Class 12 NEET Physics Electromagnetic Waves Questions

Formula used: Momentum of photonic radiation $p=\frac{E}{c}$Momentum transferred $\Delta p=p_i-p_f$Complete step by step answer:From the mass-energy equivalence, we have the energy of the photon given as $E=m c^2$, where $m$ is the energetic mass of the photon and $c$ is the velocity of light.We know that the momentum of the photon can be expressed as $p=m c$$\Rightarrow E=(m c) c=p c$This means that the momentum possessed by a photon with energy E is:$p=\frac{E}{c}$There is a photon incident on a reflecting surface. The initial momentum of the photon is given as:$p_i=\frac{E}{c}$Now, the photon falls normally on a perfectly reflecting surface. This means that the photon is reflected back with the same magnitude of incident momentum but is directionally opposite.The final momentum of the reflected photon is given as:$p_f=-\frac{E}{c}$We understand that the photon is able to reflect back after incidence because it imparts its momentum to the reflecting surface upon incidence. This momentum transferred between the photon and the reflecting surface will be equivalent to the change in momentum of the photon.Therefore, the change in momentum of the photon is given as:$\Delta p=p_i-p_f=\frac{E}{c}- \left(-\frac{E}{c}\right) =\frac{E}{c}+\frac{E}{c}$$\Rightarrow \Delta p=\frac{2 E}{c}$This is equivalent to the momentum transferred by the incident photon to the reflecting surface which consequently brings about perfect reflection.