Optics Question 108
Question: The diameter of moon is $ 3.5\times 10^{3} $ km and its distance from the earth is $ 3.8\times 10^{5} $ km. If it is seen through a telescope whose focal length for objective and eye lens are 4 m and 10 cm respectively, then the angle subtended by the moon on the eye will be approximately
[NCERT 1982; CPMT 1991]
Options:
15o
20o
30o
35o
Show Answer
Answer:
Correct Answer: B
Solution:
$ |m|\ =\frac{f _{o}}{f _{e}}=\frac{400}{10}=40 $ .
Angle subtended by moon on the objective of telescope $ \propto \ =\frac{3.5\times 10^{8}}{3.8\times 10^{8}}=\frac{3.5}{3.8}\times {{10}^{-2}}rad $ .
Also $ |m|\ =\frac{\beta }{\alpha }\Rightarrow $ Angular size of final image $ \beta =|m|\ \times \alpha $
$ =40\times \frac{3.5}{3.8}\times {{10}^{-2}} $ = 0.36 rad $ =0.3\times \frac{180}{\pi }\approx 21^{o} $
BETA
