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Light with an enargy flux of 25xx10^(4) ...

Light with an enargy flux of `25xx10^(4) Wm^(-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

A

`1.25xx106(-6)N`

B

`2.50xx10^(-6)N`

C

`1.20xx10^(-6)N`

D

`3.0xx10^(-6)N`

Text Solution

Verified by Experts

The correct Answer is:
B
Here, Energy flux, `I=25xx10^(4)Wm^(-2)`
Area, `A=15cm^(2)=15xx10^(-4)m^(2)`
Speed of light, `c=3xx10^(8)ms^(-1)`
For a perfectly reflecting surface, the average force exerted on the surface is
`F=(2IA)/(c)=(2xx25xx10^(4)Wm^(-2)xx15xx10^(-4)m^(3))/(3xx10^(8)ms^(-1))`
`=250xx10^(-5)N=2.50xx10^(-6)N`
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