Three sheets of plastic have unknown indices of refraction. Sheet 1 is placed on

Question

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 27.1? with the normal, and refracts in sheet 2 at an angle of 32.8? with 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 36.1? 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.

Answer in units of ?

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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