Question

. A diverging lens (f = –11.0 cm) is located 19.0 cm to the left of a converging lens (f = 30.0 cm). A 2.90-cm-tall object stands to the left of the diverging lens, exactly at its focal point. (a) Determine the distance of the final image relative to the converging lens. (b) What is the height of the final image (including the proper algebraic sign)?

Answer 1

(a).

1/s + 1/s' = 1/f

1/11 + 1/s' = 1/(-11)

s' = -5.5 cm

u = 19 - s' = 19 - (-5.5) = 24.5 cm

1/u + 1/v = 1/F

1/24.5 + 1/v = 1/30

v = -133.6 cm (in front of converging lens)

(b).

M₁ = s'/s = -5.5/11 = -0.5

M₂ = v/u = (-133.6)/24.5 = -5.45

M_total = M₁ M₂ = (-0.5)(-5.45) = 2.725

height of final image is

h' = (M_total) h = (2.725)(2.90) = 7.9 cm