JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Geometrical Optics

A radius of curvature of $40\text{cm}$ gives the concave mirror a focal length of magnitude $20\text{cm}$. The floating particle sits on the water surface, a distance $5\text{cm}$ above the mirror, so take $u=-5\text{cm}$ and $f=-20\text{cm}$. The mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ gives $\frac{1}{v}=-\frac{1}{20}+\frac{1}{5}=\frac{1}{20}$, so the mirror forms an image $v=\frac{20}{3}\text{cm}$ from it, that is at a depth $5+\frac{20}{3}=\frac{35}{3}\text{cm}$ below the water surface. Finally the water raises this image by refraction, so the depth seen from above is reduced by the refractive index: $d=\frac{35/3}{1.33}=\frac{35}{4}\approx 8.8\text{cm}$.