An object is placed at a distance of 60 cm from a concave lens of focal length 30 cm.(i) Use the lens formula to find the distance of the image from the lens.(ii) List four characteristics of the image (nature, position, size, erect/inverted) formed by the lens in this case.(iii) Draw ray diagram to justify your answer of the part (ii).

Object distance = $u$ = $-$ 60 cm 

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

To find: Distance of the image from the lens, $v$.

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-

$\frac{1}{v}=\frac{1}{(-30)}+\frac{1}{(-60)}$

$\frac{1}{v}=-\frac{1}{30}-\frac{1}{60}$

$\frac{1}{v}=\frac{-2-1}{60}$

$\frac{1}{v}=-\frac{-3}{60}$

$\frac{1}{v}=-\frac{1}{20}$

$v=-20cm$

Thus, the distance of the image from the lens, $v$ is 20cm.

(ii) The four characteristics of the image (nature, position, size, erect/inverted) formed by the lens in this case 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 20 cm from the optical centre of the concave lens on the same side of the object.

(iii) The ray diagram to justify the answer of the part (ii) is given below: