25. Three sheets of plastic have unknown indices of refraction. Sheet 1 is place

Question

25. Three sheets of plastic have unknown indices of refraction. Sheet 1 is placed on top of sheet 2, and a laser beam is directed onto the sheets from above so that it strikes the interface at an angle of 26.5 degrees with the normal. The refracted beam in sheet 2 makes an angle of 31.7 degrees withn the normal. The experiment is repeated with sheet 3 on top of sheet 2, and with the same angle of incidence the refracted beam makes an angle of 36.7 degrees with the normal. If the experiment is repeated again with sheet 1 on top of sheet 3, what is the expected angle of refraction in sheet 3? Assume the same angle of incidence.

Explanation / Answer

to solve this problem, we make successive use of the equation describing the law of refraction in the first experiment, where sheet 1 is on sheet 2, we write n1 sin x1 = n2 sin x2 where n1, n2 are the indices of refraction of sheets 1 and 2, and I use x to represent angles so in the first expt, we have: n1 sin 26 = n2 sin 31.7 plugging in values gives you the relationship: n1=1.19 n2 (eq 1) for the second expt, we have layer 3 on layer 2, so we have n3 sin x3 = n2 sin x2 or n3 sin 26 = n2 sin 36.2 which yields n3 = 1.34 n2 (eq 2) the third expt gives us: n1 sin x1 = n3 sin x3 where we need to solve for x3 we have: (n1/n3) sin 26=sin x3 but we know from eqs 1 and 2 that if we divide the expression for n1 by the expression for n3, we get that n/1n3 = 0.89 so we have: 0.89 sin 26 = sin x3 or x3 = 22.9 deg

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