Question

An object is located at a distance $30\text{cm}$ on the principal axis from the pole of a concave mirror of focal length $20\text{cm}$. Suddenly the mirror is displaced upwards by a distance $1.5\text{cm}$ in the direction normal to its principal axis. Calculate the displacement of the image produced by the mirror from the original axis.

• A. $1.5\text{cm}$
• B. $3\text{cm}$
• C. $4.5\text{cm}$
• D. $6\text{cm}$

Answer 1

The correct option is C $4.5\text{cm}$

Given that
$u=-30\text{cm};f=-20\text{cm}$
Displacement of mirror $\left(h\right)=1.5\text{cm}$

Position of the image formed by mirror is given by
$v=\frac{uf}{u-f}$
$\Rightarrow v=\frac{(-30)\times (-20)}{-30-(-20)}$
$\Rightarrow v=-60\text{cm}$

Magnification $\left(m\right)=\frac{-v}{u}$
$\Rightarrow m=\frac{-(-60)}{-30}=-2$
$\Rightarrow m=-2$(inverted image)

Now, mirror is displaced by $1.5\text{cm}$


Now w.r.t axis $P^{\prime }{O}^{\prime }$,
Magnification $m=\frac{h_i}{h_o}$
where, $h_i=$ height of image
$h_o=$height of extended object
$\Rightarrow m=-2=\frac{h_i}{-1.5}$
$\Rightarrow h_i=3\text{cm}$ (above the axis $P^{\prime }{O}^{\prime }$)

Therefore, total displacement of image from original axis $PO$ is $3+1.5=4.5\text{cm}$.
Hence, $\left(c\right)$ is the correct option.