question_answer

A pencil (AB) of length 20 cm is moving along the principal axis of a concave mirror MM?. with a velocity \[5\text{ }m/s\]approaching the mirror. The mirror itself is moving away from the pencil at a speed of\[2\text{ }m/s\]. Find the rate of change of length of the image of the pencil at the instant end A is at a distance of 60 cm from the mirror.

A) \[\frac{9}{4}m/s\]                

B) \[\frac{5}{3}m/s\]              

C) \[\frac{8}{3}m/s\]               

D) \[\frac{7}{2}m/s\]

Correct Answer: A

Solution :

[a] Point B is at the centre of curvature of the mirror and us image will be at the same position. Image of point A is formed at distance \[{{v}_{A}}\] given by mirror formula. \[\frac{1}{{{v}_{A}}}+\frac{1}{-60}=\frac{1}{-20}\] \[\frac{1}{{{v}_{A}}}=\frac{1}{60}-\frac{1}{20}\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,{{v}_{A}}=-30\,cm\] And   \[\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\,\,\,\Rightarrow \,\,\,-\frac{1}{{{v}^{2}}}\frac{dv}{dt}-\frac{1}{{{u}^{2}}}\frac{du}{dt}=0\] \[\frac{dv}{dt}=-{{\left( \frac{v}{u} \right)}^{2}}\frac{du}{dt}\] \[{{V}_{1m}}=-{{\left( \frac{v}{u} \right)}^{2}}.{{V}_{Om}}\] \[{{V}_{B'}}-{{V}_{m}}=-{{\left( \frac{40}{40} \right)}^{2}}.\left[ {{V}_{B}}-{{V}_{m}} \right]\] \[\left[ \begin{align}   & {{V}_{1m}}=velocity\,of\,image\,wrt\,mirror \\  & {{V}_{Om}}=velocity\,of\,object\,wrt\,mirror \\ \end{align} \right]\] \[{{V}_{B'}}-2=-\,[5-2]\] \[{{V}_{B'}}=-1\,\,m\text{/}s=1\,\,m\text{/}s\,\,(\to )\] \[{{V}_{A'}}-{{V}_{m}}=-{{\left( \frac{30}{60} \right)}^{2}}[{{V}_{A}}-{{V}_{m}}]\] \[{{V}_{A'}}-2=-\frac{1}{4}[5-2]\] \[{{V}_{A'}}=+\frac{5}{4}m/s=\frac{5}{4}m/s(\to )\] \[\therefore \]Length A 'B' is increasing at the rate of \[1+\frac{5}{4}=\frac{9}{4}m/s\]is