Lenses - Solutions

Multiple Choice problems from the handout:
1. B
2. E
3. C

Questions:
20. If the object distance is very large, then by the thin lens equation, 1/d0 approaches zero. This means the image distance is equal to the focal length. This is where the film must be placed to form an image.

Problems:
48b. 390 mm

50. Since the image is on the other side of the lens, it must be a real image. This means that the image distance is positive. Using the thin lens equation, the focal length must be 39.8 cm.

52. 72 cm on the same side of the lens as the stamp (virtual image, negative image distance). Magnification = -4

60.
a) 15.2 cm, real image (opposite side of the lens as object), height is -2.78 mm
b) -12.1 cm, virtual image (negative image distance), height is +2.22 mm

83. Since a diverging lens has a negative focal length, by the thin lens equation, 1/di can never be positive. If di is negative, the image is on the same side as the object, so it will be virtual. The only way for di to be positive is if d0 for the object is negative. This can occur, and we will discuss HOW it can occur in class Monday!



AP Problems:
1989B6.
a) real
b) di = 30 cm
c) M = -0.5
d) See the image to the left.
e) focal length decreases - by Snell's law, the rays will be bent more as they enter and exit the lens.













1986B5
a) 7.4%
b) 1.250 x 10⁶ W
c) We'll get to this problem in a couple weeks....
d)