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प्रश्न
A sphere of radius 1.00 cm is placed in the path of a parallel beam of light of large aperture. The intensity of the light is 0.5 W cm−2. If the sphere completely absorbs the radiation falling on it, Show that the force on the sphere due to the light falling on it is the same even if the sphere is not perfectly absorbing.
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उत्तर
Consider a sphere of centre O and radius OP. As shown in the figure, the radius OP of the sphere is making an angle θ with OZ. Let us rotate the radius about OZ to get another circle on the sphere. The part of the sphere between the circle is a ring of area `2pir^2sin θdθ`.
Consider a small part of area `ΔA` of the ring at point P.
Energy of the light falling on this part in time `Δt` ,
`ΔU = I Δ t (ΔA cos θ)`
As the light is reflected by the sphere along PR, the change in momentum ,
`Δp = 2 (ΔU)/c cos θ = 2/c I Δ t (ΔA cos^2 θ)`
Therefore , the force will be
`(Δp)/(Δt) = 2/c I ΔA cos^2 θ`
The Component of force on ΔA , along ZO , is
`(Δp)/(Δt) cos θ = 2/c I ΔA cos^3 θ`
Now , force action on the ring,
`dF = 2/c I (2pir^2 sin θ dθ) cos^3 θ`
The force on the entire sphere ,
`F = ∫_0^(pi/2) (4pir^2I)/c cos^3 θ sin θ dθ`
= `- ∫_0^(pi/2) (4pir^2I)/c cos^3 θd(cos θ)`
= `(pir^2I)/c`
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