JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Geometrical Optics

The liquid fills the concave upper face, so it acts as a plano-convex lens whose flat top has infinite radius and whose lower face matches the glass with radius $30\text{ cm}$. Its power is $(1.3-1)\left(\frac{1}{\infty }-\frac{1}{-30}\right)=\frac{0.3}{30}=\frac{1}{100}{\text{cm}}^{-1}$. The glass piece is a meniscus with its upper surface of radius $30\text{ cm}$ and lower surface of radius $20\text{ cm}$, both centred above the surfaces, giving a power $(1.5-1)\left(\frac{1}{-30}-\frac{1}{-20}\right)=0.5\times \frac{1}{60}=\frac{1}{120}{\text{cm}}^{-1}$. For thin lenses in contact the powers add: $\frac{1}{f}=\frac{1}{100}+\frac{1}{120}=\frac{11}{600}{\text{cm}}^{-1}$. The focal length of the liquid–glass combination is $\frac{600}{11}\text{cm}$.