An object 10.0 cm tall is placed at x=0.0 cm and a converging lens of focal length 30.0 cm is placed at x=40.0 cm. a diverging lens of focal length 30.0 cm is then placed at x=140.0 cm. (a) find the location of the image (along the x-axis) and the height of the image formed by the converging lens. (b) find the location of the image (along the x-axis) and the height of the image formed by the diverging lens.

a ) 120 cm

size = 30 cm

b ) 12 cm from diverging lens

size = 12 cm

Explanation:

For converging lens

object distance u = - 40 cm

Focal length f = 30 cm

From lens formula

m=

m=

=3

Size of image = size of object x Magnification

= 3 x 10 = 30cm

For diverging lens

object distance u = 140 - 120 = 20 cm

focal length f = -30 cm

frac{1}{v}+\frac{1}{20}=-\frac{1}{30}

v = -12 cm

m = 12/20 = 0.6

size of final image = .6 x 30 = 18 cm

because growing process had stoped after 4-5 year

i believe the answer is electric potential.

explanation: