Question

A small ideal mirror of mass `m = 10 mg` is suspended by w weightless theard of length `l = 10 cm`. Find the angle through which the thread will be deflected when a short laser pulse angles enegry `E = 13J` is shot in the horizontal direction at right angles to the mirror. Where does the mirror get its kinetic energy?

Answer 1

When light falls on a small mirror and is reflected by it, the mirror recoils. The enegry of recoil is obtained from the incident beam photon and the frequency of reflected photons is less than the frequency of the incident photons. This shify of frequency can however be neglected in calculating quantities related to recoil (to a first approximation.) Thus, the momentum acquired by the mirror as a result of the laser pulse is
`|oversetrarr(P_(f))-oversetrarr(P_(1))| = (2E)/(c)`
Or assuming `oversetrarr(P_(1)) = 0`, we get `|overset rarr(P_(f))| = (2E)/(c)`
Hence the kinetic enegry of the mirror is
`(p_(f)^(2))/(2m) = (2E^(2))/(mc^(2))`
Suppose the mirror is deflected by an angle `theta`. Then by conservation of energy fi9nal `P.E. = mgl(1-cos theta) =`Initial `K.E. = (2E^(2))/(mc^(2))`
or `mgl2 sin^(2).(theta)/(2) = (2E^(2))/(mc^(2))`
or `sin.((theta)/(2)) = ((E)/(mc))(1)/(sqrt(gl))`
Using the data. `sin.((theta)/(2)) = (13)/(10^(-5)xx3xx10^(8)sqrt(9.8xx.1)) = 4.377 xx 10^(-3)`
This given `theta = 0.502` degreess.