The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a light of wavelength 2440 $\mathop A\limits^o$ is used to see th object will be :
Solution
<p>$R.P. = {1 \over {1.22\,\lambda /a}}$</p>
<p>$= {{24.4 \times {{10}^{ - 2}}} \over {1.22 \times 2440 \times {{10}^{ - 10}}}}$</p>
<p>$= 8.2 \times {10^5}$</p>
About this question
Subject: Physics · Chapter: Optics · Topic: Diffraction and Polarization
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