A totally reflecting, small plane mirror placed horizontally faces a parallel beam of light as shown in the Fig. The mass of the mirror is 20 g. Assume that there is no absorption in in the lens and that `30%` of the light emitted by the source goes through the lens. Find the power (in`xx10^(8)`W) of the source needed to support the weight of the mirror. Take `g=10ms^(-2)`

Explanation

(a) Let P: power of lens. `30%` of light passes through lens and light is completely reflected.
Force exerted by lilght `=(2(0.3P))/(c) =(0.6P)/(c)`
Weight of mirror = mg

In equilibrium
`(0.6P)/(c) =mg`
`P=(mgc)/(0.6) =(20xx10^(-3)xx10xx3xx10^(8))/(0.6) =10^(8) W`
(b)
Intensity of light at surface
`I =(0.6P)/(4 pi r^(2))`
Pressure exerted by light
`p=(I)/(c) =(0.6 P)/(4 pi r^(2)c) =(110xx0.6)/(4pi(0.2)^(2)xx3xx10^(8))`
`=4.4xx10^(-6) N//m^(2)`
(c)
(i) Force exerted by light `F =P//c`
`tan theta =(P//c)/(mg) =(P)/(mgs)`
`theta =tan^(-1) ((P)/(mgc))`
(ii) Time period of simple pendulum
`T=2 pi sqrt((L)/(a))`
a: effective acceleration.
`a=(sqrt((mg)^(2) + (P//c)^(2)))/(m)`
`sqrt(g^(2) +((P)/(mc))^(2))`
`T =(2pi sqrt(L))/([g^(2) +((P)/(mc))^(2)]^(1//4))`