# Interference in thin films

1. Dec 18, 2006

### 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.

2. Dec 18, 2006

### 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 $\pi$. So in order to have destructive interference (a phase difference of $\pi$), 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 $\lambda_2/2$.

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

AM

3. Dec 18, 2006

### Staff: Mentor

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

4. Dec 18, 2006

Thank You!