Question

A concave lens of focal length 60 cm is used to form an image of an object of length 9 cm kept at a distance of 30 cm from it use lens formula to determine the nature , position , and length of the image formed . Also draw labelled diagram to show the image formation

Answer 1

focal length of concave lens, f = -60 cm

distance of object from the concave lens , u = -30 cm

use lens formula,

1/v - 1/u = 1/f

or, 1/v - 1/-60 = 1/-30

or, 1/v = 1/-60 + 1/-30 = 1/-60 + 2/-60

or, 1/v = 3/-60 = 1/-20 => v =-20cm

hence, v = 20 cm

now, magnification , m = v/u

or, m = (-20)/(-30) = 2/3 = length of image/length of object

2/3 = length of image/9cm

length of image = 6cm

hence, nature of image is virtual and diminished , image is placed between object and lens and length of image is 6cm.

Answer 2

Answer: See the picture from the abhi...

Explanation:focal length of concave lens, f = -60 cm

distance of object from the concave lens , u = -30 cm

use lens formula,

1/v - 1/u = 1/f

or, 1/v - 1/-60 = 1/-30

or, 1/v = 1/-60 + 1/-30 = 1/-60 + 2/-60

or, 1/v = 3/-60 = 1/-20 => v =-20cm

hence, v = 20 cm

now, magnification , m = v/u

or, m = (-20)/(-30) = 2/3 = length of image/length of object

2/3 = length of image/9cm

length of image = 6cm

hence, nature of image is virtual and diminished , image is placed between object and lens and length of image is 6cm.