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

Use the ideal detection convention of one collected electron per detected photon; otherwise replace photon totals by detected electron totals using the quantum efficiency. Let $A$ be the summed measurement on the star patch and $B^{\prime }$ the independent background-only patch. After additive offsets are removed, their means are $Q+B$ and $B$. Independent Poisson distribution counts have variance equal to their mean, and a single read of $n$ detector pixels contributes $n{r}^2$ to each patch's variance. ThereforeThe two factors of two arise from the independently measured background and the second patch's read noise, not from doubling the stellar signal. The formula assumes equal background means, independent detector pixels and patch measurements, one read per patch, and negligible dark current or other noise. Several exposure reads or correlated detector pixels require their own variance and covariance terms.

With constant detected rates, $Q=R_{Q}T$, $B=R_{B}T$. Squaring the previous signal-to-noise ratio in photon counting givesFor $R_Q>0$, nonnegative background and target $Z>0$, choose the positive root:The other root is nonpositive. With no read noise, this reduces to $T=Z^2(R_Q+2R_B)/R_Q^2$; in a read-dominated short exposure it approaches $Z\sqrt{2n}r/R_Q$. 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.