IIT JEE , AIEEE Forum - Physics , Mechanics

trisha

A small mirror of mass 'M' is suspended by a light thread of length L.Find the angle through which the thread will be deflected when a short pulse of laser of energy E falls normally on the perfectly absorbing mirror.(If 'C' is the velocity of light find the angle (in radian) in terms of E,M,C&L)

sudeep.kumar

p = E/c

So, change in momentum, = 2E/c

Just conserve momentum of mirror + beam, rest is just mechanics.

tell me if you are unable to solve upon this hint, i will post the solution.

trisha

sir, if u dont mind please give the complete solution.

bhupesh

dear

first apply conservation of momentum then we can apply conservation of energy

Its like a bullet striking a bob and bob moves up

so initial momentum is beacause of EMW paulse only which can be shown tobe

2E/c

2ER/c = mv ( 2 is multiplied because the mirorr reflects )

mgl (1-cos theta ) = 1/2 mv^2

sudeep.kumar

Bhupes is all correct....

After you get v = 2E/mc,

Just conserve energy to get

in radians.

Here, note that once the motion of the mirror starts, it will continue doing angular SHM (if we neglect damping), and the lowest position will be the mean position. Hence, 2E/mc is the velocity at mean position, and hence if you know w, you can get angular amplitude directly.

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