Light in a funnel intensity
Part D
https://isaacphysics.org/questions/collecting_light_in_a_funnel?board=e976568d-610a-4f8a-99d8-46bf75ff7f05
Honestly I’m not even sure how to get started
https://isaacphysics.org/questions/collecting_light_in_a_funnel?board=e976568d-610a-4f8a-99d8-46bf75ff7f05
Honestly I’m not even sure how to get started
Original post by Jsmithx
Part D
https://isaacphysics.org/questions/collecting_light_in_a_funnel?board=e976568d-610a-4f8a-99d8-46bf75ff7f05
Honestly I’m not even sure how to get started
https://isaacphysics.org/questions/collecting_light_in_a_funnel?board=e976568d-610a-4f8a-99d8-46bf75ff7f05
Honestly I’m not even sure how to get started
The light follows this path, as given by the diagram in the final hint.
If you trace it carefully, you see that the first angle of incidence is 75 degrees. (The light hits the mirror at 15 deg which is half the angle at the apex.)
The next reflection hits the opposite mirror with an angle of incidence of 45 degrees. (75 less the 30 deg mirror angle. You can prove this with triangle geometry)
It then bounces again and hits the opposite mirror at an angle of incidence of 15 degrees. At which point it bounces straight down (show using triangle geometry) and goes back out following an identical path as it came in with angle now increasing to 45 and 75 before it leaves parallel to its initial path.
The angles can all be worked out, as I say, with geometry.
Original post by Stonebridge
The light follows this path, as given by the diagram in the final hint.
If you trace it carefully, you see that the first angle of incidence is 75 degrees. (The light hits the mirror at 15 deg which is half the angle at the apex.)
The next reflection hits the opposite mirror with an angle of incidence of 45 degrees. (75 less the 30 deg mirror angle. You can prove this with triangle geometry)
It then bounces again and hits the opposite mirror at an angle of incidence of 15 degrees. At which point it bounces straight down (show using triangle geometry) and goes back out following an identical path as it came in with angle now increasing to 45 and 75 before it leaves parallel to its initial path.
The angles can all be worked out, as I say, with geometry.
If you trace it carefully, you see that the first angle of incidence is 75 degrees. (The light hits the mirror at 15 deg which is half the angle at the apex.)
The next reflection hits the opposite mirror with an angle of incidence of 45 degrees. (75 less the 30 deg mirror angle. You can prove this with triangle geometry)
It then bounces again and hits the opposite mirror at an angle of incidence of 15 degrees. At which point it bounces straight down (show using triangle geometry) and goes back out following an identical path as it came in with angle now increasing to 45 and 75 before it leaves parallel to its initial path.
The angles can all be worked out, as I say, with geometry.
For part D? Isn’t this a?
Sorry. Missed that bit of your post. I just saw the ' I’m not even sure how to get started'.
Original post by Stonebridge
Sorry. Missed that bit of your post. I just saw the ' I’m not even sure how to get started'.
Sorry. Any ideas for D?
Original post by Jsmithx
Sorry. Any ideas for D?
My 1st thought for D is that they are asking for intensity, which is a measure of 'concentration' of light. In this case the final aperture is 1/4 the original, so the light has been concentrated into an area (1/4)^2 =1/16 of the original. So the intensity is increased by a factor of 16. However, this has to be balanced by the fact that in the earlier part, you calculated what proportion of the light gets through. So the increased intensity applies only to the % of light that got through.
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