Timeline for Fabry-Perot Interferometer
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 10, 2020 at 18:12 | vote | accept | Kinka-Byo | ||
| May 10, 2020 at 16:36 | answer | added | Gilgamesh | timeline score: 1 | |
| May 10, 2020 at 16:10 | comment | added | Gilgamesh | I will post it as an answer if that helped you. | |
| May 10, 2020 at 16:09 | comment | added | Gilgamesh | To be clear, for each $\theta$ maps into a circle. I agree that if $\theta$ is om the same plane than the y-axis (screen axis) you will get the point on that plane. Turning on the x-axis (perpendicular to the screen) with the same $\theta$ you will get all the circle. Repeat por each angle of incidence. | |
| May 10, 2020 at 10:45 | comment | added | Kinka-Byo | @MarcoCiafa Thank you! I think I have understood: fixed the angle theta of incidence, we have a circle (theta corresponds to 1 point on the screen) whose intensity depends on theta and on the superposition of all wavelenghts coming from the cavity. If we repeat this analysis for each value of theta, we get many points on the screen, and for each of them there is a circle. So the resulting image is a set of circles with same center. Correct? | |
| May 10, 2020 at 8:29 | comment | added | Gilgamesh | Edited: Each point on the screen is the image of each angle $ \theta $ of incidence of the light. That particular point A' (it's ring) corresponds to some particular $ \theta $. Each wavelenght will have a ring pattern (long wavelenghts will have shoter one). When you add up those patterns you should get a light (when all wavelenghts peaks match) / dark (when they doesn't). I'm not familiar with the notation on those graphics. I saw them on Wikipedia but I haven't read the article. | |
| May 10, 2020 at 7:30 | history | asked | Kinka-Byo | CC BY-SA 4.0 |