index of refraction Thin Films

1. The problem statement, all variables and given/known data
The diagram shows light incident from above on a film of thickness d. Each of the three materials in the figure can be chosen to be air, with index of refraction n=1.00, water, with index n=1.33, or glass, with index n=1.50. Under which of the following conditions will the reflected light be completely or nearly eliminated by interference?
(lambda2 refers to the wavelength of the light inside the thin film.)
(Note: In the case that d<<lambda2, you can assume the thickness is so small that the travel distance in material 2 has negligible affect on the phase of the wave. Treat this as if the two interfaces are almost right on top of each other.)
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1) T/F d=lambda2/4, material 1 is air, 2 is water, 3 is glass.
2) T/F d=lambda2/2, material 1 is air, 2 is water, 3 is glass.
3) T/F d=lambda2/4, material 1 is glass, 2 is air, 3 is glass.
4) T/F d=lambda2/4, material 1 is water, 2 is glass, 3 is air.
5) T/F d<<lambda2, material 1 is water, 2 is glass, 3 is air.

2. Relevant equations

2*n*d*sinθ = (m-(1/2))*λ

m=integer
λ=wavelength
n=index of refraction = speed of light in vacuum/speed of light in medium
d= thickness

3. The attempt at a solution
We are looking for places where the refracted light is eliminated by interference, so we are looking for destructive interference.
Destructive if Path-length Difference(PLD) 2nd= (m + (1/2))λ when the sources are exactly in phase.
Destructive if Path-length Difference(PLD) 2nd = mλ when the sources are exactly out of phase.
There are phase reversals when it is reflecting of a higher n.

1) d=lambda2/4, material 1 is air(n=1.00), 2 is water, 3 is glass.
True because

There would be 2 phase reversals [1 between air and water and the other between water and glass] So the wavelength would be exactly in phase which means it needs to have PLD of (1/2)λ
d=2λ/4
(2λ/4) = (1/2)λ
(2/4)λ= (1/2)λ
(1/2)λ= (1/2)λ

2) d=lambda2/2, material 1 is air(n=1.00), 2 is water(n=1.33), 3 is glass(n=1.50).
False because

There would be 2 phase reversals [1 between air and water and the other between water and glass] So the wavelength would be exactly in phase which means it needs to have PLD of (1/2)λ
d=2λ/2
2λ/2 = (1/2)λ
(2/2)λ = (1/2)λ

3) d=lambda2/4, material 1 is glass(n=1.50), 2 is air(n=1.00), 3 is glass(n=1.50).
True because

There would be 2 phase reversals [1 between air and glass and the other between air and glass] So the wavelength would be exactly in phase which means it needs to have PLD of (1/2)λ.
d=2λ/4
(2λ/4) = (1/2)λ
(2/4)λ= (1/2)λ
(1/2)λ= (1/2)λ

4) d=lambda2/4, material 1 is water(n=1.33), 2 is glass(n=1.50), 3 is air(n=1.00).
True because

There would be 2 phase reversals [1 between air and water and the other between the water and glass] So the wavelength would be exactly in phase which means it needs to have PLD of (1/2)λ.
d=2λ/4
(2λ/4) = (1/2)λ
(2/4)λ= (1/2)λ
(1/2)λ= (1/2)λ

5) d<<lambda2, material 1 is water(n=1.33), 2 is glass(n=1.50), 3 is air(n=1.00).
False because

There would be 2 phase reversals [1 between air and water and the other between the water and glass] So the wavelength would be exactly in phase which means it needs to have PLD of (1/2)λ.
d<<lambda2
λ2= (1/2)λ

I thought i was doing right but the answer is incorrect.
Any help would be greatly appreciated.

Attached Images
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