## Monday, April 11, 2022

### Solutions To Optical Resolution Problems (Basics Of Fraunhofer Diffraction Part 3)

1) Taking 384,000 km as the mean distance from the Earth to the Moon, find the smallest distance for the separation of two objects on the Moon’s surface that can just be resolved by the Harry Bayley Observatory C-14 telescope.

Solution:  Diameter of telescope D = 14"  =  0.35 m

Visible light wavelength: l  = 5.5 x 10 -7 m

The optical resolution is defined by the Rayleigh criterion:

q 1 = 1.22  l/ D

Then:  q 1 = 1.22  ( 5.5 x 10 -7 m ) /  0.35m   =   1.91 X 10 -6 rad = 0.40 arc sec

The smallest distance d for linear separation of two objects-  that can just be resolved by the C-14 telescope on the lunar surface - will be:

d =   (3.84 x 10 8 m)( 1.91 X 10 -6 rad) =  729.6 m

2) Find the diameter of an Airy disk in the focal plane of a refracting telescope featuring an objective with focal length of 1.0 m  and diameter of 10.0 cm.  Assume the effective observed wavelength is  5.5 x 10-5  cm.

Solution:   Similar to the solution for (1)

q 1 = 1.22  ( 5.5 x 10 -5 cm ) /  10 cm   =  6.71  X 10 -6 rad

But the linear radius r is just   q 1  x  f.l.  where f.l. = 1.0 m = 100 cm

So:  (6.71  X 10 -6 rad )  x  (100 cm)  =  6.71  X 10 -4 cm

Diameter:  = 2r  =   2( 6.71  X 10 -4 cm) =   1.342  X 10 - 3 cm = 0.013 mm

3) The objective of a telescope has a diameter of 12.0 cm.   At what distance would two small green objects - 30.0 cm apart - be barely resolved by the telescope?  (Assume the resolution to be limited by diffraction for the objective only, and an effective wavelength of   5.4 x 10-5  cm.

Solution:  Note 30.0 cm is the linear size (radius)  of the just barely resolved diffraction pattern.  Let q  be the corresponding angular radius.  The diagram below shows the geometry:

Then:

q  = 1.22  l/ D  =  1.22  ( 5.4 x 10 -5 cm ) /  12.0 cm  = 5.46 x 10 -6 rad

The distance d may be found from:   tan q  =  15 cm/ d

So:  d  =   15 cm/  tan q

But for very small  q  we have:   tan q  »    q

d  =   15 cm/ 5.46 x 10 -6 rad  »    2.75 x 10 6 cm   =  27.5 km