A small mirror of mass m is suspended by a massless thread of length ℓ. A short laser pulse of energy E falls normally on the mirror and is reflected. For a small deflection angle θ, the angle is:
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Question

A small mirror of mass m is suspended by a massless thread of length ℓ. A short laser pulse of energy E falls normally on the mirror and is reflected. For a small deflection angle θ, the angle is:

G1 Physics · Accepted Answer

A normally reflected light pulse transfers momentum p = 2E/c to the mirror. Immediately after the impulse, the mirror speed is v = p/m = 2E/(mc). Its kinetic energy is K = p²/(2m) = (2E/c)²/(2m) = 2E²/(mc²). At maximum deflection θ, this kinetic energy becomes gravitational potential energy: 2E²/(mc²) = mgℓ(1-cosθ). For small θ, 1-cosθ ≈ θ²/2. Hence, 2E²/(mc²) = mgℓ(θ²/2). Therefore, θ² = 4E²/(m²c²gℓ) ⇒ θ = 2E/(mc√(gℓ)). Correct option: C.