CBSE 2023 · Region 2 · Set 2 · Q25 · 2 marks
A point object in air is placed symmetrically at a distance of $\displaystyle 60$ cm in front of a concave spherical surface of refractive index 1.5. If the radius of curvature of the surface is $\displaystyle 20$ cm , find the position of the image formed.
Marking-scheme solution
\[\begin{aligned}
& \mu=1.5, u=-60 \mathrm{~cm}, R=-20 \mathrm{~cm}, \mathrm{v}=? \\
& \frac{\mu}{\mathrm{v}}-\frac{1}{u}=\frac{\mu-1}{R} \\
& \frac{1.5}{\mathrm{v}}-\frac{1}{(-60)}=\frac{1.5-1}{(-20)} \\
& \mathrm{v}=-36 \mathrm{~cm}
\end{aligned}
\] &
\[\begin{array}{c|l}
1 / 2 & \\
1 & \\
1 / 2 & 2
\end{array}
\]
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