Question

A point object is located at a distance $15cm$ in front of a concave mirror of radius of curvature $20cm$. It has a velocity $2mm/s$ perpendicular to the principal axis. What will be the velocity of the image at that instant?

Answer 1

Hint: The velocity of the image is formed by a concave mirror. The radius of curvature is double the focal length $R=2f$. The velocity of the image depends on the direction that the mirror is moving and the position of the object with respect to the mirror.

Formula Used:
We will be using the relation of the mirror formula ${\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$.

Complete step by step answer:
Given: The object is at the distance in front of the concave mirror $u=-15cm$. Velocity perpendicular to the principal axis $v_0=2mm/s$. The radius of curvature $R=20cm$. To find the velocity of the image at that instant.
The focal length of a spherical mirror (both concave mirror and convex mirror) is half of its radius of curvature. The relationship is given by,
$R=2f$
$f=R/2$
$f=20/2=10cm$
$f=-10cm$
$(v{)}_x=-{\displaystyle \frac{v^2}{u^2}}(v_0{)}_x$;
$(v{)}_y=-{\displaystyle \frac{v}{u}}(v_0{)}_y$

The Mirror formula is defined by relating the object distance and image distance with the focal length. The distance between the object and the pole of the mirror is known as the object distance $$\left(u\right)$$ . The distance between the image and the pole of the mirror is called image distance $(v)$. The distance between the principal focus and the pole of the mirror is called the focal length $$\left(f\right)$$.

The mirror formula applies to the spherical mirror that is concave and convex mirror
Mirror formula is related by the image distance, object distance and focal length is given by:
${\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$.
${\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{-15}}={\displaystyle \frac{1}{10}}$
${\displaystyle \frac{1}{v}}={\displaystyle \frac{-1}{10}}+{\displaystyle \frac{1}{15}}={\displaystyle \frac{-3+2}{30}}$
$v=-30cm$
$(v{)}_y=-{\displaystyle \frac{v}{u}}(v_0{)}_y$
$=-({\displaystyle \frac{-30}{-15}})(2mm/s)$
$(v{)}_y=-2\times 2=4mm/s$

The velocity of the image is $v=4mm/s$.

Notes: German chemist Justus von Liebig in $1835$ gave the mirror formula. The focal length of the concave mirror is positive and the focal length of the convex mirror is negative. The radius of the curvature R is the reciprocal of the curvature.