# Interference in Thin Films?

1. Apr 22, 2008

### premed

1. The problem statement, all variables and given/known data
Light of wavelength $$\lambda$$ strikes a pane of glass of thickness T and refractice index n, as shown in figure(I attached a diagram I made). Part of the beam is reflected off the upper surface of the glass, and part is transmitted and then reflected off the lower surface of the glass. Destructive interference between these two beams will occur if :
a) T = $$\lambda$$ / 2
b) 2T = $$\lambda$$ / 2
c) T = $$\lambda$$ / 2n
d) 2T = $$\lambda$$ / 2n

2. Relevant equations

2T= m$$\lambda$$

3. The attempt at a solution
According to this equation, destructive interference occurs when 2T = $$\lambda$$ /2. Right? Since $$\lambda$$ = $$\lambda$$ $$_{0}$$ / n, then 2T= $$\lambda$$ / 2n right? The correct answer is C though. I got answer D. What am i doing wrong. Thanks

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2. Apr 22, 2008

### alphysicist

Hi premed,

In section 3 you have remarked that destructive interference occurs when $2 T = \lambda/2$. How did you find this? In your relevant equations section, you had already stated that the conditions for destructive interference was given by $2 T = m \lambda$.

In that equation, m is an integer (with appropriate restrictions) and $\lambda$ is the wavelength in the film; I don't think you can get $2 T = \lambda/2$ from that unless m=1/2 which is not allowed.

3. Apr 23, 2008

### premed

Sorry. I'll clarify. The book gives the equation for constructice interfence as 2T=m$$\lambda$$ where m is an integer. For destructice interference 2T = (m+1/2)$$\lambda$$ . thats why i put in 2T = $$\lambda$$ /2. The m in this case doesn't matter as long as 2T is not an integral multiple of $$\lambda$$ because it is destructive interference. I set m = 0 so thats how I got 2T = $$\lambda$$ /2. Does that make sense?

4. Apr 23, 2008

### alphysicist

Hi premed,

With thin film interference you have two things you have to keep track of: the effects of the path length difference of the two light rays, and the effects of reflection on the phase of the light rays.

When a light ray reflects off a material with a higher index of refraction, it experiences a phase reversal. Whether $2T = m\lambda_n$ (where $\lambda_n$ is the wavelength in the film) is the constructive or destructive condition depends on how many phase reversals there are in the problem (0, 1 or 2).

So how many phase reversals occur? What does that by itself do to the relative phase of the light rays? Once you have that information, then the condition

$$2 T = m \lambda_n$$

will leave their relative phase (from the reflection effects) unchanged, and

$$2 T = (m+\frac{1}{2}) \lambda_n$$

moves them out of phase an additional 180 degrees.