Question

An object is kept on the principal axis of a convex mirror of focal length $10\text{cm}$ at a distance of $10\text{cm}$ from the pole. The object starts moving at a velocity $20\text{mm/s}$ towards the mirror at an angle ${30}^{\circ }$ with the principal axis. What will be the velocity of its image and the angle it makes with the principal axis at that instant?

• A. $\text{Velocity of image}=\frac{2\sqrt{7}}{5}\text{mm/s};\tan \theta =\frac{\sqrt{3}}{2}$
• B. $\text{Velocity of image}=\frac{5\sqrt{7}}{2}\text{mm/s};\tan \theta =\frac{\sqrt{3}}{2}$
• C. $\text{Velocity of image}=\frac{5\sqrt{7}}{2}\text{mm/s};\tan \theta =\frac{2}{\sqrt{3}}$
• D. $\text{Velocity of image}=\frac{2\sqrt{7}}{5}\text{mm/s};\tan \theta =\frac{2}{\sqrt{3}}$

Answer 1

The correct option is C $\text{Velocity of image}=\frac{5\sqrt{7}}{2}\text{mm/s};\tan \theta =\frac{2}{\sqrt{3}}$


Given:
$u=-10\text{cm};f=10\text{cm};V_O=20\text{mm/s}$

Using mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow v=\frac{uf}{u-f}=\frac{(-10)\left(10\right)}{-10-10}=5\text{cm}$

Let $V_I$ be the speed of the image.
We know that,
$V_{Ix}=-m^2\times V_{Ox}$
$\Rightarrow$ Speed of image in $x-$ direction
$V_{Ix}=-(\frac{-v}{u}{)}^2\times V_{Ox}$
$\Rightarrow V_{Ix}=-{\left(\frac{-5}{-10}\right)}^2\times 20\times \frac{\sqrt{3}}{2}=-\frac{5\sqrt{3}}{2}\text{mm/s}$

Speed in $y-$ direction
$V_{Iy}=m\times V_{Oy}$
$\therefore V_{Iy}=(\frac{-v}{u})\times V_{Oy}$

$\Rightarrow V_{Iy}=\left(\frac{-5}{-10}\right)\times 20\times \frac{1}{2}=5\text{mm/s}$.

Hence, $\tan \theta =\left|\frac{V_{Iy}}{V_{Ix}}\right|=\frac{5}{\frac{5\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}$
and
$V_I=\sqrt{{\left(\frac{5\sqrt{3}}{2}\right)}^2+(5{)}^2}=\frac{5\sqrt{7}}{2}\text{mm/s}$.
Hence, option $\left(c\right)$ is the correct answer.