Double slit experiment with a glass-covered slit of unknown n

[Nerrad]

Homework Statement

A double-slit experiment uses a helium-neon laser with a wavelength of 633 nm and a slit separation of 12mm. When a thin sheet of glass is placed in front of one of the slits, the interference pattern shifts by 5 fringes. When the experiment is repeated under water, the shift is only 3 fringes. If the the refractive index for the water-air interface is 1.33, find:
a) the thickness of the glass sheet
b) its refractive index

Homework Equations

None

The Attempt at a Solution

I understand that the light beam is being diffracted through the sheet of glass, thus having to travel a longer path and that's why the interference pattern is shifted, but I don't know how to formulate the whole problem. Help?

 

[Charles Link]

One hint I can give you is that without the glass in front of one slit, straight ahead the two sources are in phase. The statement "shifts by 5 fringes" tells you straight ahead the two sources are now 5 wavelengths out of phase because of the extra path through the glass (the glass has higher index than air, accounting for a phase difference from that of the path with no glass). The part where it says the same experiment was performed under water, is somewhat unclear to me=I think they are simply referring to the slit and the glass being put in a container with water with the screen where the pattern is observed is in air, with basically a transparent container with water holding the slit with the glass in front of it. In this case, the path without glass contains water, (instead of air), so the phase difference isn't nearly as much, and results in 3 wavelengths of shift. Try writing out equations for the phase shift in both cases as a function of the index $n$ of the glass and thickness $d$ of the glass. $\\$ Editing: The second part can also be worked with the entire apparatus under water=perhaps your instructor can clarify the complete details.

 

[Nerrad]

One hint I can give you is that without the glass in front of one slit, straight ahead the two sources are in phase. The statement "shifts by 5 fringes" tells you straight ahead the two sources are now 5 wavelengths out of phase because of the extra path through the glass (the glass has higher index than air, accounting for a phase difference from that of the path with no glass). The part where it says the same experiment was performed under water, is somewhat unclear to me=I think they are simply referring to the slit and the glass being put in a container with water with the screen where the pattern is observed is in air, with basically a transparent container with water holding the slit with the glass in front of it. In this case, the path without glass contains water, (instead of air), so the phase difference isn't nearly as much, and results in 3 wavelengths of shift. Try writing out equations for the phase shift in both cases as a function of the index $n$ of the glass and thickness $d$ of the glass. $\\$ Editing: The second part can also be worked with the entire apparatus under water=perhaps your instructor can clarify the complete details.

So for the first bit you're saying that the thickness of the mirror is equal to 5 times the wavelength of the light?

 

[Charles Link]

So for the first bit you're saying that the thickness of the mirror is equal to 5 times the wavelength of the light?

It's a thin glass platelet that is only in front of one slit. In optics, there is a concept called optical path length (OPL). The OPL through the glass is $nd$ and the OPL through the air for the adjacent slit over the same distance is $1d$ since the index of air is "1.0". Thereby, coming from the laser to the front of the slit, the two OPL's differ by $(n-1)d$, since there is no phase difference when the beam is traveling from the laser through the air, until one portion of the beam encounters glass. I gave you the left side of the first equation=you should be able to write what $(n-1)d$ must be equal to (in regards to the interference between the two slits). Once you do that, you should also be able to write out the (second) equation with water as the alternative material (rather than air).