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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-338/3/b/ii/solution

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Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 338 / 3 / b / ii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-06

With constant detected rates,

Q = R_{Q} T

,

B = R_{B} T

. Squaring the previous signal-to-noise ratio in photon counting gives

R_{Q}^{2} T^{2} - Z^{2} (R_{Q} + 2 R_{B}) T - 2 n Z^{2} r^{2} = 0.

For

R_{Q} > 0

, nonnegative background and target

Z > 0

, choose the positive root:

T = \frac{Z ^{2} ( R _{Q} + 2 R _{B} ) + Z ^{4} ( R _{Q} + 2 R _{B} ) ^{2} + 8 n R _{Q}^{2} Z ^{2} r ^{2}}{2 R _{Q}^{2}} .

The other root is nonpositive. With no read noise, this reduces to

T = Z^{2} (R_{Q} + 2 R_{B}) / R_{Q}^{2}

; in a read-dominated short exposure it approaches

Z 2 n r / R_{Q}

. If

R_{Q} = 0

, no positive target stellar signal-to-noise is attainable. A zero target can be assigned zero exposure, with its limiting interpretation when all noise terms also vanish. Real quantum efficiency or multiple reads modifies the rates or read-variance term before solving the quadratic.

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