SATHEE: Optics Question 103

Optics Question 103

Question: The average distance between the earth and moon is $38.6 \times 10^{4}$ km. The minimum separation between the two points on the surface of the moon that can be resolved by a telescope whose objective lens has a diameter of 5 m with $λ = 6000 A A$ is

[MP PMT 1993]

Options:

A) 5.65 m

B) 28.25 m

C) 11.30 m

D) 56.51 m

Show Answer

Answer:

Correct Answer: D

Solution:

Resolving power $= \frac{1.22 λ}{a} = \frac{1.22 \times 6000 \times 10^{- 10}}{5}$ Also resolving power $= \frac{d}{D} = \frac{d}{38.6 \times 10^{7}}$

$∴ \frac{1.22 \times 6 \times 10^{- 7}}{5} = \frac{d}{38.6 \times 10^{7}}$

Therefore $d = \frac{1.22 \times 6 \times 10^{- 7} \times 38.6 \times 10^{7}}{5} m = 56.51 m$

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Optics Question 103

Question: The average distance between the earth and moon is 38.6×104 km. The minimum separation between the two points on the surface of the moon that can be resolved by a telescope whose objective lens has a diameter of 5 m with λ=6000 AA is

Options:

Answer:

Solution:

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Question: The average distance between the earth and moon is $38.6 \times 10^{4}$ km. The minimum separation between the two points on the surface of the moon that can be resolved by a telescope whose objective lens has a diameter of 5 m with $λ = 6000 A A$ is