Question

The drawing shows four different situations in which a light ray is traveling from one medium into another. In some of the cases, the refraction is not shown correctly. For cases (a), (b), and (c), the angle of incidence is 56° for case (d), the angle of incidence is 0° n2 = 1.6 n 1.4 n2 = 1.6 1.6 n2= 1.4 1 n2 = 1.41 n1 = 1.6 Determine the angle of refraction in each case (a) θ2 = (b) θ2 = (c) θ2 = | (d) θ2 = Select ▼ Select ▼ Select ▼ Select ▼

Answer 1

using snells law we have, n1sini1=n2sinr1

where i1 =angle of incidence and r1 is the angle of refraction

n1 the refractive index in the first medium and n2 is the refractive index of the second medium

μ (refractive index of first material with respect to theother) = n₂/n₁

In each case, depending onwhich material the light ray starts in, n₁ and n₂ switch

μ_a = 1.6/1.4  =1.142

similarly, μ_b = 1.066,  μ_c = 0.875,  μ_d = 0.875

a) μ = sin i/sin r

r = sin^-1(sin56°/1.142) = 46.55 degrees

b) r = sin^-1(sin56° /1.067) = 50.98 degrees

c) r = sin^-1(sin56° /0.875) = 71.35 degrees

d) r = sin^-1(sin0/0.875) = 0 degrees

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