Question

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?

Answer 1

The thinnest the sheet can be is 8.33 * 10^-8

Step-by-step explanation:

λ = 500 nm = 500 * 10^-9

n = 1.50

The thinnest can be calculated using the formula;

2nd = (m+1/2)λ

thinnest, m = 0

2d (1.50) = (0+ 1/2) (500 x 10^-9 )

3d = 2.5 * 10^-7

d = 2.5 * 10^-7/3

d= 8.33 * 10^-8

Therefore, the thinnest sheet will be 8.33 * 10^-8

Answer 2

t = 1.6667 10⁻⁷ m

Step-by-step explanation:

When light falls on a thin film on the surface we have two phenomena

- a change in phase when the beam passes to a medium with a higher index

- on the other hand, in the wavelength, due to the refractive index of the medium

λ = λ₀ / n

The second ray reflected from the other surface has no phase change

So the path difference between the two rays for destructive interference is

2t = m λ₀ / n

t = m λ₀ / 2n

Let's calculate

t = 1 500 10⁻⁹ / 2 1.50

t = 166.67 10⁻⁹ m

t = 1.6667 10⁻⁷ m