Ray Optics

JEE Main level, single correct, medium

51 of 295

Question 51 of 295

A car is fitted with a convex side-view mirror of focal length 20 cm20 \mathrm{~cm}. A second car 2.8 m2.8 \mathrm{~m} behind the first car is overtaking the first car at relative speed of 15 m/s15 \mathrm{~m} / \mathrm{s}. The speed of the image of the second car as seen in the mirror of the first one is :

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Answer: (A) 115 m/s\frac{1}{15} \mathrm{~m} / \mathrm{s}

1v+1u=1f\frac{1}{v}+\frac{1}{u}=\frac{1}{f}
−1v2dvdt−1u2dudt=0-\frac{1}{v^2} \frac{d v}{d t}-\frac{1}{u^2} \frac{d u}{d t}=0
dvdt=−v2u2(dudt)\frac{d v}{d t}=-\frac{v^2}{u^2}\left(\frac{d u}{d t}\right)
f=20 cmf=20 \mathrm{~cm}
1u+1−280=120\frac{1}{u}+\frac{1}{-280}=\frac{1}{20}
⇒v=28015 cm\Rightarrow v=\frac{280}{15} \mathrm{~cm}
v1=−(28015×280)2×15v_1=-\left(\frac{280}{15 \times 280}\right)^2 \times 15
=115 m/s=\frac{1}{15} \mathrm{~m} / \mathrm{s}

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