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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 26.50 with the normal. The refracted beam in sheet 2 makes an angle of 31.70 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 36.70 with the normal. If the experiment is repeated again with sheet 1 on top of sheet 3, determine the expected angle of refraction in sheet 3? Assume the same angle of incidence.

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Answer:

The angle of refraction of sheet 3 when sheet 1 is on top of it is
\theta_(r_s ) = 23.1 ^o

Step-by-step explanation:

From the question we are told that

The angle of incidence is
\theta _i = 26.50 ^o

The angle of refraction angle for sheet 1 is
\theta _(r_1)} = 31.70 ^o

The angle of refraction for sheet 3 is
\theta _(r_3)} = 36.70 ^o

According to Snell's law


(n_2)/(n_1) = (sin (\theta_1))/(sin (\theta_(r_1)))

Where
n_1 \ and \ n_2 are refractive index of sheet 1 and sheet 2

=>
n_2 = n_1 (sin(\theta_i))/(sin (\theta _(r_1)))

Also when sheet 3 in on top of sheet 2


(n_2)/(n_3) = (sin \theta_i)/(sin \theta_(r_3))

substituting for
n_2


n_1 (sin(\theta_i))/(sin (\theta _(r_1))) = n_3 (sin \theta_i)/(sin \theta_(r_3))


n_1 (sin(\theta_i))/(sin (\theta _(r_1))) = n_3 (sin \theta_i)/(sin \theta_(r_3))

=>
n_3 = n_1 * (sin(\theta_(r_3)))/(sin(\theta_(r_1)))

when sheet 1 in on top of sheet 3


(n_3)/(n_1) = (sin(\theta_i))/(\theta_(r_s))

where
r_s is the angle of refraction when sheet 1 is on top of sheet 3

substituting for
n_3


( n_1 * (sin(\theta_(r_3)))/(sin(\theta_(r_1))))/(n_1) = (sin(\theta_i))/(\theta_(r_s))

=>
sin (\theta _(r_s)) = n_1 * sin (\theta_i) * (sin (\theta_(r_1)))/( n_1 * sin(\theta_(r_3)))

substituting values


sin (\theta _(r_s)) = n_1 * sin (26.50) * (sin (31.70))/( n_1 * sin(36.70))

=>
\theta_(r_s ) = sin^(-1) (0.3923)

=>
\theta_(r_s ) = 23.1 ^o

User IanH
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