Question

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Q.43. The distance of moon form the earth is 3.8$\times {10}^5$ km. Supposing that the eye is most sensitive to the light of wavelength 550 nm, the separation of two points on the moon that can be resolved by a 500 cm telescope is

Answer 1

Dear Student ,
$$\begin{gathered} Thepowerofresolutionofthetelescopeisgivenby: \\ ∆\theta \approx \frac{1.22\lambda }{2a} \\ where,a=radiusofthelense \\ therefore, \\ ∆\theta =\frac{1.22\times 5000\times {10}^{-10}}{2\times 5\times {10}^{-2}}(sincethediameterisgiveinthequestion) \\ ∆\theta =6.1\times {10}^{-6}radians \\ Now, \\ ∆\theta =\frac{arc}{radius}=\frac{minimumdis\tan cebetweenthetwopoints}{dis\tan cebetweenthetelescopeandthelocationoftwopoints} \\ 6.1\times {10}^{-6}=\frac{minimumdistancebetweenthetwopoints}{3·8\times {10}^5} \\ minimumdistancebetweenthetwopoints=2·318m \\ 2·318mistheminimumdis\tan cethatthetelescopewillbeabletoresolveatthatdis\tan ce. \end{gathered}$$

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