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If light refracts into diamond (refractive index = 2.42) at ${12}^{\circ}$ to the normal, what was the angle of incidence?
in Grade 12 Physical Sciences by Diamond (50,571 points) | 38 views

2 Answers

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n1Sin &1 = n2sine &2

1sin ? = 2.42 sin 12

Sin? = 2.42 sin 12

Sin?=.5031

Angle of incidence = 30.20 degrees
by Wooden (1,121 points)
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Light travel through vacuum assumed as Refractive index 1 (n1) = 1 ; Refractive index 2 (n2) =2.42 ; Angle of incidence (θ1) = ? ; Angle of refraction (θ2) = 12 degrees

n1 Sin θ1 = n2 Sin θ2

Sin θ1 / Sin θ2 = n2 / n1

Sin θ1 / Sin 12 = 2.42 / 1

Sin θ1 = (2.42) (Sin 12)

Sin θ1 = 2.42 x 0.21

Sin θ1 = 0.5031

θ1 = Sin^-1(0.5031)

θ1 = 30.21 degrees
by Wooden (698 points)

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