White light, with a uniform intensity across the visible wavelength 400 - 690 nm, is perpendicularly incident on a water film, having index of refraction n2 = 1.33 and a thickness of 320 nm, that is suspended in air. At what wavelength is the light reflected by the film brightest to an observer? Leave your final answer in nanometers.

Wavelength at which the light reflected by the film is brightest = 567.5 nm

Explanation:

We are given;

index of refraction n2 = 1.33

Thickness; (t) = 320 nm

Now the wavelength at which the light reflected by the film is brightest is gotten from the formula for path difference in critical interference as;

Path difference = (m + ½) (λ/n)

Where;

path difference = 2 x thickness = 2 (320) = 640 nm

λ = Wavelength at which the light reflected by the film is brightest

n is Refractive index

m is an integer = 0,1,2,3 ...

Thus; at m = 0;

We have;

640 = (0 + ½) (λ/1.33)

640 = (λ/2.66)

λ = 640 x 2.66

λ = 1702.4 nm

at m = 1;

We have;

640 = (1 + ½) (λ/1.33)

640 = (3/2) (λ/1.33)

λ = 640 x 1.33 x 2/3

λ = 567.5 nm

at m = 2;

We have;

640 = (2 + ½) (λ/1.33)

640 = (5/2) (λ/1.33)

λ = 640 x 2 x 1.33/5

λ = 340.5 nm

Since we are told that the wavelength is between 400 - 690 nm.

Thus, the wavelength at which the light reflected by the film is brightest is the higher value gotten that is between 400nm and 690nm.

Thus, Wavelength at which the light reflected by the film is brightest = 567.5 nm