Question

An object of height 5 cm is placed at the distance of 15 cm in front of a concave mirror of focal length 10 cm. Find the position nature and size of image formed.

Answer 1

GIVEN:-

OBJECT'S,

=>Type of mirror = Concave

=>Height = 5cm

=>Distance = 15cm

=>Focal length = 10cm

TO FIND:-

IMAGE'S

=>Distance

=>Height

=>Nature

FORMULA USED:-

For distance,

=> 1/U + 1/V = 1/F

For height,

M => -V/U = H`/H

UNDERSTANDING THE CONCEPT:-

According to the question, we know all the given information. So, from the information and using the formula we can find the distance and height.

LET'S DO IT!!

DISTANCE = 1/U + 1/V = 1/F

$= > \dfrac{1}{ - 15} + \dfrac{1}{v} = \dfrac{1}{ - 10}$

$= > \dfrac{1}{v} = \dfrac{ - 1}{10} + \dfrac{1}{15}$

$= > \dfrac{1}{v} = \dfrac{ - 3 + 2}{30}$

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

Therefore, the distance of image is -30cm.

HEIGHT = -V/U = H`/H

$= > \dfrac{ - ( - 30)}{ - 15} = \dfrac{h` }{h}$

=> -2 = h`/5

=> H` = -10

Therefore, the height of the image is -10cm.

NATURE OF THE IMAGE = Real and enlarged.