An object is placed at a distance of 15 cm from a concave lens of focal length 30 cm. List four characteristics (nature, position, etc.) of the image formed by the lens.

Object distance = $u$ = $-$ 15 cm 

Focal length = $f$ = $-$ 30 cm

To find: Four characteristics (nature, position, etc.) of the image formed by the lens.

Solution:

Using lens formula, we get-

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

It can be rearranged as-

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$

Substituting the given values we get-

$\frac{1}{v}=\frac{1}{(-30)}+\frac{1}{(-15)}$

$\frac{1}{v}=-\frac{1}{30}-\frac{1}{15}$

$\frac{1}{v}=\frac{-1-2}{30}$

$\frac{1}{v}=-\frac{3}{30}$

$\frac{1}{v}=-\frac{1}{10}$

$v=-10cm$

The four characteristics (nature, position, etc.) of the image formed by the lens are:

1. Image formed is virtual.

2. Image is erect.

3. Image is diminished (smaller than the object).

4. Image is formed at a distance of 10 cm from the optical centre of the concave lens on the same side of the object.

Explanation

Diverging Lens or Concave Lens - It is a lens that possesses at least one surface that curves inwards in the middle. In other words, it is thin across the middle and thick at the upper and lower edges, because of which the light that enters the lens, get spread out, or diverge, which results in forming a smaller image. Due to this effect, it is also called a negative lens or a diverging lens.

The image formed by a concave lens is virtual & erect, which means it will appear to be farther away than it actually is, and therefore smaller than the object itself.