SATHEE: Optics Question 108

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 $\propto = \frac{3.5 \times 10^{3}}{3.8 \times 10^{3}} = \frac{3.5}{3.8} \times 10^{- 2} r a d$ .

Also $| m | = \frac{β}{α} \Rightarrow$ Angular size of final image $β = | m | \times α$

$= 40 \times \frac{3.5}{3.8} \times 10^{- 2}$ = 0.36 rad $= 0.3 \times \frac{180}{π} \approx 21^{o}$

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Optics Question 108

Question: The diameter of moon is 3.5×103 km and its distance from the earth is 3.8×105 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

Options:

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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