JEE Advanced 2019 Paper 2 Physics: Paragraph 2

By the lens-maker's formula, $\dfrac1f=(n-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)=0.5\times\dfrac{2}{20}$, giving $f=20$ cm, so P to (2). In the liquid, $\dfrac1f=\left(\dfrac{1.5}{4/3}-1\right)\dfrac2{20}=0.125\times0.1$, so $f=80$ cm and Q to (6); the lens loses most of its power because the refractive indices are close. For two thin lenses in contact the powers add, $\dfrac1f=\dfrac1{20}+\dfrac1{20}$, so $f=10$ cm and R to (1). For the object at $u=-30$ cm, $\dfrac1v=\dfrac1{20}-\dfrac1{30}=\dfrac1{60}$, so $v=60$ cm and S to (5). Correct code: P to 2, Q to 6, R to 1, S to 5. Distractors typically swap the liquid and contact cases or use $f=40$ cm for the combination by adding focal lengths rather than powers.

P: a cut along the axis does not change the radii of curvature, only the aperture, so each half still has $f=20$ cm; P to (2). Q: a cut perpendicular to the axis through the centre leaves a plano-convex lens with one curved face of $R=20$ cm, so $\dfrac1f=(1.5-1)\dfrac1{20}$, giving $f=40$ cm; Q to (4). R: $u=-60$ cm gives $\dfrac1v=\dfrac1{20}-\dfrac1{60}=\dfrac1{30}$, so $v=30$ cm; R to (3). S: $u=-15$ cm lies inside the focal length, $\dfrac1v=\dfrac1{20}-\dfrac1{15}=-\dfrac1{60}$, a virtual image at $60$ cm on the object side; S to (5). Correct code: P to 2, Q to 4, R to 3, S to 5.