## Tuesday, June 21, 2011

### Solutions to Part 19 Problems (Diverging lenses) The problems were:

1. Using a graphical construction find the image formed by a diverging lens with a focal length of 30 cm when the object is 60 cm from the lens.

The solution is given in the accompanying diagram, with the image seen at a position of s' = -20 cm.

2. A diverging lens has a focal length of 20 cm.

(a) Where is the image if the object is 20 cm from the lens?

we need s':

We have: f = - 20 cm

s = 20 cm

The thin lens equation for diverging lenses is:

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

and in this case, we will write:

1/s' = -1/f -1/s = - 1/20 - 1/20

1/s' = - 1/ 12

s' = - 12 cm

(b) If the object is 4 cm high, how high is the image? (Recall the lateral magnification of a thin lens - with sign conventions in place is:

M = h'/ h = -s'/s

We need h' and know h = 4cm, s' = -12 cm and s = 20cm.

h' = (-s'/s) h

h' = (12/20)4 cm = 48/20 cm = 2.4 cm

3. A diverging lens placed 10 cm from an object produces an image which is half the size of the object. Find the focal length of the lens.

We have: h'/h = 1/2 = 0.5

s = 10 cm

But:

h'/ h = -s'/s or -s' = 0.5(s) = 0.5 (10 cm)

s' = - 0.5(10 cm) = - 2 cm