Question

if the angular diameter of the moon be 30' , how far from the eye should a coin of diameter 2.2 cm be kept to hide the moon.

Answer 1

There’s a simple way to tell the approximate distance of an object if the angle it subtends at your eye is less than about 20 degrees.

The diameter is its distance times the angle in degrees divided by 60.

Alternatively, the distance is diameter times 60 divided by the angle in degrees.

So the Sun and Moon, which both subtend an angle of half a degree are each about 120 times their diameter from the Earth.

So your 2.2cm object would need to be 120 x 2.1 cm away to exactly cover either (don’t try this with the Sun). That’s 2.64 metres or 8 feet 8 inches.

I did the Open University science foundation course, S101, about 35 years ago and one of our home experiments was to measure just his angle using a small supplied disc and a metre stick. we were allsurprised at just how small the disc was.

Answer 2

AnswEr:

Suppose the coin is kept at a distance r from the eye to hide the moon completely.

Let E be the eye of the observer and let AB be the diameter of the coin.

Then, arc AB = diameter AB = 2.2 cm.

We have,

$\\ \tt \: \: \: \: \theta = 30' = ( \frac{30}{60} ) \degree = ( \frac{1}{2} \times \frac{\pi}{180} ) {}^{c} \\ \\ \tt \: \: \: \: = ( \frac{\pi}{360)} {}^{c} \\ \\ \\ \therefore \sf \: \theta \: = \frac{arc}{radius} \\ \\ \\ \implies \tt \: \frac{\pi}{360} = \frac{2.2}{r} \\ \\ \\ \implies \tt \: r = \frac{2.2 \times 360}{\pi} \: cm \\ \\ \\ \implies \tt \: r = \frac{2.2 \times 360 \times 7}{22} \\ \\ \\ \implies \tt \orange {252 \: cm} \:$