JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Geometrical Optics

This is a silvered lens system: light refracts through the lens, reflects from the silvered second surface (radius $R_2$), and refracts back through the lens — equivalent to a single mirror of focal length $f_{eq}$. A real, inverted image forming at the object's own location requires the object to sit at the centre of curvature of this equivalent mirror, i.e. $u=2f_{eq}$.

Power of the equivalent mirror:

$\frac{1}{f_{eq}}=\frac{2}{f_L}-\frac{1}{f_m},f_m=-\frac{|R_2|}{2}$

Focal length of the lens sitting in the liquid of index ${\mu }_1$:

$\frac{1}{f_L}=\left(\frac{{\mu }_2}{{\mu }_1}-1\right)\left(\frac{1}{|R_1|}+\frac{1}{|R_2|}\right)$

Combining:

$\frac{1}{f_{eq}}=\frac{2({\mu }_2-{\mu }_1)}{{\mu }_1}\cdot \frac{|R_1|+|R_2|}{|R_1||R_2|}+\frac{2}{|R_2|}=\frac{2[({\mu }_2-{\mu }_1)(|R_1|+|R_2|)+{\mu }_1|R_1|]}{{\mu }_1|R_1||R_2|}=\frac{2[{\mu }_2|R_1|+{\mu }_2|R_2|-{\mu }_1|R_2|]}{{\mu }_1|R_1||R_2|}$

Setting $u=2f_{eq}$:

$u=\frac{{\mu }_1|R_1|\cdot |R_2|}{{\mu }_2\left(|R_1|+|R_2|\right)-{\mu }_1|R_2|}$