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  • The aperture of the objective is The resolving power of this telescope, if a light of wavelength <img alt="2440 \ma

The aperture of the objective is 24.4 \mathrm{~cm}. The resolving power of this telescope, if a light of wavelength 2440 \mathrm{\AA } is used to see the object will be :
 

Option: 1

8.1 \times 10^{6}
 


Option: 2

10.0 \times 10^{7}
 


Option: 3

8.2 \times 10^{5}
 


Option: 4

1.0 \times 10^{-8}


Answers (1)

best_answer

\text{Resolving power (R.P.)}

\mathrm{\text { R.P. }=\frac{1}{1.22 \frac{\lambda}{a}}=\frac{a}{1.22 \lambda}} \\

\mathrm{=\frac{24.4 \times 10^{-2}}{1.22 \times 2440 \times 10^{-10}}} \\

\mathrm{=8.2 \times 10^{5} }

Hence the correct option is 3

Posted by

rishi.raj

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