Double Slit Experiment glass thickness

AI Thread Summary
In the double-slit experiment with a helium-neon laser, a thin glass piece with a refractive index of 1.50 is placed over one slit, causing the central point on the screen to align with the m=15.0 dark fringe. The discussion highlights the need to understand the relationship between destructive interference and the effect of the glass on the wavelength due to its refractive index. To find the glass thickness, one must modify the interference formula to account for the change in wavelength as it passes through the glass. The key is to derive a new equation that incorporates the refractive index and the shift in fringe positions. Understanding these concepts is essential for solving the problem effectively.
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A double-slit experiment is set up using a helium-neon laser (lambda =633nm) . Then a very thin piece of glass ( n=1.50 ) is placed over one of the slits. Afterward, the central point on the screen is occupied by what had been the m = 15.0 dark fringe. How thick is the glass?

I really need help on this question. I understand that for the double-slit experiment, destructive interference occurs when d sin theta= (m-0.5)lambda. However, I do not understand how to incorporate the information about the central point being occupied by what had been the m=15.0 dark fringe. Also, how does the piece of glass play a role in the question? Thanks so much for all your help!

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You have a http://groups.google.com/group/Gmail-ABCs/browse_thread/thread/1e89d941615e3d35"

As for how the glass plays a role, instead of just using some formula you will need to understand conceptually how that formula is derived (so that you can derive an appropriately modified version). Recall that the wavelength depends on the refractive index.
 
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