An opaque sphere of radius R lies on a horizontal plane. On the perpendicular through the point of contact there is a point source of light a distance R above the sphere.
is filled above the plane such that the sphere is just covered with the liquid. Show that the area of shadow now becomes 2 π R 2 .
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. Radius of shadow on the ground is MP
Fig.
Let ∠ MSP = θ ⇒ ∠ SPM = 90º – θ
In Δ OSQ, ∠ SOQ = 90º – θ
triangles OSQ and PSM are similar
hence 
= 
or r = 
R
Area of the shadow = π ( 
R) 2 = 3 π R 2 .
From the geometry of figure, we have
Fig.
P'N = P'Q' + NQ'
= R' + NT
= R' + R tan i
cos r = 
= 
or R' = 2R sec r – R tan i … (1)
But R' = P'N' + N'M = 2R tan r + R tan i … (2)
From eqns.(1) and (2),
2 sec r – tan i = 2 tan r + tan i
sec r – tan r = tan i
⇒ tan i = tan 
(using trigonometry)
2i = 
– 
… (3)
Also from Snell's law,
sin i = 
sin r
From eqns. (3) and (4), we get
sin i = 
sin r = 
… (4)
⇒ tan i = 
, tan r = 
Putting in (2), we get R' = 
R
Area of the shadow = π R 2 = 2 π R 2 .
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