SATHEE: Optics Question 104

Optics Question 104

Question: The distance of the moon from earth is $3.8 \times 10^{5} k m .$ 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 $Δ θ = \frac{1}{R e s o l v i n g P o w e r (R P)}$ ;

and if x is the distance between points on the surface of moon which is at a distance r from the telescope. $Δ θ = \frac{x}{r}$

So $Δ θ = \frac{1}{R P} = \frac{x}{r} i . e . x = \frac{r}{R P} = \frac{r}{d / 1.22 λ}$

$\Rightarrow x = \frac{1.22 λ r}{d}$

$= \frac{1.22 \times 5500 \times 10^{- 10} \times (3.8 \times 10^{8})}{500 \times 10^{- 2}} = 51 m$

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Optics Question 104

Question: The distance of the moon from earth is 3.8×105km. 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

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Question: The distance of the moon from earth is $3.8 \times 10^{5} k m .$ 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