Question

A linear object $AB$ is placed along the axis of a concave mirror. The object is moving towards the mirror with speed $V$ . The speed of the image of the point $A$ is $4V$ and the speed of the image of point $B$ is also $4V$ . Let the centre of the line $AB$ be at a distance $L$ from the mirror. Then what will be the length of the object $AB$ ?

• A. $\frac{3L}{2}$
• B. $\frac{5L}{3}$
• C. $L$
• D. $\frac{4L}{3}$

Answer 1

Answer: (C)

Lets take Velocity of Object to be $V_O$ and Velocity of the Image be $V_I$ .
Differentiating Mirror formula, we get :
$\left(\frac{-1}{v^2}\right)\cdot V_I+\left(\frac{-1}{u^2}\right)\cdot V_O=0\Rightarrow V_I=-\left(\frac{v^2}{u^2}\right)\cdot V_O$
We know, Magnification :
$m=-v/u$
Using both we get :
$V_I=-m^2.V_O$
From the question, we Know : Speed of object is 4 times object of image . so we get,
$m=\pm 2\Rightarrow v=\pm 2u$
Substituting this is the mirror formula,, we get :
$1/u+1/\pm 2u=1/f\Rightarrow u=f/2or3f/2$
The distance of midpoint of AB from the mirror is :
$\frac{3f/2+f/2}{2}=f\equiv L$
The seperation between the two ends of the object is :
$\Rightarrow 3f/2-f/2=f$
So, the length of the object is $L$ .