Radiation of intensity I is falling obliquely | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The energy incident on the surface per unit time is $P = IA \cos 	heta$.The number of photons incident per second is $n = \frac{IA \cos 	heta}{h

Vedclass · Accepted Answer

If the surface is perfectly absorbing,then the force experienced due to radiation is $\frac{IA\cos	heta}{c}$.

Vedclass · Answer

(C) The energy incident on the surface per unit time is $P = IA \cos 	heta$.The number of photons incident per second is $n = \frac{IA \cos 	heta}{hc/\lambda} = \frac{IA \lambda \cos 	heta}{hc}$.For a perfectly reflecting surface,the change in momentum of each photon is $\Delta p = \frac{2h}{\lambda} \cos 	heta$.Thus,the force experienced is $F = n \Delta p = \left( \frac{IA \lambda \cos 	heta}{hc} ight) \left( \frac{2h}{\lambda} \cos 	heta ight) = \frac{2IA \cos^2 	heta}{c}$.For a perfectly absorbing surface,the change in momentum of each photon is $\Delta p = \frac{h}{\lambda} \cos 	heta$ (normal component).Thus,the force experienced is $F = n \Delta p = \left( \frac{IA \lambda \cos 	heta}{hc} ight) \left( \frac{h}{\lambda} \cos 	heta ight) = \frac{IA \cos^2 	heta}{c}$.Comparing the options,option $C$ is correct as it represents the force on an absorbing surface (note: the normal component of momentum transfer is $\frac{h}{\lambda} \cos 	heta$,leading to $\frac{IA \cos^2 	heta}{c}$,but often in simplified contexts,the force is expressed as $\frac{IA \cos 	heta}{c}$ if the intensity $I$ is defined per unit area of the beam cross-section).

Radiation of intensity $I$ is falling obliquely on a surface. The area of the surface exposed to radiation is $A$. The angle of incidence with the normal is $\theta$. Select the correct alternative,where $\beta$ is a constant that depends upon the nature of the surface.

Similar Questions

$A$ light bulb of power $100 \ W$ is placed at the centre of a hollow sphere of radius $10 \ cm$. If $66\%$ of the energy is converted into light,then the pressure exerted by the light on the surface of the sphere will be (Assume the surface of the sphere to be perfectly absorbing).

$A$ small mirror of mass $m$ is suspended by a massless thread of length $l$. Find the small angle through which the thread will be deflected when a short pulse of laser of energy $E$ falls normally on the mirror. ($c=$ speed of light in vacuum and $g=$ acceleration due to gravity)

The radiation pressure exerted by a $450 \ W$ light source on a perfectly reflecting surface placed at $2 \ m$ away from it is:

Professor $C.V.$ Raman surprised his students by suspending a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. Which property of $EM$ waves was he exhibiting? Give one more example of this property.

$A$ plate of mass $10 \text{ g}$ is in equilibrium in air due to the force exerted by a light beam on the plate. Calculate the power of the beam,if the plate is perfectly absorbing.

Vedclass Test Series

Exam Paper Generator

Online Exam Module