JEE Main · 2011 · 10 Apr 2011 · Offline · Geometrical Optics

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A car is fitted with a convex side-view mirror of focal length 20 cm . A second car 2.8 m behind the first car is overtaking the first car at relative speed of

15 m / s

. The speed of the image of the second car as seen in the mirror of the first one is :

(A) $\frac{1}{15} m / s$ ✓ Correct
(B) $10 m / s$
(C) $15 m / s$
(D) $\frac{1}{10} m / s$

Answer: (A)

Solution

Use the mirror relation

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

with

f = + 20 cm

for the convex mirror and the second car at

u = - 280 cm

. Then

\frac{1}{v} = \frac{1}{20} - \frac{1}{- 280} = \frac{15}{280}

, so

v = \frac{280}{15} cm

. Differentiating the relation in time gives

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

, so the image speed is

(\frac{v}{u})^{2}

times the object's relative speed,

(\frac{280/15}{280})^{2} \times 15 = \frac{1}{1 5 ^{2}} \times 15 = \frac{1}{15} m s^{- 1}

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option

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JEE Main · 2011 · 10 Apr 2011 · Offline · Geometrical Optics

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