Double-slit experiment, one slit covered with glass

[cep]

Homework Statement

In a double-slit experiment, He-Ne laser light of wavelength 633 nm produced an interference pattern on a screen placed at some distance from the slits. When one of the slits was covered with a thin glass slide of thickness 12.0 um, the central fringe shifted to the point occupied earlier by the 10th dark fringe. What is the refractive index of the glass slide?

Here's a link to a similar problem with a figure: www.physics.ohio-state.edu/~gohlke/pedagogy/Phys133I_Diffraction.pdf[/URL]

[h2]Homework Equations[/h2]

n=c/v; $\Delta$x (constructive) = m$\lambda$ = dsin$\Theta$; $\Delta$x (destructive) = (m+1/2)$\lambda$ = dsin$\Theta$; Snell's law (maybe?)

[h2]The Attempt at a Solution[/h2]

I understand that the glass slide will slow the passage of light, thus effectively increasing $\Delta$x between the two light paths. I guess i need to figure out how many wavelengths the light going through the glass is "behind" the other beam. However, I don't understand how to incorporate this into the problem. It seems to get very complicated very quickly-- since the light isn't traveling straight through the glass, the distance is not equal to the thickness of the glass. Then, the light is refracted when leaving the glass, according to Snell's law. The only examples in my textbook were very simplistic, we didn't cover anything like this in class, and I'm really not sure what to do. I tried working out the problem as indicated in the link (though they solve for the thickness of the slide, and n is given), but got like -0.5 for n (and I was suspicious of their answer anyway, because it seems to neglect so many complicating factors). Can anyone help me think about this? Thanks a lot, sorry if this post is tl;dr :).

-CEP

 

[vela]

It's easiest to think of these problems in terms of optical path length, which is just the number of wavelengths between two points. The interference pattern is caused by light reaching a point along two different paths. For the central fringe, by definition, the optical path lengths are equal. For the first-order fringe, the difference in optical path length is ±1, and so on.

Come up with expressions for the optical path length for the two rays when the slide is present. Then set them equal to each other because you're working with the central fringe. The only unknown left should be n.

 

[PeterO]

Homework Statement

In a double-slit experiment, He-Ne laser light of wavelength 633 nm produced an interference pattern on a screen placed at some distance from the slits. When one of the slits was covered with a thin glass slide of thickness 12.0 um, the central fringe shifted to the point occupied earlier by the 10th dark fringe. What is the refractive index of the glass slide?

Here's a link to a similar problem with a figure: www.physics.ohio-state.edu/~gohlke/pedagogy/Phys133I_Diffraction.pdf[/URL]

[h2]Homework Equations[/h2]

n=c/v; $\Delta$x (constructive) = m$\lambda$ = dsin$\Theta$; $\Delta$x (destructive) = (m+1/2)$\lambda$ = dsin$\Theta$; Snell's law (maybe?)

[h2]The Attempt at a Solution[/h2]

I understand that the glass slide will slow the passage of light, thus effectively increasing $\Delta$x between the two light paths. I guess i need to figure out how many wavelengths the light going through the glass is "behind" the other beam. However, I don't understand how to incorporate this into the problem. It seems to get very complicated very quickly-- since the light isn't traveling straight through the glass, the distance is not equal to the thickness of the glass. Then, the light is refracted when leaving the glass, according to Snell's law. The only examples in my textbook were very simplistic, we didn't cover anything like this in class, and I'm really not sure what to do. I tried working out the problem as indicated in the link (though they solve for the thickness of the slide, and n is given), but got like -0.5 for n (and I was suspicious of their answer anyway, because it seems to neglect so many complicating factors). Can anyone help me think about this? Thanks a lot, sorry if this post is tl;dr :).

-CEP[/QUOTE]

The problem as stated in the attachment is different to what you have written here?

Problem 1. (Knight, Chapter 22, problem 67.) A double-slit experiment is set up using a helium-neon laser ( = 633nm). Then, a very thin piece of glass (n = 1.5) is placed over one of the slits. Afterward, the central point on the screen is occupied by what had been the m = 10 dark fringe. How thick is the glass?

This does not say the central fringe has moved to where the 10th dark fringe was, it says that the central point is occupied by what had been the 10th dark fringe.

The 10th dark fringe occurred where the path difference was 10.5 wavelengths.

The physical path difference at the central point is zero. The addition of the glass merely increases the optical path length by 10.5 wavelengths.

n = 1.5 for this glass, so the wavelength in the glass is 2/3 the wavelength in air.

I will let you deal with the real wavelengths, but let's use simple numbers for example

If the wavelength in air was 3 mm, so the wavelength in glass was 2 mm, and the piece of glass was 12 mm thick - those 12 mm would represent 6 wavelengths for light in the glass, but only 4 wavelengths for light in the air.

Thus, even though the light goes straight through - a path difference of 2 wavelengths is created.

Your task now is to use the nanometre wavelength given, and the requirement for a 10.5 wavelength path difference to generate the real thickness of this glass.
Note: the printed answer on the attachment [yours? someone else's?] has the mis-interpretation also, as shown by the diagram showing the central fringe moved.
EDIT: the m = 10 dark fringe should be a path difference of 9.5 wavelengths I think - my error.

 

[cep]

Also, Peter, my question is different as I am supposed to determine the refractive index (n) of the glass-- not the thickness of the slide. Thank you both for the responses-- very helpful. To clarify, I shouldn't be worried about the path length THROUGH the glass, just the change in wavelength? I'll keep working on it :)

 

[PeterO]

Also, Peter, my question is different as I am supposed to determine the refractive index (n) of the glass-- not the thickness of the slide. Thank you both for the responses-- very helpful. To clarify, I shouldn't be worried about the path length THROUGH the glass, just the change in wavelength? I'll keep working on it :)

Yes that's it. The piece of glass is a certain number of "wavelengths in air" thick, but 9.5 more "wavelengths in glass" thick. That enables you to get the wavelength in glass, and so the refractive index.

 

[vela]

So is your problem the way you stated or the way the problem PeterO found is stated?

 

[cep]

The question in the original post is copied, verbatim, from my textbook. The attachment is a solution to a similar problem, with a figure identical to the one in the book.