Question

An erect image 2.0 cm high is formed 12 cm from a lens, the object being 0.5 cm high . Find the focal length of the lens.​

Answer 1

✬ Focal Length = 4 cm ✬

Explanation:

Given:

• Height of image 2 cm.
• Distance of image from lens is 12 cm.
• Height of object is 0.5 cm.

To Find:

• What is the focal length of lens ?

Formula to be used:

• m = v/u = h'/h
• 1/f = 1/v – 1/u

Solution: Here we have

• h' = 2 cm
• h = 0.5 cm
• v = –12 cm {as image is erect}
• u = ?

Substituting the values on formula

$\implies{\rm }$ –12/u = 2/0.5

$\implies{\rm }$ –12 × 0.5 = 2 × u

$\implies{\rm }$ –6 = 2u

$\implies{\rm }$ –6/2 = u

$\implies{\rm }$ –3 = u

So, the object distance is u = –3 cm.

Now by lens formula

$\implies{\rm }$ 1/f = (1/–12) – (1/–3)

$\implies{\rm }$ 1/f = (–1/12) – (–1/3)

$\implies{\rm }$ 1/f = –1/12 + 1/3

$\implies{\rm }$ 1/f = –1 + 4/12

$\implies{\rm }$ 1/f = 3/12

$\implies{\rm }$ 1/f = 1/4

$\implies{\rm }$ f = 4 cm

Hence, focal length of the lens is 4 cm.

• Also this lens is concave lens.

• Nature of images formed by these lens are virtual and erect.

Answer 2

Answer:

$\huge\rm{\fbox\pink{Solution}}$

→ Image is erect

Height of image(h2) = 2.0cm

Height of Object(h1) = 0.5cm

Distance of the image = 12cm Becomes -12cm (Because the left side of convex lens is always negative)

f = ?

→ m = v/u = h2/h1

→ -12/4 = 2/0.5 (Cross multiply)

→  -12 x 0.5 = 24

→  24 = -6

→  4 = -6/2

= -3cm

So the required answer is -3cm