Question

An anti-reflection film coating of thickness 0.1 μm is to be deposited on a glass plate for normal incidence of light of wavelength 0.5 μm. What should be the refractive index of the film?

Hint:

For an anti-reflection film, the refractive index is calculated using the formula \( n = \frac{\lambda}{4d} \), where \( d \) is the thickness of the film and \( \lambda \) is the wavelength of light.

Answer 1

Correct Answer: 1.25

Step 1: Understanding the concept of anti-reflection coating.
For minimal reflection, the thickness of the anti-reflection coating should be \( \frac{\lambda}{4n} \), where \( \lambda \) is the wavelength of light in air, and \( n \) is the refractive index of the film. Step 2: Calculate the refractive index.
The thickness of the film is given as 0.1 μm and the wavelength of the light is 0.5 μm. Using the formula \( d = \frac{\lambda}{4n} \), solve for \( n \) to get: \[ n = \frac{\lambda}{4d} = \frac{0.5}{4 \times 0.1} = 1.25 \]