by $y = \dfrac{(\mu - 1)tD}{d}$.
2. Setup 3 (screen at $2D$): $\beta' = \dfrac{\lambda(2D)}{d}$.
3. $y = \beta' \Rightarrow \lambda = \dfrac{(\mu - 1)t}{2} = \dfrac{0.6 \times 1.8\ \mu\text{m}}{2} = 0.54\ \mu\text{m}$
4. So the answer is (b) 540 nm.