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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Centre of Mass, Linear momentum, Collision

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Centre of Mass, Linear momentum, Collision

In certain cases, light may be viewed as a stream of particles...

In certain cases, light may be viewed as a stream of particles termed photons. Each photon carries a linear momentum h/λh/\lambda, where hh is Planck's constant and λ\lambda is the wavelength of the light. A beam of light having wavelength λ\lambda strikes a plane mirror at an angle of incidence θ\theta. Determine the change in a photon's linear momentum when the beam undergoes reflection by the mirror.

Options

  1. A2hcos⁡θλ\dfrac{2h\cos\theta}{\lambda}
  2. Bhsin⁡θλ\dfrac{h\sin\theta}{\lambda}
  3. Chcos⁡θλ\dfrac{h\cos\theta}{\lambda}
  4. D2hsin⁡θλ\dfrac{2h\sin\theta}{\lambda}

Correct answer

A. 2hcos⁡θλ\dfrac{2h\cos\theta}{\lambda}

Step-by-step solution

The magnitude of the initial momentum of the photon is p=hλp = \dfrac{h}{\lambda}.

Let the normal to the mirror be along the y-axis. The incident photon makes an angle of incidence θ\theta with the normal.

The components of the initial momentum p⃗i\vec{p}_i are:

pix=psin⁡θp_{ix} = p \sin\theta

piy=−pcos⁡θp_{iy} = -p \cos\theta

After reflection, the angle of reflection is equal to the angle of incidence θ\theta. The components of the final momentum p⃗f\vec{p}_f are:

pfx=psin⁡θp_{fx} = p \sin\theta

pfy=pcos⁡θp_{fy} = p \cos\theta

The change in momentum is given by Δp⃗=p⃗f−p⃗i\Delta \vec{p} = \vec{p}_f - \vec{p}_i.

Δpx=psin⁡θ−psin⁡θ=0\Delta p_x = p \sin\theta - p \sin\theta = 0

Δpy=pcos⁡θ−(−pcos⁡θ)=2pcos⁡θ\Delta p_y = p \cos\theta - (-p \cos\theta) = 2p \cos\theta

The magnitude of the change in linear momentum is:

∣Δp⃗∣=2pcos⁡θ=2hcos⁡θλ|\Delta \vec{p}| = 2p \cos\theta = \dfrac{2h\cos\theta}{\lambda}
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