A small mirror of mass m is suspended by a massless thread of length l. 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 = speed of light in vacuum and g = acceleration due to gravity)
- = Emc gl
- = E2mc gl
- = 3E4mc gl
- = 2Emc gl
Answer
(4) = 2Emc gl
Solution
Momentum of the light pulse: p = Ec
The mirror reflects the pulse back, so the momentum given to the mirror: p = Ec - (- Ec ) = 2Ec
Speed of the mirror just after the pulse: v = 2Emc
Energy conservation as it swings up: 12mv2 = mgl(1 - )
For small : 1 - 22 v2 = gl 2 v = gl
= v gl = 2Emc gl
Answer: (4)