JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Geometrical Optics

Refraction occurs at the single spherical surface between air and the lens material. With light travelling from the object toward the surface, $u=-40$ cm; the centre of curvature lies $20$ cm in front of the surface, so $R=-20$ cm, with ${\mu }_1=1$ and ${\mu }_2=1.54$. Applying $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$: $\frac{1.54}{v}+\frac{1}{40}=\frac{0.54}{-20}$, giving $\frac{1.54}{v}=-0.052$ and $v=-29.6$ cm. The transverse magnification at a refracting surface is $m=\frac{{\mu }_{1}v}{{\mu }_{2}u}=\frac{(1)(-29.6)}{(1.54)(-40)}=0.48$. For an object $2$ cm tall the image height is $0.48\times 2\approx 1$ cm. The height of the image is $1$ cm.