Light Reflection/Refraction Problem

[firyace]

I have been given this problem in my first week of classes in Physics:

The McMath-Pierce Solar Telescope at Kitt Peak National Observatory in Arizona is a reflecting telescope (i.e., a concave mirror) of diameter 1.61 m and a focal length of 87.0 m. There is a screen in the focal plane of the mirror, so the image of the Sun can be viewed. Assume for this problem that there are no other mirrors, and the screen and image are located at the actual focal plane of the mirror. (In reality, two other mirrors are used to reflect the image to a fixed location in an observing room.)

a. The angular diameter of the Sun in our sky is 0.535 degree. (This is the angle taken up by the diameter of the Sun, as seen from the Earth.) What is the diameter of the image of the Sun, in metres? Drawing a diagram of the situation might be helpful.

b. At a time when the telescope is pointing at a blank patch of sky (not at the Sun), a Red-tailed Hawk glides towards the west across the optic axis of the telescope at a distance of 410 m from the mirror. The hawk is 56.0 cm in length (beak to tail), and is moving perpendicularly to the optic axis at a speed of 3.60 m/s.

How far from the mirror would the hawk's image form if the screen weren't there (so that the image actually could form)?
How long would the image be (beak to tail)?

What would be the speed of the image, in m/s?

iv. In what direction would the image be traveling?

v. Since the screen prevents the image from forming, what does an astronomer watching the screen really see? (e.g., nothing? a sharp-edged shape bigger than the image? a blur? or…?)


I can't figure out how to do this, can somebody help?

Thanks

 

[berkeman]

You need to show us your work so far. What equations should you use for a reflecting telescope?

 

[piercas]

for reaching out for assistance with this problem. I am happy to help you understand the concepts of light reflection and refraction in relation to this specific scenario.

First, let's address part a. In order to determine the diameter of the image of the Sun, we need to use the formula for angular magnification: M = -f/fe. In this case, the focal length (f) is given as 87.0 m and the angular diameter (fe) is 0.535 degree. Plugging these values into the formula, we get M = -87.0/0.535 = -162.62. This means that the image of the Sun will be 162.62 times larger than its actual size. To find the diameter of the image, we multiply the actual diameter of the Sun (1.61 m) by the magnification factor, giving us a diameter of 261.15 m.

Now, let's move on to part b. To determine the distance from the mirror where the hawk's image would form, we can use the formula for the magnification of a real image: M = -di/do, where di is the distance of the image from the mirror and do is the distance of the object (in this case, the hawk) from the mirror. We know that M = -162.62, do = 410 m, and the length of the hawk (lh) is 56.0 cm = 0.56 m. Plugging these values into the formula, we get -162.62 = -di/410, which gives us di = 70.91 m. This means that the hawk's image would form 70.91 m from the mirror if the screen were not there.

To find the length of the image, we can use the formula for linear magnification: M = -di/do = -lh/li, where li is the length of the image. Solving for li, we get li = lh/do * di = (0.56/410) * 70.91 = 0.0967 m. This means that the hawk's image would be 9.67 cm long.

To determine the speed of the image, we can use the formula for image velocity: vi = M * vo, where vi is the image velocity, M is the magnification, and vo is the object velocity. In this case, M = -162.