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 $
BETA
