Astronauts look down on earth surface from a space ship parked at an altitude of $500$ km. They can resolve object of the earth of the size (It can be assumed that the pupils diameter is 5 mm and wavelength of light is $500$ nm)

0.5 m

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5 m

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50 m

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500 m

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Solution

The correct option is C $50 m$
Resolving power of eye $= λ / a$

$= \frac{500 \times 10^{- 9}}{5 \times 10^{- 3}} = 10^{- 4}$ radians

Now, $a r c = a n g l e \times r a d i u s = 10^{- 4} \times (500 \times 10^{3}) m = 50 m$

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Astronauts look down on earth surface from a space ship parked at an altitude of 500 km. They can resolve object of the earth of the size (It can be assumed that the pupils diameter is 5 mm and wavelength of light is 500 nm)

The correct option is C 50mResolving power of eye =λ/a=500×10−95×10−3=10−4 radiansNow, arc=angle×radius=10−4×(500×103)m=50m

Astronauts look down on earth surface from a space ship parked at an altitude of $500$ km. They can resolve object of the earth of the size (It can be assumed that the pupils diameter is 5 mm and wavelength of light is $500$ nm)

The correct option is C $50 m$
Resolving power of eye $= λ / a$

$= \frac{500 \times 10^{- 9}}{5 \times 10^{- 3}} = 10^{- 4}$ radians

Now, $a r c = a n g l e \times r a d i u s = 10^{- 4} \times (500 \times 10^{3}) m = 50 m$