Optics Question 104

Question: The distance of the moon from earth is $ 3.8\times 10^{5}km. $ The eye is most sensitive to light of wavelength 5500 Å. The separation of two points on the moon that can be resolved by a 500 cm telescope will be

[AMU (Med.) 2002]

Options:

A) 51 m

B) 60 m

C) 70 m

D) All the above

Show Answer

Answer:

Correct Answer: A

Solution:

As limit of resolution $ \Delta \theta =\frac{1}{Resolving\ Power(RP)} $ ;

and if x is the distance between points on the surface of moon which is at a distance r from the telescope. $ \Delta \theta =\frac{x}{r} $

So $ \Delta \theta =\frac{1}{RP}=\frac{x}{r}i.e.\ x=\frac{r}{RP}=\frac{r}{d/1.22\ \lambda } $

$ \Rightarrow x=\frac{1.22\ \lambda r}{d} $

$ =\frac{1.22\times 5500\times {{10}^{-10}}\times (3.8\times 10^{8})}{500\times {{10}^{-2}}}=51\ m $



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