Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 9 / a

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- Solution a

Solution

The Fermat principle makes the optical path length stationary. For a graph

y = ξ (x)

the variational integrand is

L (x, ξ, ξ^{'}) = n (x, ξ) 1 + ξ^{'2} .

The Euler-Lagrange equation is

\frac{d}{d x} [\frac{n ( x , ξ ) ξ ^{'}}{1 + ξ ^{'2}}] - n_{y} (x, ξ) 1 + ξ^{'2} = 0.

Use the chain rule

d n (x, ξ (x)) / d x = n_{x} + n_{y} ξ^{'}

and differentiate

ξ^{'} / 1 + ξ^{'2}

to obtain

\frac{( n _{x} + n _{y} ξ ^{'} ) ξ ^{'}}{1 + ξ ^{'2}} + \frac{n ξ ^{''}}{( 1 + ξ ^{'2} ) ^{3/2}} - n_{y} 1 + ξ^{'2} = 0.

Multiplying through and combining the

n_{y}

terms gives the required second-order ordinary differential equation with no derivatives in denominators:

n (x, ξ) ξ^{''} = (1 + ξ^{'2}) [n_{y} (x, ξ) - ξ^{'} n_{x} (x, ξ)] .

All partial derivatives of the refractive index are evaluated along the ray.

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