Question

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.

Answer 1

Answer:

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