 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2014/paper-2/section-a/2/solution

For a lossless single interface at normal incidence,

R = 1 - T = 0.04

R = [(n - 1) / (n + 1)]^{2}

, taking air's refractive index as one. Selecting the glass branch

n > 1

gives

n = \frac{1 + R}{1 - R} = 1.50 .

The reciprocal root

2/3

is the mathematical lower-index branch, not ordinary glass.

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