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:

A) 15o

B) 20o

C) 30o

D) 35o

Show Answer

Answer:

Correct Answer: B

Solution:

$ |m|\ =\frac{f _{o}}{f _{e}}=\frac{400}{10}=40 $ .

Angle subtented by moon on the objective of telescope $ \propto \ =\frac{3.5\times 10^{3}}{3.8\times 10^{3}}=\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} $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक