JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Geometrical Optics

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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 5

m / s

, the speed and direction of the image will be:

(A) $2.26 \times 1 0^{- 3} m / s$ away from the lens
(B) $0.92 \times 1 0^{- 3} m / s$ away from the lens
(C) $3.22 \times 1 0^{- 3} m / s$ towards the lens
(D) $1.16 \times 1 0^{- 3} m / s$ towards the lens✓ Correct

Answer: (D)

Solution

Use the lens equation

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

with

u = - 20 m

and

f = + 0.3 m

\frac{1}{v} = \frac{1}{0.3} - \frac{1}{20} = \frac{197}{60}

, so

v = \frac{60}{197} \approx 0.305 m

, just outside the focus. Differentiating the lens relation gives

\frac{d v}{d t} = \frac{v ^{2}}{u ^{2}} \frac{d u}{d t}

. The object recedes at

5 m/s

, so

\frac{d u}{d t} = - 5 m/s

, and with

\frac{v}{u} = \frac{3}{197}

we get

\frac{d v}{d t} = (\frac{3}{197})^{2} (- 5) \approx - 1.16 \times 1 0^{- 3} m/s

. The negative sign shows

v

is shrinking, so the image creeps towards the lens at

1.16 \times 1 0^{- 3} m/s

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Geometrical Optics

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