A small mirror of mass m is suspended by a massless thread of length \(l\). Then the small angle through which the threa…

A laser pulse of energy E is incident normally and reflected from the mirror, so momentum change is twice that of incident light.
\(p=\frac{E}{c}\)
Change in momentum transferred to mirror is:
\(\mathrm{Δp}=\frac{2E}{c}\)
Using impulse–momentum relation for the mirror:
\(mv=\frac{2E}{c}\)
Horizontal velocity acquired by the mirror is:
\(v=\frac{2E}{mc}\)
The mirror then behaves as a pendulum of length l performing small oscillations.
Maximum angular deflection for small angles is given by:
\(\theta =\frac{v}{\sqrt{gl}}\)
Substituting the value of v:
\(\theta =\frac{2E}{mc\sqrt{gl}}\)
Final answer:
\(\theta =\frac{2E}{mc\sqrt{gl}}\)