Light with an energy flux of 25 times 10^4 , | vedclass

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(B) The energy flux (intensity) is given by $I = 25 	imes 10^4 \, W/m^2$.The surface area is $A = 15 \, cm^2 = 15 	imes 10^{-4} \, m^2$.The speed of

Vedclass · Accepted Answer

$2.50 	imes 10^{-6} \, N$

Vedclass · Answer

(B) The energy flux (intensity) is given by $I = 25 	imes 10^4 \, W/m^2$.The surface area is $A = 15 \, cm^2 = 15 	imes 10^{-4} \, m^2$.The speed of light is $c = 3 	imes 10^8 \, m/s$.For a perfectly reflecting surface,the momentum transferred per unit time (force) is given by the formula $F = \frac{2IA}{c}$.Substituting the values:$F = \frac{2 	imes (25 	imes 10^4) 	imes (15 	imes 10^{-4})}{3 	imes 10^8}$$F = \frac{2 	imes 25 	imes 15 	imes 10^0}{3 	imes 10^8}$$F = \frac{750}{3} 	imes 10^{-8} \, N$$F = 250 	imes 10^{-8} \, N = 2.50 	imes 10^{-6} \, N$.

Light with an energy flux of $25 \times 10^4 \, W/m^2$ falls on a perfectly reflecting surface at normal incidence. If the surface area is $15 \, cm^2$,the average force exerted on the surface is:

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