A converging lens L1 has a focal length of 10 cm. A 5cm tall object is located 1

Question

A converging lens L1 has a focal length of 10 cm. A 5cm tall object is located 15cm to the left of L1.

(a) Construct a ray diagram indicating the position and vertical orientation of the image produced by lens L1. Calculate this position and see that it matches your diagram.

(b) A diverging lens L2 with a focal length of -20 cm is placed 40cm on the right of lens L1.

The image in (a) above now serves as the object for lens L2. Construct a ray diagram to find the image (final image) produced by lens L2. Calculate its position and see that it matches your diagram.

Is this final image real or virtual? Can it be shown on a screen?

(c) Move the diverging lens L2 to a new location which is 20 cm to the right of lens L1. In the space below, sketch the two lenses, the original object, and the image produced by lens L1. Calculate the location of the final image. (Ray diagram is not required). Is this final image real or virtual? Can it be shown on a screen?

Explanation / Answer

a) 1/u =+1/v = 1/f
For the first lens
1/v = 1/ 10- 1/ 15
v = 30 cm.

For the second lens
1/v'= 1/f' - 1/u'

1/v' = -1/20 -1/ 40

v' = 13.33 cm behind L2

real inverted image
2.
1/v'= 1/f' - 1/u'

u' = 80 -20 = 60 cm
f = -20cm

1/v' = - 1/20 -1/ 60

v' = - 15 cm infront of L2
virtual upright image is formed.
-----------------------------
Diverging lens alone cannot produce real image and hence
Cannot form image on a screed.

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