How large would a telescope lens have to be to resolve something that stretches 0.000000000000011111steradians?
Saturday, February 19th, 2011 at
11:42 pm
0.000000000000011111steradians of arc in the sky?
US $.01
Do I look like einstein to u?
I don’t know, but for 0.00000000000001111 steradians an aperture of 100 inches would do it.
The moment you said ‘Steradians’ you’ve committed that you are talking of squares like Square degrees (°²).
Since a Radian= (180/TT)° ,
Steradian = [(180/TT)²]°
= 3282.80 °².
Your
0.000000000000011111steradians
= 1.11111 X 10^-14 steradians
= 3.647558964 X 10^-11 °².
Since this is an area and the dimension across is square root of it, that dimension is
6.03950 X 10^-6 °
= 0° 0′ 0″.02 across.
the famous formula for the telescope aperture (lens dimension) in inches
= 4.56 / arc seconds (for resolution)
= 755,029.816 ” {three quarter million inches}
= 62,919.15133 ‘
= 11 miles 4839′ 1.085721 “. If you want it in meters
755029.816 “= 19177757.33 mm
= 19177.75733 m
= 19.17775733 km {I merely shifted the decimal place}.
Try rigging up a telescope with a lens 20 km across either here on ground or in an Earth orbit (ensuring that the rigging, nuts and bolts don’t come lose and falling apart).
Just point out that poornakumar b made a slight mistake in using his figure for the angle in degrees (6 x 10^-6) instead of arcseconds (0.02″). His answer is thus 3600 times too big. You actually need a mirror or lens about 210 inches or 5.3 metres in diameter to get the desired resolution. Mind you, the mirror needs to be in space or have a good adaptive optics system to reach the theoretical 0.02″ figure. Normal atmospheric turbulence would limit it to the 0.5″ – 1″ range.