 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/9/ii/c/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 9 / ii / c / Solution by

Codex  0 2026-10-18

For

c > 1

, the catenary has a positive minimum height

x_{0} = (c - 1) / β

above the road. Its derivatives are

x^{'} (y) = sinh [\frac{β}{c} (y - y_{0})], x^{''} (y) = \frac{β}{c} cosh [\frac{β}{c} (y - y_{0})] > 0.

(1)

Thus the light ray is symmetric about

y = y_{0}

, descends towards a horizontal tangent at its minimum, and then rises. It never intersects the road. This smooth upward bending in the refractive index gradient produces an inferior mirage, without reflection at the road.

Figure 1.
A turning light ray above a hot road in a linearly increasing refractive index
. **Details**.

 Read the full article

 New to topics? Read the docs here!