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Electromagnetic Waves question

2024 · 27 Jan · Shift 2 · Q76
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  5. /2024 · 27 Jan · Shift 2 · Q76

Electromagnetic Waves question

2024 · 27 Jan · Shift 2 · Q76

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
An object is placed in a medium of refractive index 3 . An electromagnetic wave of intensity 6×108 W/m26 \times 10^8 \mathrm{~W} / \mathrm{m}^26×108 W/m2 falls normally on the object and it is absorbed completely. The radiation pressure on the object would be (speed of light in free space =3×108 m/s=3 \times 10^8 \mathrm{~m} / \mathrm{s}=3×108 m/s ) :
  1. A
    6 Nm−26 \mathrm{~Nm}^{-2}6 Nm−2
  2. B
    36 Nm−236 \mathrm{~Nm}^{-2}36 Nm−2
  3. C
    18 Nm−218 \mathrm{~Nm}^{-2}18 Nm−2
  4. D
    2 Nm−22 \mathrm{~Nm}^{-2}2 Nm−2
View written solutionFree

Correct answer: A

  1. Radiation pressure formula

For an electromagnetic wave incident normally on a surface and completely absorbed, the radiation pressure is

p=Ivp = \frac{I}{v}p=vI​

where:

  • III = intensity of the wave in the medium
  • vvv = speed of light in that medium
  1. Find speed of light in the medium

Given refractive index,

n=3n = 3n=3

We know,

v=cn=3×1083=1×108 m/sv = \frac{c}{n} = \frac{3 \times 10^8}{3} = 1 \times 10^8\ \text{m/s}v=nc​=33×108​=1×108 m/s

  1. Substitute the given intensity

Given,

I=6×108 W/m2I = 6 \times 10^8\ \text{W/m}^2I=6×108 W/m2

Therefore,

p=6×1081×108=6 N/m2p = \frac{6 \times 10^8}{1 \times 10^8} = 6\ \text{N/m}^2p=1×1086×108​=6 N/m2

  1. Match with options

p=6 N/m2p = 6\ \text{N/m}^2p=6 N/m2

So the correct option is:

A: 6 N m−26\ \text{N m}^{-2}6 N m−2

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