Interference in thin films

samdiah

1. The problem statement, all variables and given/known data

Calculate the minimum thickness of a layer of manesium flouride(n=1.38) on flint glass (n=1.66) in a lens system, if light wavelength 5.50x10^2 nm in air undergoes destructive interference.

Given:
n1=1.38
n2=1.66
λ1=5.5*10^2 nm

2. Relevant equations

Δx=L(λ/2t)
n2/n1 = λ1/λ2


3. The attempt at a solution

Destructive interference is at λ/4, 3λ/4, 5λ/4 ....

So the λ in flint flass is λ2=(n1*λ)/n2
= 4.57 * 10^-7

Therefore the shortest thicknss should be λ/4
4.57*10^-7/4
1.14*10^-7

The answer however is wrong the answer is suppose to be 199 nm.

Can someone please help! Thanku

Andrew Mason

The interference is between the incident wave at the air (n1=1) and wave reflecting from the first layer (n2=1.38). Because the second layer has a higher index of refraction, the reflection at the first/second layer surface has a phase shift of [itex]\pi[/itex]. So in order to have destructive interference (a phase difference of [itex]\pi[/itex]), the path length through the first layer to the reflecting surface and back to the air has to be a full wavelength. So the thickness has to be [itex]\lambda_2/2[/itex].

The wavelength in the magnesium fluoride layer is [itex]\lambda_2 = n_1\lambda_1/n_2 = 5.5\times 10^{-7}/1.38 = 3.98\times 10^{-7}m = 398 nm[/itex]

AM

berkeman

I get 99.6nm, not 199nm. Remember to use the n of air, not the flint glass.

samdiah

Thank You!