JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Geometrical Optics

Let the two concave surfaces, each of radius of curvature $R$ and refractive index 1.5, share a common center of curvature $P$ midway between their vertices $A$ and $B$, so that $AP=PB=R$ and $AB=2R$. The object $O$ lies midway between $P$ and $B$, so $OP=OB=\frac{R}{2}$ and $OA=OP+PA=\frac{3R}{2}$.

Each surface forms its own image of $O$ directly, using the single-surface refraction formula

with light going from air (${\mu }_1=1$) into glass (${\mu }_2=1.5$).

For surface B, $u=-\frac{R}{2}$, with the center of curvature on the incident side so $R\to -R$:



For surface A, $u=-\frac{3R}{2}$, with the same sign convention:



Both images form on the object side of their respective surfaces, at distances $0.6R$ from $B$ and $\frac{9R}{7}$ from $A$, measured back into the gap between the surfaces. The separation between the two images is therefore