Question

A 6cm tall object is placed perpendicular to the principal axis of a concave mirror of focal length 30 cm the distance of the object from the mirror is 45 CM use mirror formula to determine the position nature and size of the image formed also draw labelled diagram

Answer 1

height of object=6cm
f= -30 cm
u=-45cm
v=?
nature=?
height of image=?
By using mirror formula
1/f=1/v+1/u

Answer 2

Answer:

The distance of the image is 90 cm, size of the image is 12 cm and enlarged. The image is real and inverted enlarged.

Explanation:

Given that,

Height of the object h = 6 cm

Focal length f = -30 cm

The distance of the object u = -45 cm

Using mirror formula,

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

$\dfrac{1}{-30}=\dfrac{1}{v}+\dfrac{1}{-45}$

$\dfrac{1}{v}=-\dfrac{1}{90}$

$v = -90\ cm$

The distance of the image  is 90 cm.

The size of the image is

$m= -\dfrac{h'}{h}=-\dfrac{v}{u}$

$\dfrac{h'}{6}=\dfrac{-90}{-45}$

$h'=12\ cm$

The magnification is

$m=-\dfrac{h'}{h}$

$m=-\dfrac{12}{6}$

$m=-2$

The magnification is 2.

Hence, The distance of the image is 90 cm, size of the image is 12 cm and enlarged. The image is real and inverted enlarged.