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JEE Main 2025 (4 Apr, Shift 1) · Photons: Energy, Momentum & Radiation Pressure

Photons: Energy, Momentum & Radiation Pressure question from JEE Main 2025 (4 Apr, Shift 1)

Q10JEE Main 2025 · 4 APR · SHIFT 1Difficult

Photons: Energy, Momentum & Radiation Pressure

A small mirror of mass mm is suspended by a massless thread of length ll. Then the small angle through which the thread will be deflected when a short pulse of laser of energy E falls normal on the mirror

(c=c = speed of light in vacuum and g=g = acceleration due to gravity)

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Answer: (d)

  1. Mirror reflects the pulse, so momentum given =2Ec= \dfrac{2E}{c} and v=2Emcv = \dfrac{2E}{mc}
  2. Energy conservation: 12mv2=mgl(1−cos⁡θ)≈mglθ22\tfrac12 mv^2 = mgl(1-\cos\theta) \approx mgl\dfrac{\theta^2}{2}, so v=θglv = \theta\sqrt{gl}
  3. θ=vgl=2Emcgl\theta = \dfrac{v}{\sqrt{gl}} = \dfrac{2E}{mc\sqrt{gl}}
  4. So the answer is (d).
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