Question

An object 3cm high is placed at a distance of 9cm in front of concave mirror of focal length 18cm. What is the positions nature and size of the image formed?​

Answer 1

behind the mirror,virtual,errect and magnified

Answer 2

Answer:

$\boxed{\sf The\: image\: is \:virtual,magnified \:and\: is \:formed\: behind\: the\: mirror.}$

$\boxed{The \:size \:of \:the \:image \: is \: 6 cm.}$

Step-by-step explanation:

Given

Size of the object(h) = 3 cm

Object distance, (u)= - 9 cm

Focal Length (f) = -18

Let the image distance be  v

Then by mirror formula,

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

Therefore,

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

$\frac{1}{v} = \frac{-1}{18} -\frac{-1}{9}$

LCM of 18 and 9 = 18

So,

 $\frac{1}{v} = \frac{-1+2}{18} = \frac{1}{18}$

∵ 1/v = 1/18

v = 18 cm

∵ The value of v is positive the image is virtual and is formed behind the mirror.

Magnification of lens is the ratio of the image distance and the object distance .

It can also be told as the ratio of height of the Size of object and the Size of image.

i.e, $m = \frac{-v}{u} or \frac{h'}{h}$

$m = \frac{-18}{-9} = 2$

$2 = \frac{h'}{h} \:\:\: \sf where\: h \:is \:the \:size \:of \:object \:and \:h'\: is \:the\: size\: of \:the\: image$

h  , is given as 3 cm in the question.

$2 = \frac{h'}{3}$

6 = h'

∴ h' or size of the image = 6cm