JEE PYQ: Ray Optics Question 48
Question 48 - 2019 (11 Jan Shift 1)
An object is at a distance of 20 m from a convex lens of focal length 0.3 m. The lens forms an image of the object. If the object moves away from the lens at a speed of 5m/s, the speed and direction of the image will be:
(1) $2.26 \times 10^{-3}$ m/s away from the lens
(2) $0.92 \times 10^{-3}$ m/s away from the lens
(3) $3.22 \times 10^{-3}$ m/s towards the lens
(4) $1.16 \times 10^{-3}$ m/s towards the lens
Show Answer
Answer: (4)
Solution
By lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v} - \frac{1}{(-20)} = \frac{10}{3}$ $\frac{1}{v} = \frac{10}{3} - \frac{1}{20}$, $v = \frac{60}{197}$ $\frac{1}{v} = \frac{197}{60}$ Magnification of lens (m) is given by $m = \left(\frac{60}{197}\right) \times \frac{1}{20}$ velocity of image wrt. to lens is given by $v_{OL} = m^2 v_{OL}$ direction of velocity of image is same as that of object $v_{OL} = 5$ m/s $v_{I/L} = \left(\frac{60 \times 1}{197 \times 20}\right)^2 (5)$
$= 1.16 \times 10^{-3}$ m/s towards the lens
BETA
