Question

A transparent dielectric coating is applied to glass ($\epsilon_r = 4, \mu_r = 1$) to eliminate the reflection of red light ($\lambda_0 = 0.75 \mu m$). The minimum thickness of the dielectric coating, in $\mu$m, that can be used is ___________ (rounded off to two decimal places).

Hint:

For a standard single-layer anti-reflection (AR) coating between two media (e.g., air and glass), remember the two ideal conditions: $n_{coating} = \sqrt{n_{air} n_{glass}}$ and thickness $d = \lambda_0 / (4n_{coating})$. This is called a quarter-wave transformer.

Answer 1

Correct Answer: 0.12

This question from the official GATE 2023 paper is flawed as it does not specify the refractive index of the coating material, and was marked as 'Marks to All'. To arrive at a solution, we must make a standard assumption for an ideal single-layer anti-reflection coating.
The standard design for a single-layer anti-reflection coating requires two conditions:
1. Amplitude Condition: The refractive index of the coating ($n_c$) should be the geometric mean of the refractive indices of the surrounding media. 2. Phase Condition: The thickness of the coating should be a quarter of the wavelength of light within the coating material ($d = \lambda_c / 4$).
Step 1: Calculate the refractive indices.
Refractive index of air (assumed), $n_{air} \approx 1$.
Refractive index of glass, $n_{glass} = \sqrt{\epsilon_r \mu_r} = \sqrt{4 \times 1} = 2$.
Step 2: Apply the amplitude condition to find the ideal refractive index of the coating.
$n_c = \sqrt{n_{air} \times n_{glass}} = \sqrt{1 \times 2} = \sqrt{2} \approx 1.414$.
Step 3: Apply the phase condition to find the minimum thickness.
The thickness should be an odd multiple of a quarter-wavelength in the coating material for destructive interference (assuming $n_{air}<n_c<n_{glass}$). $d = (2m+1) \frac{\lambda_c}{4}$, where $m=0, 1, 2, ...$ The wavelength in the coating is $\lambda_c = \frac{\lambda_0}{n_c}$.
$d = (2m+1) \frac{\lambda_0}{4n_c}$.
The minimum thickness occurs for $m=0$:
$d_{min} = \frac{\lambda_0}{4n_c}$.
Step 4: Substitute the values.
Given $\lambda_0 = 0.75 \mu$m. $d_{min} = \frac{0.75 \mu m}{4 \times \sqrt{2}} \approx \frac{0.75}{5.657} \approx 0.13258 \mu$m.
Rounding off to two decimal places, the minimum thickness is 0.13 $\mu$m.