 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/20/b/iii/solution

The acute angles follow from part (ii) and the Snell's law relation:

θ = π /6 = 3 0^{\circ}, ϕ = π /4 = 4 5^{\circ} .

The path lengths below and above the interface are respectively

ℓ_{1} = \frac{3}{c o s ( π /6 )} = 2, ℓ_{2} = \frac{1}{c o s ( π /4 )} = 2 .

Hence the total distance and minimum travel time are

ℓ = 2 + 2, T_{m i n} = \frac{2}{c /2} + \frac{2}{c / 2} = \frac{6}{c} .

For a sketch to scale, choose the interface as

y = 0

, with

A = (0, - 3)

and

B = (2, 1)

. The crossing is

C = (1, 0)

because

d_{1} tan θ = d_{2} tan ϕ = 1

. The figure uses equal axis scales and shows the two normal-relative angles.

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