# An object of length 2cm is placed at a distance of 2.5*f from a concave mirror where f is its focal length. what is the length of the IMAGE?

## length is provided and not the hieght.just need the length of image. PLease as fast as possible.

Nov 12, 2016

The conjugate foci relation of spherical mirror is

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \ldots \ldots \left(1\right)$

Where

v->"Image distance"=?

$u \to \text{Object distance} = - 2.5 f$

$f \to \text{Focal length} = - f$

So inserting the values in equation (1)

$\frac{1}{v} - \frac{1}{2.5 f} = - \frac{1}{f}$

$\implies \frac{1}{v} - \frac{1}{2.5 f} = - \frac{1}{f}$

$\implies \frac{1}{v} = - \frac{1}{f} + \frac{1}{2.5 f}$

$\implies \frac{1}{v} = - \frac{1}{f} + \frac{2}{5 f} = \frac{- 5 + 2}{5 f} = - \frac{3}{5 f}$

$\implies v = - \frac{5 f}{3}$

Negative means real and inverted image

Now Magnification $m \equiv \frac{v}{u} = \frac{\frac{5 f}{3}}{\frac{5 f}{2}} = \frac{2}{3}$

So $\text{size of image"=mxx"size of object} = \frac{2}{3} \times 2 c m = \frac{4}{3} c m$