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 27.4 ° with the normal. The refracted beam in sheet 2 makes an angle of 32.1 ° 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 37.4 ° 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 27.4 = n2 sin 32.1

plugging in values gives you the relationship:

n1=1.155 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 27.4 = n2 sin 37.4 which yields

n3 = 1.32 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 27.4=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
n1/n3 = 0.876 so we have:

0.876 sin 27.4 = sin x3 or x3 = 23.8 deg

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