Question

A radiation of energy 'E' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is
(c = velocity of light)

$straight E over straight c$
$fraction numerator 2 straight E over denominator straight c end fraction$
$fraction numerator 2 straight E over denominator straight c squared end fraction$
$straight E over straight c squared$

Solution

fraction numerator 2 straight E over denominator straight c end fraction

The radiation energy is given by
$straight E equals space hc over straight lambda Initial space momentum space of space the space radiation space is space straight P subscript straight i space equals space minus straight h over straight lambda space equals space minus straight E over straight c The space reflected space momentum space is space straight P subscript straight r space equals space minus straight h over straight lambda space equals space minus straight E over straight c So comma space the space change space in space momentum space of space light space is space increment straight P subscript light space equals space straight P subscript straight r space minus straight P subscript straight i space equals negative fraction numerator 2 straight E over denominator straight c end fraction Thus comma space the space momentum space transferred space to space the space surface space is space increment straight P subscript light space equals space fraction numerator 2 straight E over denominator straight c end fraction$

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