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

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SUMMARY

The discussion focuses on deriving the expression for the vertical shift in the mth maximum in Young's Experiment when a thin parallel sheet of glass with refractive index n and thickness d is placed over one of the slits. The key formula derived is x = (mλL/δ) - ((n-1)Ld/δ), which accounts for the additional optical path length introduced by the glass. Assumptions include negligible diffraction effects and minimal impact on light from the other slit. The discussion emphasizes the importance of understanding optical path length changes in interference patterns.

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  • Understanding of Young's Experiment and interference patterns
  • Familiarity with optical path length and refractive index concepts
  • Basic knowledge of trigonometry for deriving relationships
  • Experience with wavelength calculations in optics
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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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