InterviewSolution
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InterviewSolution
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A point source of light is placed at the centre of curvature of a hemispherical surface. The radius of curvature is r and the inner surface is completelyreflecting. Find the force on the hemisphere due to the light falling on its if the source emits a power W. |
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Answer» Solution : The energy emitted by the source per unit time, i.e., W falls on an area ` 4PI (r^2)` at a distance r in unit time. Thus, the energy falling per  unit area per unit time is `(W/(4pi (r^2))).` Consider a small area dA at the POINT P of the hemisphere . The energy falling per unit time on it is `(WdA/ (4 pi(r^2)))`. The corresponding momentum INCIDENT on this area per unit time is ` (WdA/(4pi(r^2)c)). As the light is reflected BACK, the change in momentum per unit time, i.e.,the force on dA is (dF= (2 Wd A/ (4 pi (r^2) c)). ` Suppose the radius OP through the area dA makes an angle theta with the symmetry axis OX. The force on dA is along this radius. By symmetry, the resultant force on the hemisphere is along OX. The component of dF along OX is ` dF cos theta= (2 WdA/ (4 pi (r^2)c)) cos theta.` If we project the area dA on the plane containing the rim, the projectin is dA cos theta. Thus, thecomponent of dF along OX is, ` dF cos theta = (2W/ (4pi (r^2)c))(projection of dA).` The net force along OX ` F= (2W/ (4pi(r^2)c))(sum projection of dA).` When all the small areas dA are projected, we get the area enclosed by the rim which is (`pi (r^2). Thus, ` (F= (2W/ (4pi (r^2)c)) xx pi (r^2)= (W/2c).)` |
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