Question

Q. A telescope has an objective lens of 10 cm. diameter and a situated at a distance of one kilometre from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000 $\stackrel{\circ }{A}$, is the order of : -

1. 5 m
2. 5 mm.
3. 5 cm.
4. 0.5 m

Answer 1

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

Regards