Question

1. A converging lens of focal length 20 cm is at the 0 cm mark of a long ruler. An object is placed 30 cm to the left of the lens. The object is 2 cm tall.

1. Where is the image formed and what is its height and it is inverted or upright?
2. Is the image from part a real or virtual? Explain
3. A second lens of focal length 5.0 cm is added at the 70 cm mark. Using the first image at its object, where is the final image formed?

Answer 1

a) using lens equation

1/f =1/p +1/q

1/p=1/f -1/q

Here f=20cm , q=30 cm

1/p =1/20 -1/30 =(3-2)/60 =1/60

p=60 cm (image location ) right side of the lens.

We know that the magnification

M=h'/h =-p/q

h' =-ph/q

h' = - 60×2/30

h' =-4 cm (image height)

Since h' is negative so the image is inverted .

b) the image is real . Since the image is right side of the lens . Or the image distance p is positive so the image is real.

c) similarly

1/p=1/f -1/q

Here f=5cm , q=(70-60) =10 cm ( since the first image is the object for the 2nd lens)

1/p=1/5 -1/10 = (2-1)/10

p=10 cm

p=10cm ( 10cm right of the 2nd lens).