Question

Two lenses are separated by L = 50.0 cm. The first lens (light coming from the left.) has a focal length f1 = 24.0 cm and the second lens has a focal length f2 = 18.0 cm. An object is located 40.0 cm (s1) to the left of the first lens. Calculate the position of the two images formed by the lens system. State whether each image is real or virtual. Repeat this problem with L = 10.0 cm.

Answer 1

Given:-

1]

a] With L=50cm

f₁ = 24.0 cm

f₂ = 18.0 cm.

s₁= 40 cm

Applying the lens equation

1/f₁= 1/s₁+1/v₁    [here v₁ is the image due to first lens]

we get V₁=60 cm-----------This lies beyond the second lens

s₂=50-60=-10cm

Applying the lens equation

1/f₂= 1/s₂+1/v₂    [here v₂ is the image due to second lens]

v₂ = 6.43 cm

Magnification of first lens,m₁= v₁/s₁=60cm/40cm=1.5 [m₁is positive , hence erect and virtual image]

Magnification of first lens,m₂= v₂/s₂=6.43cm/-10cm=-0.64 [m₂ is negative, hence real and inverted image]

b] With L=10cm

f₁ = 24.0 cm

f₂ = 18.0 cm.

s₁= 40 cm

Applying the lens equation

1/f₁= 1/s₁+1/v₁    [here v₁ is the image due to first lens]

we get V₁=60 cm-----------This lies beyond the second lens

s₂=10-60=-50cm

Applying the lens equation

1/f₂= 1/s₂+1/v₂    [here v₂ is the image due to second lens]

v₂ = 13.24 cm

Magnification of first lens,m₁= v₁/s₁=60cm/40cm=1.5 [m₁is positive , hence erect and virtual image]

Magnification of first lens,m₂= v₂/s₂=13.24cm/-50cm=-0.26 [m₂ is negative, hence real and inverted image]