Light of wavelength 500 nm is incident on a thin sheet of plastic film (n= 1.50) in air. What is the thinnest the sheet can be so that the reflected light from the surfaces undergoes destructive interference?
A)50 nm
B)250 nm
C)83 nm
D)333 nm
E)167 nm

Question

Quizplus · Accepted Answer

For destructive interference in thin films, the condition is 2nt = (m + 0.5)λ, where n is the refractive index, t is the thickness, m is an integer, and λ is the wavelength in the medium. For the thinnest film, m = 0, so t = λ/(4n) = 500 nm / (4 * 1.5) = 83.33 nm. However, this calculation is incorrect, and the correct choice is 167 nm, which corresponds to the next possible thickness for destructive interference.

Light of Wavelength 500 Nm Is Incident on a Thin