A tank of gasoline (n = 1.40) is open to the air (n = 1.00). A thin film of liqu

Question

A tank of gasoline (n = 1.40) is open to the air (n = 1.00). A thin film of liquid floats on the gasoline and has a refractive index that is between 1.00 and 1.40. Light that has a wavelength of 626 nm (in vacuum) shines perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive interference. The thickness of the film is 298 nm and is the minimum nonzero thickness for which constructive interference can occur. What is the refractive index of the film?

Explanation / Answer

Please note that, we will get constructive interference when the optical thickness of the film is a multiple of 1/2 wavelength

This means, out and back through the film is some number of complete wavelengths.

Suppose, 'n' is the refractive index of the film.

Therefore, for the bright fringe with the longest wavelength, we should have -

(298 nm) * n = (626 nm/2)

=> n = 313 / 298 = 1.05

So, the refractive index of the film = 1.05

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.