Question

determine the force exerted on perfectly reflecting surface when 47.0mW beam of radius 1.50mm is reflected from it

Answer 1

Explanation:

Area of the beam = πr^2 = (3.14)(1.5)^2 = 7.065 m^2

Now we have to find the intensity:

I = P/A = 47m/7.065 = 6.65 mWm^(-2)

Now, pressure will be:

P = 2I/c

By putting values we get,

P = 4.43 × 10^(-11 )

And

F = PA

F = ( 4.43 × 10^(-11) )( 7.065 )

F = 3.13 × 10^(-10) N

Answer 2

Answer:

3.13*10^-7 N.

Explanation:

Since, the diameter of the beam is given as 1.5 mm, hence the radius will be 0.0015/2 m. So, the area will be πR^2 or 3.14*0.00075 which on solving we will get 0.002355 m^2.

So, the intensity of the beam will be I=P/A =  47/0.002355 = 19957.53 W/m^2. So, the radiation pressure will be 2I/c or the force exerted will be given as pressure*area or 2IA/c = 2* 19957.53 * 0.002355 /3*10^8 = 3.13*10^-7 N.