Question

An object of height 10 cm is placed 60 cm from a 30 cm focal length of a diverging lens. find the nature of image, height of the image ,magnification , image distance

Answer 1

Answer:

Image is virtual, erect and diminished.

Height of image is 10/3 cm

Magnification is 1/3

Image distance is 20 cm

Explanation:

For lens,

m=f/f+u = - 30/-30+(-60) = 1/3

m is positive means image is erect, erect image of real object means image is virtual

height of image = 10*1/3=10/3 cm

m = v/u =1/3=v/-60 => v= - 20 cm, v is negative it confirms image is virtual.