Confused about this Optics problems Solution

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AI Thread Summary
The discussion revolves around calculating the radii of bright rings surrounding the Airy disc in a diffraction pattern from a large refracting telescope. A key issue raised is the absence of a specified wavelength for the calculations, which is critical for determining the diffraction pattern. The solution assumes a wavelength of 550nm, which corresponds to green light in the visible spectrum. Participants clarify that in real-world scenarios, making reasonable assumptions about missing data is common practice. The conversation concludes with an understanding that such approximations are necessary for practical calculations in optics.
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Homework Statement


The Lick Observatory has one of the largest refracting telescopes, with an aperture diameter of 36 in. and a focal length of 56 ft. Determine the radii of the first, second, and third bright rings surrounding the Airy disc in the diffraction pattern formed by a star on the focal plane of the objective.

Homework Equations


\frac{\pi}{\lambda}Dsin\theta=\gamma
Where \gamma is either the maxima (bright fringes) or minima (dark fringes) of the bessel function which occurs in the solution to the diffraction through a circular aperture.
r=f \theta
Where f is the focal length.
Maxima of Gamma function: 5.136,8.417,...

The Attempt at a Solution


This is not an attempt, this is part of the actual solution. I do not understand a certain aspect of it though.

So using the first equation above and solved for sine. I came into a problem though. The problem doesn't give the wavelength! The solution just says..
"Thus for wavelength of 550nm..."
And then they plug it in. Where did this come from?!

Thanks :D
 
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550nm is green light - right in the middle of the visible spectrum and thus in the middle of the fringes/rings.

This sort of thing is common - IRL you are unlikely to have all the information you need to do a calculation right then and there. It may be difficult to get a figure too. So you have to make a reasonable guess. If you knew the spectra of the star in question, you would pick a different wavelength perhaps but you are only told that it is visible.
 
Thank you very much! That makes perfect sense :]
 
No worries then - have fun :)
 
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