Can you Lens me a dollar? Times are tough....

Multiple Choice problems from the handout:
1. B

2. E

3. C

4. D


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 tomorrow!

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, so the place where the rays intersect the principal axis will be closer to the center of the lens.

1997B5.

a) converging - the image is located behind the lens, which means image distance is positive. This can only occur with a converging lens.

b) 1/f = 1/di + 1/do so f = 22.5 mm

c) The rays will probably intersect near the 90 mm mark, but the 22.5 mm focal length makes it difficult to get it exactly. Don't worry if it isn't exactly on the 90 mm mark.

d) The image is inverted, real, and larger than the object. You can also confirm the size change by noticing that M = -di/do = -90/30 = -3.

e) The resulting image should be inverted and to the left of the mirror.