JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Geometrical Optics

For least strain on the eye the final image must be at infinity, which happens when the objective forms its image in the focal plane of the eyepiece. With the lenses $10$ cm apart and $f_e=5$ cm, that intermediate image lies $10-5=5$ cm beyond the objective, so $v=5$ cm. Applying $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_o}$ with $f_o=1$ cm, $\frac{1}{5}-\frac{1}{u}=1$, giving $\frac{1}{u}=-\frac{4}{5}$ and $|u|=\frac{5}{4}=\frac{50}{40}$ cm. Hence $n=50$.