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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2021

Paper 154

From $b_s+b^2=-\eta$ one obtains
$$
\frac d{ds}\sqrt{b^2+\eta}
=\frac{bb_s}{\sqrt{b^2+\eta}}
=-b\sqrt{b^2+\eta}.
$$
Since $\lambda_s=-b\lambda$, the ratio $\sqrt{b^2+\eta}/\lambda$ is constant. Its value at $t=-1$ is one, so
$$
b^2+\eta=\lambda^2.
$$
Using $s_t=\lambda^{-2}$ and $\lambda_s=-b\lambda$ gives
$$
\lambda_t=-\frac b\lambda,
\qquad
b_t=-\frac{b^2+\eta}{\lambda^2}=-1.
$$
The initial conditions now give
$$
\boxed{b_\eta(t)=-t,
\qquad \lambda_\eta(t)=\sqrt{t^2+\eta}}.
$$
Finally,
$$
s(t)-s(-1)
=\int_{-1}^t\frac{d\tau}{\tau^2+\eta}
=\frac1{\sqrt\eta}
\left(\arctan\frac t{\sqrt\eta}
+\arctan\frac1{\sqrt\eta}\right).
$$
These are the <explicit lens parameters>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 154 / 3 / 3 / Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 154 / 3 / 3 / Solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 154 / 3 / 3 / Solution

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