Question

An object of 5cm size is placed at a distance of 20cm from a converging mirror of focal 15cm. At what distance from the mirror should a screen be placed to get a sharp image? Also calculate the size of the image?

Answer 1

Step 1: Given that:

Size of object($h_o$)= $5cm$

Object distance(u)= $-20cm$

Focal length(f) of converging or concave mirror= $-15cm$

(All the distances to the left of the pole of s spherical mirror is taken as negative.)

Step 2: Calculation of image distance(v):

According to mirror formula, we get;

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ , we get

$\frac{1}{v}+\frac{1}{-20cm}=\frac{1}{-15cm}$

$\frac{1}{v}=-\frac{1}{15}+\frac{1}{20}$

$\frac{1}{v}=\frac{-4+3}{60}$

$\frac{1}{v}=\frac{-1}{60}$

$\frac{1}{v}=-\frac{1}{60}$

$v=-60cm$

Step 3: Calculation of size of the image:

According to magnification formula;

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

$\frac{h_i}{5cm}=-\frac{-60cm}{-20cm}$

$\frac{h_i}{5}=-3$

$h_i=-15cm$

Thus,

The image is formed at 60cm in front of the concave mirror.

The image is of the size of 15cm and it will be real and inverted.