A concave lens of focal length 30 cm forms an image of an object placed at 60 cm from it. The position and nature of the image formed will be :

Correct option is A

The correct answer is (A)   -20 cm, Virtual and erect

Given:

• f = focal length = - 30 cm
  
• u = object distance = 60 cm 
  
• v = image distance = ?
  

Formula used:  ​$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$

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

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

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

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

v = - 20 cm

Hence, the image formed at a distance - 20 cm, Virtual and erect.