A 4.0 * 102-nm thick film of kerosene (n = 1.2) is floating on water. White light is normally incident on the film. What is the visible wavelength in air that has a maximum intensity after the light is reflected? Note: the visible wavelength range is 380 nm to 750 nm.

the visible wavelength is 480 nm

Explanation:

Given data

thick film = 4.0 * 10² nm

n = 1.2

wavelength range = 380 nm to 750 nm

to find out

the visible wavelength in air

solution

we know that index of water is 1 and kerosene is 1.2

we can say that when light travel reflected path difference is = 2 n t

and for maximum intensity it will be k * wavelength

so it will be 2 n t = k * wavelength

2 * 1.2 * 4.0 * 10² = k * wavelength

wavelength = 2 * 1.2 * 4.0 * 10² / k

here k is 2 for visible

so wavelength = 2 * 1.2 * 4.0 * 10² / 2

wavelength = 480 nm

the visible wavelength is 480 nm