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Define resolving power of an astronomical refracting telescope and write expression for it in normal adjustment. Assume that light of wavelength 6000Å is coming from a star, what is the limit of resolution of a telescope whose objective has a diameter of 2.54 m? |
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Answer» Solution :The resolving power of a TELESCOPE is its ability to show as distinct (separate) two light rays coming from nearby points of an astronomical. Mathematically, resolving power of telescope is reciprocal of its limit of resolution. In normal adjustment of an astronomical telescope resolving power `=(1)/(Delta THETA)=(A)/(1.22lambda)`, whereA is the APERTURE of telescope objective and `lambda` the wavelength of light. Here `lambda=6000Å=6000xx10^(-10)m and A=2.54m` `therefore"Limit of resolution "Delta theta=(1.22lambda)/(A)=(1.22xx(6000xx10^(-10)))/(2.54)=2.88xx10^(-7)" RAD"` |
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