Question

An object 2cm high is placed at a distance of 16cm from a concave mirror which produces a real image of 3cm high.
(i) find the position of image.
(ii) find focal length. ​

Answer 1

Answer :

• Position of the image is (-24) cm.
• Focal length of the mirror is (-9.6) cm.

Explanation :

Given :

• Object distance, u = 16 cm
• Height of object, hᵢ = 2 cm
• Height of the image, hₒ = 3 cm

To find :

• Position of the image, v = ?
• Focal length of the mirror, f = ?

Knowledge required :

• Mirror formula :

⠀⠀⠀⠀⠀⠀⠀⠀⠀1/f = 1/v + 1/u⠀

[Where : f = Focal length of the mirror, v = Image distance of the mirror, u = object distance]

• Formula for magnification :

⠀⠀1)⠀m = - v/u⠀⠀⠀⠀⠀⠀⠀2) m = hᵢ/hₒ⠀

[Where : m = Magnification of the mirror, v = Image distance, u = object distance, hᵢ = Height of the image, hₒ = Height of the object]

• By sign convention, in a concave mirror the object distance and the focal length are always negative.

Solution :

To find the magnification of the mirror :

By using the second formula magnification and substituting the values in it, we get :

⠀⠀=> m = hᵢ/hₒ

⠀⠀=> m = 3/2

⠀⠀=> m = 1.5

⠀⠀⠀⠀∴ m = 1.5

Hence the magnification of the mirror is 1.5.

To find the image distance of the mirror :

By using the first formula for magnification and substituting the values in it, we get :

⠀⠀=> m = - v/u

⠀⠀=> 1.5 = - (-v)/(-16)

⠀⠀=> 1.5 = - v/16

⠀⠀=> 1.5 × 16 = - v

⠀⠀=> 24 = - v

⠀⠀=> -24 = v

⠀⠀⠀⠀∴ v = -24 cm

Hence the image distance of the mirror is (-24) cm.

To find the focal length of the mirror :

By using the mirror formula and substituting the values in it, we get :

⠀⠀=> 1/f = 1/v + 1/u

⠀⠀=> 1/f = 1/(-24) + 1/(-16)

⠀⠀=> 1/f = (-2 - 3)/48

⠀⠀=> 1/f = -5/48

⠀⠀=> 48/-5 = f

⠀⠀=> -9.6 = f

⠀⠀⠀⠀∴ f = -9.6 cm

Hence the focal of the mirror is (-9.6) cm.