Question

3: An object of height 2.6 cm is placed 26 cm in front of a diverging lens of focal length 15 cm. Behind the diverging lens, and 11 cm from it, there is a converging lens of the same focal length.

(a) Find the location of the final image, in centimeters beyond the converging lens.

s''=

(b) What is the magnification of the final image? Include its sign to indicate its orientation with respect to the object.

m_total=

Answer 1

For diverging lens

f = - 15 cm, u = - 26 cm

By lens formula

1/v - 1/u = 1/f

1/v - 1/(-26) = 1/(-15)

v = - 9.5 cm

By magnification = hi/ho = v/u

hi/2.6 = (-9.5)/(-26)

hi = 0.95 cm

This image acts as object for converging lens

So u = - (9.5+11) = -20.5 cm

f = 15 cm

By lens formula

1/v - 1/u = 1/f

1/v - 1/(-20.5) = 1/15

v = 55.94 cm

hi for 1st lens acts as ho for 2nd

By magnification = hi/ho = v/u

hi/0.95 = 55.94/(-20.5)

hi = 2.59 cm

So magnification total = 2.59/2.6 = 0.997