Timeline for Angular diameter and length of baseline
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 15 at 8:27 | comment | added | Elena Greg | What about to compute where the visibility is 0.5 instead of 0? | |
| Jan 15 at 8:20 | comment | added | Elena Greg | Is it $b = \frac{1.22 \lambda }{ \Theta}$? | |
| Jan 15 at 7:50 | comment | added | Elena Greg | I mean that I have a prediction of the angular diameter of a star and I would like to know how long should be the baseline to see the first lobe (to choose an appropriate interferometer). Basically, I am asking about the relation between visibility, baseline, angular diameter, and wavelength. | |
| Jan 13 at 8:11 | comment | added | ProfRob | What do you mean by "compute the length of the baseline"? You measure it. Then the visibility varies as a function of angular diameter, as shown. | |
| Jan 10 at 12:09 | comment | added | uhoh | Oh, see "Box 2" on page 652 in Measuring the Sizes of Stars; Fringe Benefits of Interferometry by Rajaram Nityananda I think there's your equation, with the integral over the disk. (archived) | |
| Jan 10 at 12:06 | comment | added | uhoh | while the angle from another part of the star is producing a minimum. This effect gets worse as the optical path difference increases, either by letting the star drift, or by increasing the baseline. You have to write an expression for the amplitude of the fringes for a point source, then integrate over the disk of the star to get this curve. Maybe it's an easy integral and results in an analytic expression, maybe not. Sorry I can't be more helpful, but I'll bet within a half-day you have a very nice answer. | |
| Jan 10 at 12:04 | comment | added | uhoh | Stellar interferometry is a cool way to measure the diameters of stars when we can't directly image them. cf. What equipment and techniques were used to study Betelgeuse's diameter in 1920? I can't give you a formula, but I'll be it can be derived for a uniform disk and two slits. If the start was a point, the visibility would be flat at 1.0 (perfect constructive and destructive interference) but the finite diameter has a "smearing" effect that makes the max/min ratio drop because the angle from one part of the star is producing a maximum | |
| Jan 9 at 10:28 | comment | added | Elena Greg | I quickly went through it and did not find it. | |
| Jan 9 at 8:01 | comment | added | planetmaker | Did you look at the extensive explanation material the site links? Why or how wasn't that helpful? | |
| Jan 9 at 7:33 | comment | added | Elena Greg | If they are theoretical curves, my question is what formula is used. There needs to be some relation to how to compute the length of baseline for the particular angular diameter $\theta$. | |
| Jan 9 at 7:33 | history | edited | Elena Greg | CC BY-SA 4.0 |
added 188 characters in body
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| Jan 8 at 19:46 | comment | added | planetmaker | Putting the diagramme and the question into slightly more context would be really helpful. Did you consider it being theoretical curves? | |
| Jan 8 at 15:17 | history | asked | Elena Greg | CC BY-SA 4.0 |