Question

A $1cm$ high object is placed at a distance of $32cm$ from a concave mirror. The image is real, inverted and $1.5cm$ in size. Calculate the focal length of the mirror and the position of the image.

Answer 1

Step 1: Given Data

Height of the object $h_o=1cm$

Object distance $u=-32cm$

The object is present in front of the mirror therefore the object distance is negative.

Height of the image $h_i=-1.5cm$

The image is inverted therefore the image height is negative.

Let the image distance be $v$ and the focal length be $f$.

Step 2: Calculate Image Distance

We know that the magnification of a mirror is given as,

$m=\frac{h_i}{h_o}=-\frac{v}{u}$

Upon substituting the values we get,

$\frac{-1.5}{1}=-\frac{v}{\left(-32\right)}$

$\Rightarrow v=-1.5\times 32$

$\Rightarrow v=-48cm$

Step 3: Calculate Focal Length

According to the mirror formula,

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Upon substituting the values we get,

$\frac{1}{f}=\frac{1}{-32}+\frac{1}{-48}$

$\Rightarrow \frac{1}{f}=-\frac{1}{16}\left(\frac{1}{2}+\frac{1}{3}\right)$

$\Rightarrow \frac{1}{f}=-\frac{1}{16}\left(\frac{5}{6}\right)$

$\Rightarrow \frac{1}{f}=-\frac{5}{96}$

$\therefore f=-\frac{96}{5}=-19.2cm$

Hence, the focal length of the mirror is $19.2cm$ in front of the mirror and the position of the image is $48cm$ in front of the mirror.