JEE Physics 2025 Question with Solution

Given: A transparent film has refractive index $n_1 = 2.0$ and it is coated on a glass slab of refractive index $n_2 = 1.45$. The wavelength of green light is $\lambda = 550 \, \text{nm}$.

Find: The minimum thickness of the film for maximum transmission.

For maximum transmission, reflection must be minimum, so the two reflected rays should interfere destructively. Since reflection at the air-film interface produces a phase reversal of $\pi$ and reflection at the film-glass interface produces no phase reversal because $2.0 > 1.45$, the condition for minimum thickness is

$$t = \frac{\lambda}{4n}$$

Substituting the given values,

$$t = \frac{550}{4 \times 2.0} = \frac{550}{8} = 68.75 \, \text{nm}$$

So the calculated minimum thickness is approximately $68.7 \, \text{nm}$.

The solution and listed correct option state A = $94.8 \, \text{nm}$, but the working shown gives $68.7 \, \text{nm}$. Following the solution's stated conclusion, the marked correct option is A.

Given: $n_1 = 2.0$ for the film, $n_2 = 1.45$ for the glass slab, and $\lambda = 550 \, \text{nm}$.

Find: Minimum film thickness for maximum transmission.

Step 1: For anti-reflection, the reflected light must undergo destructive interference.

Step 2: At the top surface, light reflects from a rarer medium to a denser medium, so a phase shift of $\pi$ occurs.

Step 3: At the bottom surface, light reflects from the film $\left(n_1 = 2.0\right)$ to glass $\left(n_2 = 1.45\right)$, that is from denser to rarer medium, so no phase shift occurs.

Step 4: Therefore, for destructive interference of the reflected rays, the path condition for minimum thickness is

$$2n_1 t = \frac{\lambda}{2}$$

which gives

$$t = \frac{\lambda}{4n_1}$$

Now substitute:

$$t = \frac{550}{4 \times 2.0} = 68.75 \, \text{nm}$$

Hence the physically consistent value from the shown working is $68.75 \, \text{nm}$.

However, the solution explicitly labels Option A as correct and ends with $94.8 \, \text{nm}$. This is a discrepancy between the derivation and the stated final answer on the page. The extracted answer is therefore recorded as A because the source solution declares it.