Electromagnetic Waves

JEE Main level, single correct, medium

41 of 165

Question 41 of 165

An object is placed in a medium of refractive index 33. An electromagnetic wave of intensity 6×108Wm−26\times {10}^{8}W{m}^{-2} falls normally on the object and it is absorbed completely. The radiation pressure on the object would be (speed of light in free space =3×108ms−1=3\times {10}^{8}m{s}^{-1} ):

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Answer: (C) 6Nm−26N{m}^{-2}

The formula to calculate the radiation pressure is given by
P=Iv...(1)P=\dfrac{I}{v}...\left(1\right)
Also, the velocity of the electromagnetic wave can be expressed as
μ=cv...(2)\mu =\dfrac{c}{v}...\left(2\right)
Equations (1) and (2) imply that
P=I⋅μc=6×108×33×108=6Nm−2\begin{matrix}P & = & \dfrac{I\cdot \mu }{c} \\ & = & \dfrac{6\times {10}^{8}\times 3}{3\times {10}^{8}} \\ & = & 6N{m}^{-2}\end{matrix}

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