A transparent film of refractive index, 2.0 is coated on a glass slab of refractive index, 1.45. What is the…
- 137.5 nm
- 275 nm
- 94.8 nm
- 68.7 nm
Solution

For transmitted green light to be maxima, reflected green should be minima.
$\begin{aligned}
& \Delta \mathrm{P}=2 \mu_0 \mathrm{t}=\mathrm{n} \lambda \\ & \Rightarrow \mathrm{t}=\frac{\mathrm{n} \lambda}{2 \mu_0} \therefore \mathrm{t}_{\min }=\frac{\lambda}{2 \mu_0}=\frac{550}{2 \times 2}=137.5
\end{aligned}$
Asked in: JEE Main 2025 (22 Jan Shift 2)