Question

1) where should an object be placed from a converging lens of focal length 20 cm., so as to obtain a real image of twice the magnification?

Answer 1

f = +20 cm
let (height of object) = h
so, m = 2h ( given that 2times magnified
real image )
m = 2h/h = 2
Again,
m = v/u
2 = v/u
》 v =2u
Now, 1/f = 1/v - 1/u
1/20 = 1/2u - 1/u
1/20 = 1-2/2u
1/20 = -1/2u
》 u = -10cm ( object distance )

Answer 2

Answer:U = -10,V = -20

Explanation:

Given

f =20cm

m =2

u =?

v =?

m= v/u

2=v/u

2u = v

1/f= 1/v-1/u

1/20=1/2u-1/u

1/20=1-2/u

1/20=-1/2u

2u = -20

u = -20/2

u = -10 cm

v = 2u

v =2×-10

v = -20 cm