Question

The focal length of the mirror shown in figure is $20\text{cm}$. A point object at $\text{O}$ is moving with velocity $5\text{mm/s}$ making an angle of ${37}^{\circ }$ with the principal axis as shown. Then, which of the following graphs correctly represents the position and direction of velocity of its image at this instant?

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

We know, from mirror formula
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Substituting the values,
$u=-30\text{cm}$ and $f=-20\text{cm}$ in the mirror formula, we have
$\frac{1}{v}+\frac{1}{-30}=\frac{1}{-20}$
Solving this equation, we get
$v=-60\text{cm}$

Further, mangnification $m=-\frac{v}{u}=-\frac{(-60)}{(-30)}=-2$
and $m^2=4$

Object velocity along the axis is $5\cos {37}^{\circ }=4\text{mm/s}$ (along OP). Therefore, image velocity along the principal axis direction should be $m^2$ times or $16\text{mm/s}$ in the opposite direction of object velocity.
Also, object velocity perpendicular to axis is $5\sin {37}^{\circ }=3\text{mm/s}$ (upwards). Therefore, image velocity will be $m$ times or $-6\text{mm/s}$ (downwards).

Thus, the position and velocity of the image will be as shown below.


Magnitude of velocity of image
$v_I=\sqrt{(16{)}^2+(6{)}^2}=\sqrt{292}\text{mm/s}$
Hence, option (a) is correct