1
$\begingroup$

Suppose there is point source at the focal point of parabolic mirror. The observer's eyes are on the axis of the parabola. What does the observer see as he starts to move away from the light source along the axis? The amount of light (in terms of energy) would be same if he's 1 meter or 1 light year away, ignoring dissipation. What would be the difference in the eyes of the observer?

Problem: The mirror is perfectly parabolic with 1 meter in diameter. The light source is a homogeneous spheric bulb with 1 cm in diameter that produces 100 watt light. The center of the bulb is exactly located at the focal point of the mirror. What is the maximum distance from the mirror (on its axis) that the light is visible by naked eye? What are the relevant assumptions and equations for solving this problem?

$\endgroup$
3
$\begingroup$

Let's add one more feature to your setup: Let's add a mask that blocks all of the direct rays from the point source to the eye. The only light entering the eye is light that was reflected off the mirror.

Assuming that the mirror is a perfect paraboloid and, that the light source is a true point and, that the light source is located exactly at the focus of the mirror; then the geometry tells us that all of the rays reflected off the mirror will be perfectly parallel. The light should seem equally bright from a distance of one meter or, from a distance of one light year.

That's what they geometry says about ideal rays of light. I don't know what quantum physics says about actual, physical photons.


P.S., This is a degenerate case of the inverse square law at work. A bundle of parallel rays coming off a paraboloid mirror is indistinguishable from rays emitted by an infinitely distant point source. If you are infinitely distant from the light source, and you move one light year further away, then the inverse square law predicts zero change in the brightness.


P.P.S., If you remove that mask that I added at the top, then the light reaching the eye basically is the sum of the direct rays and the reflected rays. The reflected rays are parallel, and their contribution will not be affected by distance, but the direct rays will be divergent, and they will obey the inverse square law in the more familiar way.

$\endgroup$
  • $\begingroup$ Thank you. What is confusing to me is that the observer sees a disk of light, right? What happens to the apparent size of the disk as the observer moves away? $\endgroup$ – Asmani Apr 7 '18 at 8:25
  • $\begingroup$ Wait... Is a true point source even visible? Assuming that eye is also a point, the observer receives one out of infinite number of rays. Since the energy is finite, division by infinity means that the eye receives no light. $\endgroup$ – Asmani Apr 7 '18 at 8:29
  • 1
    $\begingroup$ @Asmani, The observer does not see a disk. The observer sees only an infinitessimal point reflected in the mirror. If you have perfect vision, then your eyes will focus parallel rays entering your pupils down to single points on your retinas. $\endgroup$ – Solomon Slow Apr 7 '18 at 15:52
  • 1
    $\begingroup$ @Asmani, A true point source is not a physical possibility. There are many different ways to describe the "brightness" of a light source. You would have to deal with infinities and infinitessimals if you naively applied some of those to a theoretical point source. Probably, the best approach would be to start with an area source having some brightness, and then use high-school calculus to describe what would happen as you shrank the size of the source down toward zero. $\endgroup$ – Solomon Slow Apr 7 '18 at 16:01
  • $\begingroup$ I added a practical problem. $\endgroup$ – Transcendent Apr 7 '18 at 20:30
2
$\begingroup$

The previous answer (assuming a perfect point source and perfect parabolic mirror) is true, ignoring diffraction. Without diffraction, in the perfect geometric case with perfect collimation, you would see a perfect point source. But since light is a wave, it can never be perfectly collimated, nor could you detect a point source even with arbitrarily well-collimated light. Indeed, light can be awfully well collimated if the width of the beam is millions of times larger than the wavelength, but what you will ultimately see all depends on the length scales involved, including the size of your beam, the wavelength, the distance, and the size of your detector.

Thus, even for the case of perfect geometry, you will see a blurred spot and/or a dimmed spot. The way to counteract diffraction is to use a larger parabolic mirror (producing a larger beam waist) and a larger detector the farther away you want to measure.

So it’s clear that the beam cannot be collimated due to diffraction and will develop a wavefront curvature. In the short range, you will be able to capture all of the quasi-collimated light in your (finite size) detector. The spot will be bright (full intensity, so to speak), but the wavefront curvature will leave it slightly blurred. On the other hand, in the limit of large distance away from the emitter and small detector compared to the wavefront curvature, the beam will have diverged dramatically. In this case, the beam wavefront will look practically flat, producing a diffraction-limited point on your detector. But the brightness will diminish with distance because the farther away you are, the less light you capture (analogous to a faraway star).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.