Answer:
1. The image is real
2. 5.85
3. h' = 3.05 mm
4. The image is upright
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
Frame of reference, in simplest terms, describes the state of motion of the observer. The frame of reference may also be described by using a set of coordinates, time and motion. We formulate all our equations and solve them using the frame of reference.
Answer:
45.1/3× 3.1/2 + 50.1/2× 2.1/2
=136/3. 7/2.+ 101/2× 5/2
=952/6 + 505/2
=( 952+ 1515)/6
=411.16
Explanation:
Make sure to read and go with it
+1
An electron has a negative charge so losing a charge of -1 from an uncharged, or neutral, atom will leave an ion with a positive charge.