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A photon of wavelength $\;5030 A^{0}\;$ is incident on a totally reflecting surface. The momentum delivered by photon is equal to

$(a)\;6.63\times10^{-27} Kgm/s\qquad(b)\;2\times10^{-27} Kgm/s\qquad(c)\;10^{-27} Kgm/s\qquad(d)\;None$

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Answer : (b) $\;2\times10^{-27} Kgm/s$
Explanation :
The momentum of incident radiation is given as $\;P=\large\frac{h}{\lambda}$
When light is totally reflected normal to surface direction of ray is reversed . That means it reverses the direction of its momentum without charging its magnitude .
Charge in momentum has a magnitude
$\bigtriangleup P=2P=\large\frac{2h}{\lambda}$
$\bigtriangleup P=2 \;\large\frac{(6.63\times10^{-34}\; J-sec)}{6630 \times10^{-10} m}$
$\bigtriangleup P=2 \times10^{-27}\;Kg/s\;.$
answered Feb 25, 2014 by yamini.v

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