Two converging lenses (f1 = 9.00 cm and f2 = 6.00 cm) are separated by 18.0 cm.

Question

Two converging lenses (f1 = 9.00 cm and f2 = 6.00 cm) are separated by 18.0 cm. The lens on the left has the longer focal length. An object stands 12.0 cm to the left of the left-hand lens in the combination.

a) Graphically locate the image. Use a full sheet of paper and draw to scale. Use a ruler!

b) Calculate the final image distance relative to the lens on the right.

c) Calculate the overall magnification.

d) Is the final image real or virtual?

e) Is the final image upright or inverted?

f) Is the final image larger or smaller?

Explanation / Answer

Now for the first lens

1/12+1/v =1/9

1/v =1/9-1/12 =0.111-0.0833=0.0277

then v =36.10cm

Now object distance for the second lens is u =18-36.10 =-18.10cm

1/-18.10+1/v =1/6

1/v =1/6+1/18.10

v =(18.10)(6)/(18.10+6)=108.606/24.1=4.506cm

Therefore the final image is located at a distance of v =4.506cm to the right of the right side lens

b)

Magnification is given by

m =(36.10/12)(-4.506/18.10) =-0.7489

c)

Here the magnification is negative therefore the image is inverted and real

d)

And the magnification is <1 , therefore the final image will be smaller

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