Question

an object 2cm high produces a real image 2.5cm high when placed at a distance of 15cm from a concave mirror find the position of image and focal length of the concave mirror

Answer 1

v = +18.75cm and f = -0.12cm

Answer 2

Objects are kept on left side so u = -15cm

Let the distance of the image be v from the mirror

We know the magnification formula m = -$\frac{v}{u}$ = $\frac{h_{2}}{h_{1} }$

where $h_{2}$ and $h_{2}$ are heights of image and object respectively.

Real images are inverted and hence $h_{2}$ is negative.

$h_{2}$ = -2.5cm

So $\frac{-2.5}{2} = \frac{-v}{(-15)}\\v = \frac{-2.5}{2}$x 15

v = -18.75 cm

We know the formula for mirrors

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

$\frac{1}{f} = \frac{1}{-18.75} + \frac{1}{-15}$

Taking LCM

$\frac{1}{f} = \frac{-33.75}{281.25}$

Taking inverse

$f = -8.33 cm$