Question

The distance (V) of a virtual image formed by a lens of focal length 15 cm never exceed a certain finite value, then this value will be ______.

Options

• less than 15 cm.

• between 15 cm to 30 cm.

• less than or equal to 30 cm

• less than or equal to 15 cm.

Answer 1

The distance (V) of a virtual image formed by a lens of focal length 15 cm never exceed a certain finite value, then this value will be less than or equal to 15 cm.

Explanation:

Given: Focal length (f) = −15 cm

By using the lens formula:

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

`1/v = 1/-15 + 1/u`

Since the object is real, u can range from 0 to −∞.

Object is at infinity (u = ∞):

`1/v = 1/-15 + 1/infty`

`1/v = -1/15 + 0`

v = −15 cm

This means the image is formed at the focus. The image distance is 15 cm.

Object is at the optical centre (u = 0):

`1/ v = 1/-15 + 1/0`

= ∞

This limit is not practical, but as the object moves from infinity towards the lens, the image moves from the focus towards the optical centre.
For any real object placed in front of a concave lens, the virtual image is always formed between the optical centre (O) and the principal focus (F).
Therefore, the image distance |v| will always be less than the focal length |f|.
So, 0 < ∣v∣ ≤ ∣f∣.
Given f = 15 cm, the image distance (v) will be less than or equal to 15 cm.