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 25.9 degrees with the normal and refracts in sheet 2 at an angle of 31.5 degrees 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 37.3 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 sheeg 3? assume the same angle of incidence? anwser in units of degrees!

Explanation / Answer

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 25.9 = n2 sin 31.5

plugging in values gives you the relationship:

n1=1.196 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 25.9 = n2 sin 37.3 which yields

n3 = 1.38 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 25.9=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 = 0.866 so we have:

0.866 sin 25.9 = sin x3 or

x3 = 22.22 deg

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