# Question 7c5e0

Jul 10, 2016

Given

$f = \text{The focal length of convex mirror} = 20 c m$

$u = \text{Object distance} = - 6 m = - 600 c m$

v="Image distance"=?

${h}_{o} = \text{Height of the object} 1.6 m$

${w}_{o} = \text{Width of the object} = 2 m$

color(blue)("The Mirror Formula":#

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

$\implies \frac{1}{v} - \frac{1}{600} = \frac{1}{20}$

$\implies \frac{1}{v} = \frac{1}{20} + \frac{1}{600} = \frac{31}{600}$

$\implies v = \frac{600}{31} \approx 20 c m$

$m = \text{Magnification} = - \frac{v}{u} = \frac{1}{31}$

So

${h}_{i} = \text{Height of the image} = m \times {h}_{o} = \frac{160}{31} c m$

${h}_{i} = \text{Width of the image} = m \times {w}_{o} = \frac{200}{31} c m$

So the image of the car will be seen $\approx 20 c m$ behind the mirror and its height will be $\frac{160}{31} c m$ and its width will be $\frac{200}{31} c m$