Wire radius from interference fringes

[SL_1719]

Homework Statement

The diameters of fine fibers can be accurately measured using interference patterns. Two optically flat pieces of glass that each have a length L are arranged with the fiber between them, as shown below. The setup is illuminated by monochromatic light, and the resulting interference fringes are observed. Suppose that L is 20.0 cm and that yellow sodium light (590 nm) is used for illumination. If 19 bright fringes are seen along this 20-cm distance, what are the limits on the diameter of the fiber? Hint: The nineteenth fringe might not be right at the end, but you do not see a twentieth fringe at all.

Homework Equations

Bright Fringes at distances=n*lambda. n is an integer.

The Attempt at a Solution

So I took each fringe to represent a distance of 1/2 wavelengths, as it has to reach the other side and return, and took the minimum and maximum distance to be the distances constrained by what 19 fringes dictates, and what 20 fringes dictates. For nineteen fringes I took 590nm*9.5=5.605µm (answer to be in µm), and the maximum diameter to be 590nm*10=5.9µm. These did not work, and so I tried half for radius, double in case I for some reason had found the radius, yet none of these three combinations worked. What am I doing wrong?

 

[gneill]

Light reflecting from a surface with a higher index of refraction than the medium it's in will be phase shifted by π radians (1/2 λ). So not only the distance matters.

 

[SL_1719]

Well, if that is true to every fringe, how does that effect the end result? Or is it that everything is shifted down one half wavelength?

 

[gneill]

It matters to how many fringes appear over the given distance. You are told that 19 (and no more!) fringes appear over the 20cm distance.

 

[SL_1719]

Well we are told at least 19, in that it could be at the edge, but less than 20, as in the 20th is not there so it being less than but not able to be equal to. So in what ways does the number of fringes per wavelength effect my answer? Because I was going off of what I saw in other posts in which people were looking for similar answers, yet mine didnt work.

 

[gneill]

Well we are told at least 19, in that it could be at the edge, but less than 20, as in the 20th is not there so it being less than but not able to be equal to. So in what ways does the number of fringes per wavelength effect my answer? Because I was going off of what I saw in other posts in which people were looking for similar answers, yet mine didnt work.

Suppose the diameter of the wire is d, and λ is the wavelength of the light in air.

For the final fringe, for constructive interference, the extra path length traveled by the light ray bouncing off the surface of the bottom glass will have to be one of

2d + (1/2)λ = λ, 2λ, 3λ, ...

What's an expression that that relates 2d to the wavelength for the nth fringe?

Now, you can count the fringes that appear, so you know something about the final (nth) fringe (and no more!) and how it pertains the the above expession...

 

[SL_1719]

I am trying to understand what you are saying. The wording to me is a little confusing. So you say that 2d+(1/2λ)=λ, 2λ... Since there are 19 fringes seen, then 2d+(1/2λ)=19λ, as 19 is the n. So 2d=18.5λ if the fringe is right on the edge? I see now, I realized I made a mistake as I did on an earlier problem which won't happen again. The problem I had, which in looking at the equation you gave me, is that there is a 180 degree phase shift upon reflection, and to compensate for that, there is a subtraction of half a wavelength, so instead of 2d=19λ which is what I had tried, I needed to remove the (1/2)λ. Is this reasoning correct? I want to actually understand the concept to help me not make the mistake in the future again, plus it is nice just to know.

 

[gneill]

2d+(1/2λ)=λ, 2λ...

So

2d = λ/2, λ, 3λ/2, 5λ/2,...

2d = (n + 1/2)λ for n = 0,1,2,...

Because the 19th fringe is visible you can say

(19 - 1/2)(λ/2) < d

Because the 20th fringe is not visible you can say

(20 - 1/2)(λ/2) > d

So you have upper and lower limits on the diameter of the wire.