What wavelength monochromatic source in the visible region (390 to 710 nm)can be used to constructively reflect off a soap film (n = 1.46)if the film is 78.0 nm thick?
A)467 nm
B)386 nm
C)456 nm
D)404 nm
E)499 nm

Question

Quizplus · Accepted Answer

The condition for constructive interference due to reflection from a soap film is given by:

2nt = mλ

Where,
t = thickness of the film = 78.0 nm
n = refractive index of the soap film = 1.46
m = order of interference (m=1 for first order)
λ = wavelength of light

Rearranging the equation, we get:

λ = 2nt/m

Substituting the given values:

λ = 2 x 1.46 x 78.0 x 10^(-9) / 1

λ = 0.227 nm

However, this is the wavelength in vacuum. To find the wavelength in the soap film, we need to divide by the refractive index of the soap film:

λ (in soap film) = λ (in vacuum) / n

λ (in soap film) = 456 nm

Therefore, the monochromatic wavelength that can be used to constructively reflect off a soap film of thickness 78.0 nm is 456 nm.

What Wavelength Monochromatic Source in the Visible Region (390 to 710