Q.
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 $AB$ ?
NTA AbhyasNTA Abhyas 2022
Solution:
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,
$m=\pm2\Rightarrow v=\pm2u$
Substituting this is the mirror formula,, we get :
$ 1/u+1/\pm 2 u=1/f\Rightarrow u=f/2or3 f/2$
The distance of midpoint of AB from the mirror is :
$\frac{3 f/2 +f/2}{2}=f\equiv L$
The seperation between the two ends of the object is :
$\Rightarrow 3 f/2-f/2=f$
So, the length of the object is $L$ .
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,
$m=\pm2\Rightarrow v=\pm2u$
Substituting this is the mirror formula,, we get :
$ 1/u+1/\pm 2 u=1/f\Rightarrow u=f/2or3 f/2$
The distance of midpoint of AB from the mirror is :
$\frac{3 f/2 +f/2}{2}=f\equiv L$
The seperation between the two ends of the object is :
$\Rightarrow 3 f/2-f/2=f$
So, the length of the object is $L$ .