 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2019/7/a/solution

R_{1}

and

R_{2}

, the first surface obeys

n / v_{1} - 1/ u = (n - 1) / R_{1}

, and the second obeys

1/ v - n / v_{1} = (1 - n) / R_{2}

. The separation of the surfaces is neglected, so the intermediate image coordinate

v_{1}

cancels:

\frac{1}{v} - \frac{1}{u} = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = \frac{1}{f} .

u < 0

for a real object. For an equiconvex thin lens and light travelling left to right,

R_{1} = R

and

R_{2} = - R

. Hence

\frac{1}{f} = \frac{2 ( n - 1 )}{R}, f = \frac{R}{2 ( n - 1 )} .

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