# Homework Help: Optics: Interference and Phase

1. Apr 6, 2013

### pious&peevish

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

Two flat slides of glass are seperated at one edge by a thin wire, as shown below. The top surface of the upper slide and the bottom surface of the lower slide have special coatings on them so that they reflect no light. The system is illuminated with light of wavelength 534. nm. Looking down from above you see 28 interference minima.

a) What is the diameter of the wire? Assume that the last minima occurs at the right edge where the wire is placed.

2. Relevant equations

See below (also see attached image)

3. The attempt at a solution

Seems simple enough at first glance... combining the knowledge that there is a phase inversion with the fact that we're dealing with minima, I used the formula 2*(thickness) = (number of minima) * (wavelength). But I'm not sure how to incorporate the additional information that the coating is non reflective... I know this means the reflected waves emerge 1/2 cycle out of phase. Now I'm just confused...

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2. Apr 6, 2013

### barryj

I think what they are saying is that you do not consider the non-reflecting surfaces in the problem. The only surfaces to consider in this problem are the bottom surface of the top slide and the top surface of the bottom slide.

3. Apr 6, 2013

### pious&peevish

Thanks! But when I plugged in the values into the equation 2*(thickness) = (number of minima) * (wavelength), I didn't get the right answer... there must be something I'm still missing, but I don't know what it is.

4. Apr 6, 2013

### barryj

Hmmm.. Are you remembering that there is a 180 deg phase shift at one of the surfaces, at the second one. Remember if the index of refraction goes from low to a higher value, there is a 180 phase shift as I recall. Hig to a lower is no shift. Maybe this is the difference.

5. Apr 6, 2013

### barryj

See if you are off by 1/2 wavelength.

6. Apr 6, 2013

### pious&peevish

OK. So just to confirm, if I wrote that 2*(thickness) = [(number of minima) * (wavelength)] + [(1/2)*(wavelength)], would that be right?

7. Apr 6, 2013

### barryj

I have not worked the problem but I would think this might be the error. It might be plus or minus 1/2 wavelength.

8. Apr 7, 2013

### pious&peevish

Hmm... I tried that and it didn't work either. Is there something fundamental I'm not grasping? That's what I'm most worried about now.

9. Apr 7, 2013

### barryj

deleted

Last edited: Apr 7, 2013
10. Apr 7, 2013

### barryj

Correction correction correction because of the 180d phase shift, the first dark band will be at
The first minima will be where 2t = 0 w w= wavelength
2nd minima will be when 2t = 1w
3rd minima will be when 2t = 2w
4th minima will be when 2t =3w

so the 28th minima will be when 2t = 27w

w = 534E-9 so I would expect the 28th minima to be where t = 7209 nm

yes-no?

Last edited: Apr 7, 2013
11. Apr 7, 2013

### pious&peevish

Thanks a ton - that was right!