Question

A 10 cm tall object is placed perpendicular to the principal axis of a concave mirror of focal length 12 cm. The distance of the object from the lens is 18 cm. Find the nature, position and size of the image formed

Answer 1

Explanation:

Given = object height,h = 10cm

focal length of concave mirror = -12cm

object distance = -18cm

To find = nature, position and size of the image formed

Using the mirror formula,

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

$\boxed{ \frac{1}{ - 12} = \frac{1}{v} - \frac{1}{18} }$

$\boxed{ \frac{1}{ - 12} + \frac{1}{18} = \frac{1}{v} }$

$\boxed{ \frac{3 - 2}{ - 36} = \frac{1}{v} }$

$\boxed{v = - 36}$

Now,

Linear magnification = h'/h° = -v/u

h'/10 = -(-36)/-18

h'/10 = -2

h' = -20 cm

Now, magnification = -v/u

m = -(-36)/-18

m = -2

Since the magnification is negative hence the image formed will be real and inverted and highly diminished.