Archived Optics: Shift in y from a thin pane of glass, young's experiment.

AI Thread Summary
In Young's Experiment, introducing a thin parallel sheet of glass with refractive index n and thickness d over one slit alters the optical path length, resulting in a shift in the position of the mth maximum. The extra distance introduced by the glass is (n-1)d, modifying the constructive interference condition to include this term. The derived expression for the position of the maxima becomes x = mλL/δ - (n-1)Ld/δ, where λ is the wavelength, L is the distance to the screen, and δ is the slit separation. Assumptions include negligible diffraction effects from the plate's edges and minimal impact on light from the other slit. The discussion emphasizes the importance of these assumptions for accurate results in the far field regime.
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Homework Statement


With regard to Young’s Experiment, derive a general expression for the shift in the vertical position of the mth maximum as a result of placing a thin parallel sheet of glass of index n and thickness d directly over one of the slits. Identify your assumptions.


Homework Equations





The Attempt at a Solution


Sorry if this is very simple, I am having a brain fart or sorts.

I've found the number of wavelengths that will travel though a plane of thickness d:
N = dn/lamda_0
and with that tried to form a triangle to find the difference, delta y, however the answer's units don't make sense.

I know this should be of a similar form to

delta y = a/s * lambda

Thanks for the help.
 
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The condition for constructive interference in a standard Young's slit setup is that ##m\lambda = x\delta/L##, where x is the distance across the screen, L is the distance to the screen, ##\delta## is the slit separation, and m is an integer. The derivation of this is that the extra distance from the slit on the -x side of the axis to the point on the screen must be a whole number of wavelengths.

If we introduce a plate of refractive index n and thickness d in front of the slit on the -x side then we increase the optical path length in the volume occupied by the plate from d to nd. So the optical path distance (and hence the extra distance) from this slit to the screen increases by (n-1)d. So the criterion for constructive interference becomes ##m\lambda = x\delta/L+(n-1)d##. It's trivial to rearrange this to get the positions of the maxima: ##x=m\lambda L/\delta -(n-1)Ld/\delta##. The last term is the offset.

I am assuming that diffraction effects from the edge of the plate are negligible, and that the plate does not affect light from the other slit. I am also assuming that refraction through the plate is negligible and that distance through the plate doesn't change significantly across the interference pattern. The latter two are probably justifiable in the far field regime. The former two would depend on d being small - of similar order to ##\delta## or smaller, I suspect.
 
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