Answer:
1. The image is real
2. 5.85
3. h' = 3.05 mm
4. The image is upright
Step-by-step explanation:
1. Start with the first lens and apply 1/f = 1/p + 1/q
1/5.01 = 1/13.7 + 1/q
q = 7.90 cm
Since that distance is behind the first lens, and the second lens is 62.5 cm behind the first lens, that distance is 62.5 - 7.90 = 54.6 cm in front of the second lens, and becomes the object for that lens, thus,
1/25.9 = 1/54.6 + 1/q
q = 49.3 cm behind the second lens
Using that information, since q is positive, the image is real
2. Also, using that information, you have the second answer, which is 49.3 cm
The height can be found from the two magnifications.
m = -q/p
m1 = -7.9/13.7 = -.577
m2 = -49.3/54.6 = -.903
Net m = (-.577)(-.903) = .521
Then, m = h'/h
.521 = h'/5.85
3. h' = 3.05 mm
4. For the fourth answer, since the overall magnification is positive, the final image is upright