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Electromagnetic Waves question

2013 · Shift 1 · Q55
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  5. /2013 · Shift 1 · Q55

Electromagnetic Waves question

2013 · Shift 1 · Q55

JEE AdvancedPhysicsElectromagnetic WavesMCQ+3 / −1
A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3 ×\times× 108 ms −-− 1. The final momentum of the object is
  1. A
    0.3 ×\times× 10 −-− 17 kg-ms −-− 1
  2. B
    1.0 ×\times× 10 −-− 17 kg-ms −-− 1
  3. C
    3.0 ×\times× 10 −-− 17 kg-ms −-− 1
  4. D
    9.0 ×\times× 10 −-− 17 kg-ms −-− 1
View written solutionFree

Correct answer: B

Step-by-step Derivation:

  1. Identify the given information:

    • Duration of the light pulse, t=100 ns=100×10−9 s=10−7 st = 100 \text{ ns} = 100 \times 10^{-9} \text{ s} = 10^{-7} \text{ s}t=100 ns=100×10−9 s=10−7 s.
    • Power of the pulse, P=30 mW=30×10−3 WP = 30 \text{ mW} = 30 \times 10^{-3} \text{ W}P=30 mW=30×10−3 W.
    • Speed of light, c=3×108 m/sc = 3 \times 10^8 \text{ m/s}c=3×108 m/s.
    • The object is initially at rest, so its initial momentum is zero.
    • The pulse is completely absorbed by the object.

  2. Calculate the total energy of the light pulse: The energy E of the pulse is the product of its power P and duration t. E=P×tE = P \times tE=P×t Substituting the given values: E=(30×10−3 W)×(10−7 s)E = (30 \times 10^{-3} \text{ W}) \times (10^{-7} \text{ s})E=(30×10−3 W)×(10−7 s) E=30×10−10 J=3×10−9 JE = 30 \times 10^{-10} \text{ J} = 3 \times 10^{-9} \text{ J}E=30×10−10 J=3×10−9 J

  3. Calculate the momentum of the light pulse: For electromagnetic radiation (like a pulse of light), the momentum p is related to its energy E and the speed of light c by the formula: p=Ecp = \frac{E}{c}p=cE​ Substituting the calculated energy and the given speed of light: p=3×10−9 J3×108 m/sp = \frac{3 \times 10^{-9} \text{ J}}{3 \times 10^8 \text{ m/s}}p=3×108 m/s3×10−9 J​ p=1×10−17 kg-m/sp = 1 \times 10^{-17} \text{ kg-m/s}p=1×10−17 kg-m/s

  4. Determine the final momentum of the object: According to the law of conservation of momentum, when the object absorbs the light pulse, the momentum of the light pulse is completely transferred to the object. Since the object was initially at rest, its final momentum will be equal to the momentum of the absorbed light pulse. Initial momentum of the system (object + light) = pobject,i+plight=0+pp_{object,i} + p_{light} = 0 + ppobject,i​+plight​=0+p. Final momentum of the system (object) = pobject,fp_{object,f}pobject,f​. By conservation of momentum: pobject,f=pobject,i+plight=0+1×10−17 kg-m/sp_{object,f} = p_{object,i} + p_{light} = 0 + 1 \times 10^{-17} \text{ kg-m/s}pobject,f​=pobject,i​+plight​=0+1×10−17 kg-m/s pobject,f=1.0×10−17 kg-m/sp_{object,f} = 1.0 \times 10^{-17} \text{ kg-m/s}pobject,f​=1.0×10−17 kg-m/s

  5. Compare with the options:

    • A: 0.3×10−170.3 \times 10^{-17}0.3×10−17 kg-m/s
    • B: 1.0×10−171.0 \times 10^{-17}1.0×10−17 kg-m/s
    • C: 3.0×10−173.0 \times 10^{-17}3.0×10−17 kg-m/s
    • D: 9.0×10−179.0 \times 10^{-17}9.0×10−17 kg-m/s

    Our calculated final momentum matches option B.

Conclusion:

The final momentum of the object is 1.0×10−171.0 \times 10^{-17}1.0×10−17 kg-m/s.

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