Case $(i):$ when it is a concave mirror of focal length $28 \mathrm{cm}$
$f_{m}=\infty$
Hence $\frac{1}{28}=\frac{2}{f_{1}}+\frac{1}{f_{m}}$
$\Longrightarrow f_{1}=56 \ldots(i)$
Case $(ii):$ when it is a concave mirror of focal length $10 \mathrm{cm}$
Hence $\frac{1}{10}=\frac{2}{f_{1}}+\frac{1}{f_{m}}$
Substituting the value of $f_{1}$, we get
$\Longrightarrow \frac{1}{10}=\frac{2}{56}+\frac{1}{f_{m}}$
$\Longrightarrow \frac{1}{f_{m}}=\frac{1}{10}-\frac{1}{28}$
$\Longrightarrow \frac{1}{f_{m}}=\frac{28-10}{280}$
$\Longrightarrow \frac{1}{f_{m}}=\frac{28-10}{280}$
$\Longrightarrow f_{m}=\frac{280}{18}=\frac{140}{9} \mathrm{cm}$
Therefore, the radius of curvature, $R=2 f_{m}=2 \times \frac{140}{9}=\frac{280}{9} \mathrm{cm}$
Now,
$\frac{1}{f_{1}}=(\mu-1) \frac{1}{R}$
Substituting the respective values, we get
$\frac{1}{56}=(\mu-1) \frac{1}{\frac{280}{9}}$
$\Longrightarrow \mu-1=\frac{280}{9} \times \frac{1}{56}$
$\Longrightarrow \mu=1+\frac{5}{9}=\frac{14}{9}$
is the refractive index of the material of the lens.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
