Determine "ring" around the moon and calculate size of water


It can be shown that, except for theta = 0, a circular obstacle produces the same diffraction pattern as a circular hole of the same diameter. Furthermore, if there were many such obstacles, such as water droplets located randomly, then the interference effects vanish, leaving only the diffraction associated with a single obstacle.

(a) Explain why one sees a "ring" around the Moon on a foggy night. The ring is usually reddish in color. Explain why.

(b) Calculate the size of the water droplets in the air if the ring around the Moon appears to have a diameter 1.5 times that of the Moon. The angular diameter of the Moon in the sky is 0.5 degrees.

(c) At what distance from the Moon might a bluish ring be seen? Sometimes the rings are white. Why?

(d) The color arrangement is opposite to that in a rainbow. Why should this be so?

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Physics: Determine "ring" around the moon and calculate size of water
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