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
h'/ h = -s'/s or -s' = 0.5(s) = 0.5 (10 cm)
s' = - 0.5(10 cm) = - 2 cm