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Demonstrate that the angle theta between the propegation direction of light and the x axis transfroms on transition from the reference frame K to K' according to the formula cos theta' = (cos theta - beta)/(1-beta cos theta), where beta = V//c and V is the velocity of the frame K'with respect to the frame K. the x and x' axes of the reference frames coincide. |
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Answer» SOLUTION :We consider the invariance of the plane of a wave moving in the `x-y` plane. We write `omega't' - k'_(x)chi' - k'_(y)y' = omegat - k_(x)chi - k_(y)y` From Lorcntz transformations, `L.H.S.` `= omega' gamma (t-(vx)/(c^(2))) -k'_(x)(x-vt) gamma - k'_(y)y` so EQUATING `omega = gamma (omega' +v k'_(x))` `k_(x) = gamma (k'_(x)+(v omega')/(c^(2)))` and `k_(y) = k'_(y)` `omega' = gamma (omega - vk_(x))` so inverting `k'_(x) = gamma (k_(x) -(v omega)/(c^(2)))` `k'_(y) = k_(y)` `k'_(x) = k' costheta', k_(x) = k cos theta` weiting `k'_(y) = k' sin theta', k_(y) = k sin theta` we get on using `ck' = omega', ck = omega` `cos theta' = (cos theta - beta)/(1-beta costheta)` where `beta = (v)/(c)` and the primed frame is moving with velocity `v` in the `x-`direction w.r.t. the unprimed frame. For small `beta lt lt 1`, the situation is as SHOWN. We see that `theta' =- beta` Then `theta' =- ((pi)/(2)+sin^(-1)beta)` This is exactly what we get from ELEMENTARY nonrelativistic law of addition of VELOCITIES, 
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