Question

1. An object 3 cm high is placed 24 cm away from a convex lens of focal length 8 cm. Find by calculations, the position, height and nature of the image.
2. If the object is moved to a point only 3 cm away from the lens, what is the new position, height and nature of the image?
3. Which of the above two cases illustrates the working of a magnifying glass?

Answer 1

a. Height (h₁) = 3 cm

Object distance (u) = −24 cm

Focal length (f) = 8 cm

By using the lens formula:

`1/v - 1/u = 1/f`

⇒ `1/v - 1/-24 = 1/8`

⇒ `1/v = 1/12`

⇒ v = 12 cm

The image is formed 12 cm behind the lens. 

Magnification (m) = `v/u = h_2/h_1`

⇒ `12/-24 = h_2/3`

⇒ `1/-2 = h_2/3`

⇒ h₂ = −1.5 cm

The image is 1.5 cm high, real, and inverted.

b. Given: Height of object (h_o) = +3 cm

Object distance (u) = −3 cm

Focal length (f) = +8 cm

By using the lens formula:

`1/v - 1/u = 1/f`

⇒ `1/8 = 1/v - 1/-3`

⇒ `1/8 = 1/v + 1/3`

⇒ `1/v = 1/8 - 1/3`

⇒ `1/v = (3 - 8)/24`

⇒ `1/v = -5/24`

⇒ v = −4.8 cm

∴ The image is formed 4.8 cm in front of the lens.

Magnification (m) = `v/u`

⇒ m = `(-4.8)/-3`

⇒ m = +1.6 

Height of the object (h_i) = m × h_o

h_i = 1.6 × 3

h_i = 4.8 cm

c. Case (b) illustrates the working of a magnifying glass.