question_answer

A concave mirror has radius of curvature of 40 cm. It is at the bottom of a glass that has water filled up to 5 cm (see figure). If a small particle is floating on the surface of water, its image as seen, from directly above the glass, is at a distance d from the surface of water. The value of d is close to : (Refractive index of water = 1.33)                                                        [JEE Main Held on 12-4-2019 Morning]

A) 8.8 cm

B) 11.7 cm

C) 6.7 cm            

D) 13.4 cm

Correct Answer: A

Solution :

Light incident from particle P will be reflected at mirror \[u=-5cm,\,f=-\frac{R}{2}=-20cm\] \[\frac{1}{\text{v}}+\frac{1}{u}=\frac{1}{f}\] \[\] This image will act as object for light getting refracted at water surface So, object distance\[d=5+\frac{20}{3}=\frac{25}{3}cm\]below water surface. After refraction, final image is at\[d'=d\left( \frac{{{\mu }_{2}}}{{{\mu }_{1}}} \right)\] \[=\left( \frac{35}{3} \right)\left( \frac{1}{4/3} \right)\] \[=\frac{35}{4}=8.75cm\]                \[\approx 8.8cm\]