An object $5 c m$ is placed at a distance of $20 c m$ in front of a convex mirror of radius of curvature $30 c m$ . Find the position of the image, its nature and size.

Question

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Answer 1

Formula:

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

Given,
$\frac{1}{v} + \frac{1}{( - 2 0 )} = \frac{1}{1 5}$
$\frac{1}{v} = \frac{7}{6 0} \Rightarrow v = \frac{6 0}{7}$
$v = 8.57 \approx 8.6 c m$

The image is formed behind the mirror at a distance of $8.6 c m$
$m = \frac{h _{2}}{h _{1}} = - \frac{v}{u}$
$\to \frac{h _{2}}{5} = - \frac{8 . 5 7}{2 0}$

$∴ h_{2} = 2.175 c m$

Thus the image is $2.175 c m$ high, virtual and erect.

Hence, this is the answer.

Answer 2

Formula:

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

Given,

\frac{1}{v} + \frac{1}{( - 2 0 )} = \frac{1}{1 5}

\frac{1}{v} = \frac{7}{6 0} \Rightarrow v = \frac{6 0}{7}

v = 8.57 \approx 8.6 c m

The image is formed behind the mirror at a distance of

8.6 c m

m = \frac{h _{2}}{h _{1}} = - \frac{v}{u}

\to \frac{h _{2}}{5} = - \frac{8 . 5 7}{2 0}

∴ h_{2} = 2.175 c m

Thus the image is

2.175 c m

high, virtual and erect.

Hence, this is the answer.