If 2.5 times 10^{-6} N average force is exer | vedclass

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

Question

(D) For a non-reflecting surface (complete absorption) with normal incidence,the force $F$ exerted by light is given by $F = \frac{P}{c}$,where $P$ is

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(D) For a non-reflecting surface (complete absorption) with normal incidence,the force $F$ exerted by light is given by $F = \frac{P}{c}$,where $P$ is the power and $c$ is the speed of light.Since power $P = I 	imes A$,where $I$ is the intensity (energy flux) and $A$ is the area,the force equation becomes $F = \frac{I A}{c}$.Rearranging for intensity $I$,we get $I = \frac{F c}{A}$.Given values: $F = 2.5 	imes 10^{-6} \ N$,$A = 30 \ cm^{2}$,and $c = 3 	imes 10^{8} \ m/s$.Substituting these values: $I = \frac{2.5 	imes 10^{-6} 	imes 3 	imes 10^{8}}{30} = \frac{7.5 	imes 10^{2}}{30} = \frac{750}{30} = 25 \ W/cm^{2}$.The energy flux is $25 \ W/cm^{2}$.

Similar Questions

$A$ 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 $1 \,m$ away from a $330 \,W$ electric bulb is

Light with an intensity of $18 \ W \ m^{-2}$ is incident normally on an opaque surface. If the area of the surface is $20 \ m^2$,calculate the average force exerted on the surface in $30 \ min$.

$A$ 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).

$A$ light with an energy flux of $18 \,W \,cm^{-2}$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $20 \,cm^2$, the average force exerted on the surface during a $30$ minute time span is . . . . . . $\left(c=3 \times 10^8 \,ms^{-1}\right)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module