A thin biconvex glass lens of refractive index 1.5 has both surfaces of radius of curvature 20 cm. A point object is placed on the axis in air at 60…

The lens has different media on its two sides, so the lens-maker formula for a lens in a single medium cannot be used; apply the single-surface refraction formula $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ twice (thin lens, so distances are measured from the lens). Surface 1 (air to glass, $R=+20$, $u=-60$): $\dfrac{1.5}{v_1}=\dfrac{0.5}{20}-\dfrac1{60}=\dfrac1{120}$, so $v_1=180\ \mathrm{cm}$. This image acts as the virtual object for the second surface. Surface 2 (glass to water, $R=-20$, $u=+180$): $\dfrac{4/3}{v}=\dfrac{4/3-1.5}{-20}+\dfrac{1.5}{180}=\dfrac1{120}+\dfrac1{120}=\dfrac1{60}$, so $v=80\ \mathrm{cm}$. Option $180\ \mathrm{cm}$ is the intermediate image, and $30\ \mathrm{cm}$ is the answer if the lens were in air on both sides ($f=20\ \mathrm{cm}$).