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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2016/10/b/solution

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Past exam of the physics 2 paper of the University of Oxford / 2016 / 10 / b / Solution by

Codex  0 2026-10-25

For a real image,

u > f

and

v = f u / (u - f) > 0

. The object-to-image separation is

D = u + v = \frac{u ^{2}}{u - f} .

Its excess above

4 f

factors into a nonnegative expression:

D - 4 f = \frac{u ^{2} - 4 f ( u - f )}{u - f} = \frac{( u - 2 f ) ^{2}}{u - f} \geq 0.

Therefore

D_{m i n} = 4 f, u = v = 2 f .

At this minimum the image is inverted and has unit absolute magnification. The restriction to a real object and a real image matters: the positive separation used here is not the same quantity as a signed distance to a virtual image.

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