Question

An object 4cm high is place at distance of 27cm front of a convex lens of focal length 18cm. Find the position,nature and size of the image formed

Answer 1

Given:

• Distance = -27cm
• Focal length = 18 cm
• Height = 4 cm

To find:

• Position of image
• Size of image

Solution:

Firstly finding position

Using

- 1/u + 1/v = 1/f

Where

• v : position of image = ?
• u : Distance = -27
• f : Focal length = 18

Substituting the values we have

→ 1/v + 1/-(-27) = 1/18

→ 1/v = 1/27 - 1/18

→ 1/v = 1/54

By cross multiplication

→ v = 54 cm

Hence , position of image = 54cm

Using magnification formula

hᵢ/hₒ = v/u

Simplifying

hᵢ = v × hₒ/u

Where

• hᵢ : height of image = ?
• hₒ : height of object = 4
• v : position of image = 54
• u : Distance = -27

substituting the values we have

♦ hᵢ = 54 × 4/-27

♦ hᵢ = -2 × 4

♦ hᵢ = -8 cm

Hence, height of image = - 8cm

Answer 2

Given :

▪ Height of object = 4cm

▪ Distance of object = 27cm

▪ Focal length of lens = 18cm

▪ Type of lens : convex

To Find :

▪ Position, size and nature of image.

CalculatioN :

✴ Position of image :

$\implies\sf\:\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\ \\ \implies\sf\:\dfrac{1}{v}-\dfrac{1}{(-27)}=\dfrac{1}{18}\\ \\ \implies\sf\:\dfrac{1}{v}=\dfrac{1}{18}-\dfrac{1}{27}\\ \\ \implies\sf\:\dfrac{1}{v}=\dfrac{3-2}{54}\\ \\ \implies\underline{\boxed{\bf{\red{v=54\:cm}}}}\:\gray{\bigstar}$

✴ Size (Height) of image :

$\Rightarrow\sf\:m=\dfrac{v}{u}=\dfrac{h_i}{h_o}\\ \\ \Rightarrow\sf\:m=\dfrac{54}{(-27)}=\dfrac{h_i}{h_o}\\ \\ \Rightarrow\sf\:m=-2=\dfrac{h_i}{4}\\ \\ \Rightarrow\underline{\boxed{\bf{\blue{h_i=-8\:cm}}}}\:\gray{\bigstar}$

✴ Nature of image :

• Real
• Inverted
• Enlarged