A totally reflecting small plane mirror place | vedclass

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

Question

(B) For a perfectly reflecting mirror,the force $F$ exerted by a beam of light with power $P_{mirror}$ is given by:$F = \frac{2 P_{mirror}}{c}$For the

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(B) For a perfectly reflecting mirror,the force $F$ exerted by a beam of light with power $P_{mirror}$ is given by:$F = \frac{2 P_{mirror}}{c}$For the mirror to be in equilibrium,the radiation force must balance its weight:$F = mg$$\frac{2 P_{mirror}}{c} = mg$$P_{mirror} = \frac{mgc}{2}$Given $m = 20 \, g = 0.02 \, kg$,$g = 10 \, m/s^2$,and $c = 3 	imes 10^8 \, m/s$:$P_{mirror} = \frac{0.02 	imes 10 	imes 3 	imes 10^8}{2} = 0.01 	imes 3 	imes 10^8 = 3 	imes 10^6 \, W = 3 \, MW$Since only $30 \%$ of the total power $P_{source}$ emitted by the source reaches the mirror:$P_{mirror} = 0.30 	imes P_{source}$$3 \, MW = 0.30 	imes P_{source}$$P_{source} = \frac{3}{0.30} = 10 \, MW$Wait,re-evaluating the calculation: $P_{mirror} = \frac{0.02 	imes 10 	imes 3 	imes 10^8}{2} = 30 	imes 10^6 \, W = 30 \, MW$. Then $P_{source} = \frac{30}{0.30} = 100 \, MW$.

Similar Questions

$A$ photon of wavelength $6630 \ \mathring{A}$ is incident on a totally reflecting surface. The momentum delivered by the photon is equal to

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.

The radiation pressure $1 \,m$ away from a $330 \,W$ electric bulb is

Light having an energy flux of $40 \ Wcm^{-2}$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $20 \ cm^2$,the total momentum delivered (for complete absorption) during $10 \ min$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module