1.A small mirror of mass mmm is suspended by a light thread of length LLL. The angle through which the thread will be deflected when a short pulse of laser of energy EEE falls normally on the mirror isa.θ=2Emcg\theta = \dfrac{2E}{mc\sqrt{g}}θ=mcg2Eb.θ=2Emc2g\theta = \dfrac{2E}{mc\sqrt{2g}}θ=mc2g2Ec.θ=mc2g2E\theta = \dfrac{mc\sqrt{2g}}{2E}θ=2Emc2gd.θ=2Emcge\theta = \dfrac{2E}{mc\sqrt{ge}}θ=mcge2ELogin to continueOnly logged in users canattempt or see the solution.