Question

A glass window of index of refraction 1.3 is coated with an antireflective film of thickness 2.0 µm. Which of the following indices of refraction of the film would cause the intensity of reflected light of wavelength 800 nm from a normally-incident beam to be suppressed?

A. 1.1 B. 1.2 C. 1.4 D. 1.7 E. 1.8

Answer 1

Case 1 : index of refraction of film (n) is less than that of glass

Light reflected from both surfaces of film undergoes phase change. Hence for destructive interference of reflected light, two times optical path length of light inside film (2nt , t is thickness of film) should be odd integral multiple of half the wavelength.

A) 2nt = 4400 nm = m (800 nm /2)

m = 11 hence destructive interference.

B) 2nt = 4800 nm = multiple of 800 nm

Hence constructive interference.

Case 2 : index of refraction of film is more than that of glass.

Light reflected from film glass interface does not undergo phase change. For destructive interference of reflected light, two times optical path length of light inside film must be integral multiple of wavelength.

C) 2nt = 5600 nm = integral multiple of 800nm

Hence destructive interference.

D) 2nt = 6800 nm , not integral multiple of 800

Hence not destructive interference

E) 2nt = 7200 nm = integral multiple of 800 nm

Hence destructive interference.

Hence A C and E are correct.