Answer
\[\theta_{1} = \tan^{-1}\left(\frac{n_{2}}{n_{1}}\right)\]
Work Step by Step
From (b),
\[\frac{n^2_{2}-n^2_{1}}{n^2_{2}}\sin^2\theta_{1} = \frac{n^2_{2}-n^2_{1}}{n^2_{1}}\cos^2\theta_{1}\]
\[\implies \frac{\sin^2\theta_{1}}{\cos^2\theta_{1}} = \frac{\frac{n^2_{2}-n^2_{1}}{n^2_{1}}}{\frac{n^2_{2}-n^2_{1}}{n^2_{2}}}\]
\[\implies \frac{\sin^2\theta_{1}}{\cos^2\theta_{1}} = \frac{n^2_{2}}{n^2_{1}}\]
Taking the square root on both sides, we have that
\[\frac{\sin\theta_{1}}{\cos\theta_{1}} = \frac{n_{2}}{n_{1}}\]
\[\implies \tan\theta_{1} = \frac{n_{2}}{n_{1}}\]
\[\implies \theta_{1} = \tan^{-1}\left(\frac{n_{2}}{n_{1}}\right)\]
which is the required answer.
