A 5.85-mm-high firefly sits on the axis of, and 13.7 cm in front of, the thin lens A, whose focal length is 5.01 cm. Behind lens A there is another thin lens, lens B, with focal length 25.9 cm. The two lenses share a common axis and are 62.5 cm apart. 1. Is the image of the firefly that lens B forms real or virtual? a. Real
b. Vrtual
2. How far from lens B is this image located (expressed as a positive number)?
3. What is the height of this image (as a positive number)?
4. Is this image upright or inverted with respect to the firefly?
a. Upright
b. Inverted

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

if a star is moving toward earth, its light waves appear to be squeezed together. hope it

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

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