Question

A thin film of MgF₂ (refractive index = 1.38) is deposited on a glass so that it is antireflecting at a wavelength of 580 nm under normal incidence. What wavelength is minimally reflected when the light is incident at 45°?

Answer 1

To find the wavelength of light that is minimally reflected when incident at a 45° angle, use the formula for path length difference in thin film interference.

Step-by-step explanation:

In order to determine the wavelength of light that is minimally reflected when incident at a 45° angle, we can use the formula for the path length difference in a thin film interference:

2nt cos(θ) = mλ

where n is the refractive index of the film, t is the thickness of the film, θ is the angle of incidence, m is the order of the interference, and λ is the wavelength of light. We can rearrange this formula to solve for λ:

λ = 2nt cos(θ) / m

Given that the refractive index of the film is 1.38, the incident angle is 45°, and the wavelength of light that is minimally reflected under normal incidence is 580 nm, we can substitute these values into the formula to find the wavelength of light that is minimally reflected at a 45° angle.