# Double-Slit+ Glass Refraction

1. Jul 24, 2014

### alingy1

A double-slit experiment is set up using a helium-neon laser (l = 633 nm). 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 = 10 dark fringe. How thick is the glass?

I have the solution's attached.

My question is: why does the textbook not use snell's law? Is it making the simplistic argument that light traverses the glass piece perpendicularly? Would the problem be somehow changed otherwise?

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2. Jul 24, 2014

### Simon Bridge

Basically the text does not use Snells law because it has no effect on the outcome (well, any effect is smaller than the approximations already being used).

Recall that the two sides of the glass are parallel - so rays emerge at the same angle as the incident rays. The effect, therefore, is to displace the ray by a distance that depends on the thickness of the glass, making the slits appear closer together ... but you are told that the glass is very thin, so the displacement will be very small.

You can check this for yourself:
Repeat the calculation, taking snell's law into account, and see what difference it makes for glass thickness t and refractive index n. Then see what happens when you make the thickness very small compared with the other dimensions of the experiment.

The calculation already involves approximations, this one is small compared to the others.

3. Jul 24, 2014

### alingy1

Simon, could you draw me some form of diagram to help me visualize the variables I should take into account?