For the following object distances and focal lengths of thin
Solution
(a).Focal length f = 20 cm
Focal length is positive.So, it is convex lens
Object distance u = 40 cm
From the relation ( 1/v ) -( 1/ u ) = (1/f)
(1/v) -( 1/-40) = (1/20)
(1/v) = (1/20 ) -(1/40)
=(2-1)/40
= 1/40
v = 40 cm
Magnification m = v /u = 40 cm / -40 cm = -1
So, the image is real ,inverted and in same size.
(b).
Object distance u = 20 cm
From the relation ( 1/v ) -( 1/ u ) = (1/f)
(1/v) -( 1/-20) = (1/20)
(1/v) = (1/20 ) -(1/20)
= 0
= 1/0
v = infinity
Magnification m = v /u = infinity / -40 cm = infinity
i.e., highly magnified.
So, the image is at inifinity and highly magnified.
(c).
Object distance u = 40 cm
Focal length f = -20 cm
Focal length is negative.So it is concave lens.
From the relation ( 1/v ) -( 1/ u ) = (1/f)
(1/v) -( 1/-40) = (1/-20)
(1/v) = (1/-20 ) +(1/-40)
=(-2-1)/40
= -3/40
v = -40 /3 = -13.333cm
Magnification m = v /u = -13.333 cm / -40 cm = 1/3
So, the image is erect ,virtual and dimished.
(d).
Object distance u = 20 cm
From the relation ( 1/v ) -( 1/ u ) = (1/f)
(1/v) -( 1/-20) = (1/-20)
(1/v) = (1/-20 ) +(1/-20)
= -2/20
v = -10 cm
Magnification m = v /u = -10 / -20 cm = 1/2
So, the image is erect ,virtual and dimished.

