JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Geometrical Optics

For the first lens, $f_P=10\text{cm}$ with $u=-15\text{cm}$: $\frac{1}{v}=\frac{1}{f_P}+\frac{1}{u}=\frac{1}{10}-\frac{1}{15}=\frac{1}{30}$, so $v=30\text{cm}$ to the right of $P$. This image lies $30-15=15\text{cm}$ beyond lens $Q$ and so is a virtual object for $Q$ with $u^{\prime }=+15\text{cm}$. For $Q$, $f_Q=15\text{cm}$: $\frac{1}{v^{\prime }}=\frac{1}{f_Q}+\frac{1}{u^{\prime }}=\frac{1}{15}+\frac{1}{15}=\frac{2}{15}$, so $v^{\prime }=7.5\text{cm}$ to the right of $Q$, and being positive the image is real. The overall magnification is $m=\left(\frac{v}{u}\right)\left(\frac{v^{\prime }}{u^{\prime }}\right)=\left(\frac{30}{-15}\right)\left(\frac{7.5}{15}\right)=-1$, so the image is the same size as the object. The final image is real, formed $7.5\text{cm}$ to the right of lens $Q$, with the same size as $AB$.