JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Geometrical Optics

The two planoconvex lenses are in contact, so their powers add. From the figure the left curved surface has ${\mu }_1=1.5$, $R_1=15$ cm and the right curved surface has ${\mu }_2=1.2$, $R_2=12$ cm. By the lens maker's formula for a planoconvex lens, $\frac{1}{f_1}=({\mu }_1-1)\frac{1}{R_1}=(1.5-1)\frac{1}{15}=\frac{1}{30}$ and $\frac{1}{f_2}=({\mu }_2-1)\frac{1}{R_2}=(1.2-1)\frac{1}{12}=\frac{1}{60}$. The combination therefore has $\frac{1}{f}=\frac{1}{30}+\frac{1}{60}=\frac{1}{20}$, so $f=20$ cm. With $u=-30$ cm, the thin-lens equation gives $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$, hence $v=60$ cm. The magnification is $m=\frac{v}{u}=\frac{60}{-30}=-2$, a real, inverted image twice the size of the object.