Question

A thin film with n=1.32 is surrounded by air. What is the minimum thickness of this film such that the reflection of normally incident light with λ=500nm is minimized?

a. 151.5 nm
b. 265 nm
c. 303 nm
d. 378 nm

Answer 1

To minimize the reflection of light, we need destructive interference. This can be achieved by having a film thickness equal to an odd multiple of half the wavelength. By applying the formula, we find that the minimum thickness is 151.5 nm.

Step-by-step explanation:

To minimize the reflection of normally incident light with a wavelength of 500 nm, we need to have destructive interference between the reflected waves. In order for destructive interference to occur, the path length difference between the two reflected waves must be equal to an odd multiple of half the wavelength. This can be expressed as:

2•nt = λ/2

Where n is the index of refraction of the film, t is the thickness of the film, and λ is the wavelength of the incident light in the film. Rearranging the equation, we get:

t = (m•λ)/(4n)

Where m is an odd integer. Plugging in the given values, we get:

t = (1•500•10-9)/(4•1.32) = 151.5 nm

Therefore, the minimum thickness of the film required to minimize the reflection of light with a wavelength of 500 nm is 151.5 nm, which corresponds to option a.