Question

A 50 cm tall object is at a very large distance from a diverging lens. A virtual, erect and diminished image of the object is formed at a distance of 20 cm in front of the lens. How much is the focal length of the lens?

Answer 1

Given: Height of object (h_o​) = 50 cm

Image distance (v) = −20 cm    ...(Negative because it is virtual)

Object distance (u) = ?

By using the lens formula:

`1/f = 1/v + 1/u`

⇒ `1/f = 1/-20 + 1/u`    ...(i)

Since the object is at a very large distance, we can assume u = ∞. Thus:

`1/u` = 0

Substituting this values in equation (i), we get,

⇒ `1/f = 1/-20 + 0`

⇒ `1/f = 1/-20`

⇒ f = −20

∴ The focal length of the lens is −20 cm.