The resolution of a telescope is given by beta=wavelength/D, beta is the size of

Question

The resolution of a telescope is given by beta=wavelength/D, beta is the size of the airy disk or the smallest angle that can be resolved inr adians. Consider an earth like planet in a 1 AU orbit around a star at a distance of 30 pc. How large a telescope is required to have the star and planet be separated by 5 times the minimum angle that the telescope can resolve? Consider two wavelengths, 0.5 micrometers and 10 micormeters.

Im not sure if pc has to be converted or not and how to set up the equation?? Was thinking 5beta=0.5 micrometers/30pc but i think there is some conversions for it to be in radians?!?

Explanation / Answer

heta = 1.22lembda /D

The distancesbetween two objects a distanceraway and separated by an angle?iss=r?.
hence theta =149,597,870,700/30856775814671900 = 4.8*10^-6 radians

hence the diameter of the telescope required would be 1.22 lembda/ theta
= 1.22*10*10^-6/4.8*10^-6 (for wavelenght of 10 micro meter
= 2.58 m
and (for wavelenght of .5micro meter) D = .125 M

IN THE ORDER TO GET 5 TIMES BETTER RESOLUTION MULTIPLY IT BY 5


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