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Asked by swayamagarwal2114 | 15 Jul, 2022, 09:18: AM
answered-by-expert Expert Answer
Let P is point of luminous source. Light ray from P incident on A is getting reflected first and
then at B is getting reflected second time and then reaches the source point P as shown in figure .
 
a is the distance from P to centre O and b is the distance from centre to the point
where reflected ray meets the diamenter through P .
 
Let us use the property of triangle as applied to ΔPQR as shown in figure.
 
If the line PS divides the base QR in the ration m : n and makes angle α , β and γ as shown in figure ,
 
then we have ,  ( m + n ) cotγ = m cotα - n cotβ
 
By using this property to ΔPAB as shown in left side of figure, we get
 
( a + b ) cotφ = (a-b) cotθ
 
it can be shown that φ = 90 - θ , hence cotφ = cot(90-θ) = tanθ
 
tan2θ = (a-b) / a+b
 
begin mathsize 14px style tan theta space equals space square root of fraction numerator a minus b over denominator a plus b end fraction end root end style
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