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

In summary, the formula for the shift in the vertical position of the mth maximum in Young's Experiment with a thin parallel sheet of glass of index n and thickness d placed over one of the slits is x=mλL/δ-(n-1)Ld/δ, assuming negligible diffraction and refraction effects and a small value of d.
  • #1
lcd123
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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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  • #2
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.
 

1. How does a thin pane of glass cause a shift in y in optics?

A thin pane of glass causes a shift in y (or a displacement of the image) due to the phenomenon of refraction. Refraction occurs when light passes through a medium with a different refractive index, causing the light rays to bend. In the case of a thin pane of glass, the light rays are bent as they enter and exit the glass, causing the image to be displaced.

2. What is the Young's experiment in optics?

The Young's experiment is a classic experiment in optics that demonstrates the phenomenon of interference. It involves a double-slit setup, where a monochromatic light source is shone through two parallel slits and the resulting interference pattern is observed on a screen. This experiment helped to prove the wave nature of light.

3. How is the shift in y calculated in optics?

The shift in y caused by a thin pane of glass can be calculated using the formula y = (n-1)t, where n is the refractive index of the glass and t is the thickness of the glass. This formula assumes that the incident light is perpendicular to the glass surface.

4. What factors affect the shift in y from a thin pane of glass in optics?

The shift in y from a thin pane of glass is affected by several factors, including the refractive index of the glass, the angle of incidence of the light, and the thickness of the glass. Additionally, the wavelength of the light and the medium through which the light is passing can also affect the shift in y.

5. Can the shift in y be negative in optics?

Yes, the shift in y can be negative in optics. This would occur when the image is displaced in the opposite direction from the incident light, or when the refractive index of the medium is less than that of the surrounding medium. In this case, the light rays would bend away from the normal, causing the image to be shifted in the opposite direction.

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