Answer
$3.16\times 10^{-3} rad$
Work Step by Step
We can find the required angle as follows:
$M=(\frac{s}{f_1})(\frac{d-f_1}{(d-f_1)f_2(d-f_1-f_2)})$
We plug in the known values to obtain:
$M=(\frac{25.0cm}{2.7cm})(\frac{12.0cm-2.7cm}{(12.0cm-2.7cm)(0.49cm)(12.0cm-2.7cm-0.49cm)})$
$\implies M=166.48$
and $\theta^{\prime}=|M_{total}|\theta$
We plug in the known values to obtain:
$\theta^{\prime}=(166.48)(1.9\times 10^{-5} rad)$
$\implies \theta^{\prime}=3.16\times 10^{-3} rad$
