Double Slit Experiment glass thickness

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SUMMARY

The discussion focuses on the double-slit experiment using a helium-neon laser (λ = 633 nm) and a thin piece of glass with a refractive index of n = 1.50 placed over one of the slits. The central point on the screen, which corresponds to the m = 15.0 dark fringe, indicates a shift in the interference pattern due to the glass. To determine the thickness of the glass, one must incorporate the refractive index into the destructive interference condition, specifically how it modifies the wavelength of light in the medium.

PREREQUISITES
  • Understanding of the double-slit experiment and interference patterns.
  • Knowledge of the formula for destructive interference: d sin θ = (m - 0.5)λ.
  • Familiarity with the concept of refractive index and its effect on wavelength.
  • Basic principles of wave optics, particularly regarding phase shifts.
NEXT STEPS
  • Research how to calculate the effective wavelength in a medium with a refractive index.
  • Study the derivation of the interference formulas used in the double-slit experiment.
  • Explore the impact of phase shifts introduced by different materials on interference patterns.
  • Learn about the relationship between fringe shifts and optical path differences in wave optics.
USEFUL FOR

Students and educators in physics, particularly those studying wave optics, as well as researchers interested in experimental setups involving interference phenomena.

y_chen
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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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