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IOWA ACADEMY OP SCIENCE 
series of diffraction rings. Allowing this diffraction pattern to fall 
within a telescope we find that it gives ns an image of the illuminated 
object. Airy investigated this diffraction pattern and gave us; a formula 
for measuring the diameter of these rings. 2 Since most of the light is 
contained in the central bright spot and the first five diffraction rings, 
we shall make use of this particular number in our discussion. 
In order better to understand the last paragraph let usi fefer to Fig. 
1. Light from the monochromatic illuminator M, is focussed by lens L x 
upon the pin hole S. The lens L 2 focusses this pin hole upon the aper- 
ture A, and provides the light for an image of object 0 in the camera 
or telescope C. If we wish to use diffused light from the back, S is re- 
placed by a carbon lamp which has a ground glass in front of it. If 
we wish to illuminate 0 from in front, the lamp cam be moved to the 
position R, which is in front and a little to one side of 0. In the last 
case as a matter of convenience, it is well to remove the lens L 2 . 
Now Airy’s formula for the radius of the fifth diffraction ring is Sin 
y=2.62lVr where y is the angle of diffraction, A the wave length of the 
light used, and r the radius of the lens casting the image of the point. 
Since the radius of Mars is about 2,100 miles, we have Sin y— 2.621 x 
.0000'55/2100 where .000055 cm. is the average wave length of light in 
the visible spectrum. Let X equal the radius of the fifth minimum^ dif- 
fraction ring and we have Sin y==X /35, 050 ,000 where 35,050,000 miles 
is the closest distance Mars gets to the earth. From these two expres- 
sions for Sin y we get the value of X which is 2.406 cm., the radius of 
the fifth dark ring, or 2X equal 4.812 cm. the diameter of the fifth 
dark ring for Mars. 
If we use a 24 in. telescope, we find that the telescope diameter is 
12.66 times as great as the diameter of the diffraction pattern, out to 
and including the fifth dark ring from Mars. 
Taking as the object which represents Mars, a hole 1.05 mm. in 
diameter, I illuminated it from behind with a point source of light and 
brought this light to a focus with the lens L 2 as shown in Fig. 1. Focus- 
sing a lens on the point where this light was focussed I was able to study 
the diffraction pattern, and by means of a scale on the lens, I measured 
the diameter of the fifth ring. This observed value was 6 mm., a 
value a trifle smaller than that computed for these conditions. 
If the telescope aperture is 4 mm. in diameter, the diameter of the 
fifth diffraction ring admitted should be 4/12.66 which equals .3159 mm. 
Reasoning from this, .3159: 6 ::1.05:X and X equals 1.994 cm. which 
2 Wood’s Physical Optics, p. 237. 
