This is a glass cylinder of radius R and index of refraction N. The distance h i

Question

This is a glass cylinder of radius R and index of refraction N.

The distance h is how far the laser beam was from hitting the side of the cylinder dead on.

The distance is how far the actual exit path of the laser beam was from if it had gone straight through.

Where would I begin in getting as a function of h?

Please post anything you can think of that could be of relevance to the solution of this problem.

Any helpful hints will be regarded with extreme appreciation, and a lifesaver rating.

Thank you.

Explanation / Answer

Here is a little help. You have the angle theta. The slope of the circle at (x, h) = h/x. x^2 + y^2 = R^2 x^2 = R^2 - h^2 the normal to the circle is the negative reciprocal of the slope. The difference between the normal to the circle at that point and zero (laser horizontal) is the angle of incidence. Then you can use Snell's law to find the angle inside the glass. n1 sin incidence = n2 sin refracted. Then, if the drawing is correct, the slope of the glass is infinite where the beam comes out, so the normal is horizontal, the legs of the triangle are both radii, so it's an isosceles triangle and the angle you calculated is the same here. Then it is just Snell's law again to find the angle. (If not, you will need to find where the beam is incident on the cylinder and find the slope there) Once you are back in air it is this angle that is the difference from your horizontal laser beam (the delta in your figure) If you really want delta = distance, you don't need Snell's law the second time, you would find the y value where the beam is incident on the cylinder from the inside and subtract from h. viel Glueck, und ich versuche die andere Frage morgen zu antworten.

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