Question

A perfecting reflecting mirror has an area of 1cm² . Light energy is allowed to fall on it for an hour at the rate of 10 Wcm^-2. The force that acts on the mirror is 

A. 3.35×10^-7N 

B. 6.7×10^-7N 

C. 3.35×10^-8N 

D. 6.7×10^-8N

Answer 1

Given:

Area of the mirror = 1 cm² ,

Power transmitted to the mirror = 10 Wcm^-2 , 

Calculations: - 

The average force is given by \(\frac{Δp}{Δt}\)

Since the surface is perfectly reflecting we can write Δp = 2\(\frac{Δt}{c}\)

So F = 2 \((\frac{ΔE}{cΔt})\) = 2 \(\frac{p}c\)

=2 \(\frac{10\times10^4\times1\times10^{-4}}{3\times10^8}\)

= \(\frac{20}{3}\)x 10^-8 N

= 6.7 x 10^-8 N