A thin film of refractive index 1.40 covers a flat sheet of glass of refractive index 1.50.If light of wavelength 600 nm strikes the film from air perpendicular to its surface,the minimum film thickness needed to cause destructive interference in the reflected light is 150 nm.

Question

Quizplus · Accepted Answer

The minimum film thickness needed to cause destructive interference in the reflected light can be calculated using the formula:

t = (m+1/2)λ/n

where t is the film thickness, λ is the wavelength of light and n is the refractive index of the film.

Using this formula with m=0 (for minimum thickness), λ=600 nm and n=1.40, we get:

t = (0+1/2) x 600 nm/1.40 = 214.3 nm

This is greater than the given minimum thickness of 150 nm, which means that the reflected light will not undergo destructive interference at this thickness. Therefore, the answer is B (false).

A Thin Film of Refractive Index 1