Thin Film Interference Problem: A hair is placed at the edge .

  • #1
Thin Film Interference Problem: A hair is placed at the edge.....

Homework Statement



Thin Film Interference Problem: A hair is placed at the edge between two 10.0 cm glass plates of n= 1.52.
Light of 400 nm wavelength shines on the glass and an interference pattern is formed. If the separation between the second and fourth dark fringe is measured to be 0.800 mm, find the diameter of the hair.



Homework Equations




I think I'm supposed to use the equation 2*d = m*L_air / n_glass, where L is the wavelength , but I don't know how to solve for m, what to take from the rest of the problem.


The Attempt at a Solution



I'm having a hard time visualizing the problem and I don't know how to find m. I thought about maybe m is the length of a glass plate divided by the length of one fringe which is the distance from the second to the fourth fringe divided by 3?

Any help is appreciated.
 

Answers and Replies

  • #2
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41
m is related to the fringe number
 
  • #3
how do I find the fringe number? When I look at problems similar to this they are given a total number of fringes, like 171 dark fringes. How do I find this total number of fringes?
 
  • #4
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41
Find the thickness between the plates where the 1st dark fringe occurs.
You know the distance between fringes, .800/2
Find the angle
inish the problem
 
  • #5
Ohh, thank you very much for your help, barryj!
 
  • #6
Still very confused.

I have a question about the distance given, the distance between the second and fourth dark fringes.

The solution given in class today was

0.0008 m= ( 2*3-1)* (400* 10^-9 m * .1 m) / 4*d

rearrange and solved for d to get the diameter of the hair. I know this problem is simple, but I don't understand what's going on.

the equation used to solve for d is the equation for the location of a bright fringe

x_m = (2*m-1) * Lambda* l/ 4d with

m=3
l= .1m
Lambda= 400 *10^-9 m.

Isn't x_m supposed to be the distance/position wrt to the m=0 fringe?
Why isn't x_m for x_3= .0008m (the distance from the 2nd to the 4th dark fringe) + .0008m/2 ?
Why did they use the bright fringe equation with m=3?

I asked my TA, but I still don't understand. I can't go any further until I do.
 
  • #7
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41
I would think about it this way.

The distance between adjacent dark lines will be .0008/2 = .0004 metersThese is a dark line where the glass touches because there is a 180 shift at one glass surface and no shift at the other. Adding gives a dark line. Now, the next dark line will be when 2 X (gap thickness) = 1 wavelength of light at 400E-9 meters. So I figure that the angle of the glass, in radians, will be 200E-9/.0004. Now if the slass is 0.1 meters long, the thickness at 0.1 meters shoulod be .. recalling d = r X theta = 0.1 X 200E-9/.0004 = 5E-5 meters
 
  • #8
764
41
BTW, I don't think the index of refraction of the glass matters because the interference is due to the path delay in the air, not the glass.
 
  • #9
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Let me clarify. At a distance of .0004 meters from where the glass plates touch will be where the second dark line occurs. The first line will be where the glass touches. At that point, the 2 X (gap thickness) must be a wavelength of the light. This means the thickness must be 200E-9 nm. So, the angle, in radians will be the thickness of the gap divided by .0004 = 200E-9/.0004 = 5E-4 radians.
 

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