1.

A parallel beam of monochromatic light of frequency v is indcident on a surface. The intensity of the beam is I and area of the surface is A. Find the force exerted by light of beam on the surface is perfectly reflecting and the light beam is incident at an angle of incidence theta. (The speed of light is denoted as c.)

Answer»

`(2lAsin^(2)theta)/(pic)`
`(lAcos^(2)theta)/(c)`
`(2lAcos^(2)theta)/(c)`
`(lAcos^(2)theta)/(sqrt2c)`

Solution :When a light beam is incident at an angle of incidence ` theta ` , the intensity across the SURFACE is geometrically REDUCED by the cosion of that angle and the component of the RADIATION force against the surface will also be reduced by the cosione of`theta ` hence resulting pressure ,
` p_("incidnet") = (I) /(c) cos^(2) theta ` ,
where , c is velocity of light .
since , light wave reflects also from reflecting surface , hence hthe recoil due to the REFLECTED wave will further contribute to the radiation pressure .
` thereforep_("emitted") = (I)/(c)cos^(2) theta `
` thereforep_("net") = P_("incident" ) + p_("emitted") `
` = (I)/(c)cos^(2) theta+ (I)/(c) cos^(2) theta `
` p_("net") = (2I)/(c) cos^(2) theta `
` therefore ` Force exerted by the light beam ,
` F= P_("net") XX "Area (A) " = (2IA cos^(2) theta )/(c) `


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