Question

An object 4cm in size is placed at a distance of 25 cm from a concave mirror of focal length 15 cm. Find the position, nature and height of the image?

Answer 1

Answer:

size of object , H₀ = 4 cm

object distance , u = -25 cm

focal length of concave mirror , f = -15 cm

Use formula ,

1/v + 1/u = 1/f

⇒1/v + 1/-25 = 1/-15

⇒1/v = 1/-15 + 1/25 = (-25 + 15)/375 = 1/-37.5 cm

⇒ v = - 37.5 cm ,

∴ image is formed object side , 37.5 cm from pole of concave mirror.

Now, use formula,

Hi/H₀ = -u/v

Where Hi is the size of image .

Now, Hi/4cm = -(-37.5cm)/(-25cm)

Hi = -4 × 37.5/25 cm = -6cm

# size of image = 6cm  

the image is formed behind the mirror. Since the Distance is Negative, thus the Image is Real.

hope it will help u dear!

Answer 2

Answer:

$\sf h_i=6 \ cm$

Explanation:

Given :

Object distance u = 25 cm

Focal length f = 15 cm

We have mirror formula :

1 / f = 1 / v + 1 / u

- 1 / 15 = 1 / v - 1 / 25

1 / v = 1 / 25 - 1 / 15

5 / v = 1 / 5 - 1 / 3

5 / v = ( 3 - 5 ) / 15

5 / v = - 2 / 15

v = - 75 / 2 = - 37.5 cm

We know :

h_i / h_o = - v / u

Where i and o represent image and object

h_i = - ( - 37.5)  × 4 / 25

h_i = 6 cm

Nature of the image are as :

Real , inverted and magnified.