JEE PYQ: Ray Optics Question 43
Question 43 - 2019 (12 Apr Shift 1)
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 $= 1.33$)
(1) 6.7 cm
(2) 13.4 cm
(3) 8.8 cm
(4) 11.7 cm
Show Answer
Answer: (3)
Solution
If $v$ is the distance of image formed by mirror, then $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{v} + \frac{1}{-5} = \frac{1}{-20}$ $\therefore v = -\frac{20}{3}$ cm Distance of this image from water surface $= \frac{20}{3} + 5 = \frac{35}{3}$ cm Using, $\frac{RD}{AD} = \mu$ $\therefore AD = d = \frac{RD}{\mu} = \frac{(35/3)}{1.33} = 8.8$ cm
BETA
