Light with an energy flux of $25 imes 10^4 ext{ W m}^{-2}$ falls on a perfectly reflecting surface at norm

Question

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

Accepted Answer

1. **Identify the Formula:** The force exerted by an electromagnetic wave on a surface depends on whether the surface absorbs or reflects the radiation. For a **perfectly reflecting** surface at normal incidence, the change in momentum is double that of an absorbing surface. The force ($F$) is given by:
   $$ F = \frac{2IA}{c} $$
   where $I$ is the energy flux (intensity), $A$ is the surface area, and $c$ is the speed of light .

2. **Convert Units:**
   - Energy Flux ($I$) = $25 	imes 10^4 	ext{ W/m}^2$
   - Surface Area ($A$) = $15 	ext{ cm}^2 = 15 	imes 10^{-4} 	ext{ m}^2$
   - Speed of light ($c$) $\approx 3 	imes 10^8 	ext{ m/s}$

3. **Calculate Force:**
   Substitute the values into the formula:
   $$ F = \frac{2 	imes (25 	imes 10^4) 	imes (15 	imes 10^{-4})}{3 	imes 10^8} $$
   $$ F = \frac{2 	imes 25 	imes 15 	imes 10^0}{3 	imes 10^8} $$
   $$ F = \frac{750}{3 	imes 10^8} $$
   $$ F = 250 	imes 10^{-8} 	ext{ N} $$
   $$ F = 2.50 	imes 10^{-6} 	ext{ N} $$

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