Question 47: EM waves

Question

Average force exerted on a non-reflecting surface at normal incidence is $$2.4 	imes 10^{-4} \mathrm{~N}$$. If $$360 \mathrm{~W} / \mathrm{cm}^2$$ is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:

Accepted Answer

To solve for the area of the surface, we need to understand the relationship between the force exerted by the light, the light energy flux, and the area of the surface. The pressure exerted by the light on a non-reflecting surface is given by the formula:
$$ P = \frac{F}{A} $$
where $$P$$ is the pressure, $$F$$ is the force, and $$A$$ is the area. The pressure due to the light can also be related to the energy flux $$I$$ by the relationship:
$$ P = \frac{I}{c} $$
where $$I$$ is the light energy flux and $$c$$ is the speed of light in a vacuum ($$3 	imes 10^8 \mathrm{~m/s}$$).
Given that the average force $$F$$ is $$2.4 	imes 10^{-4} \mathrm{~N}$$ and the light energy flux $$I$$ is $$360 \mathrm{~W/cm}^2$$, we first convert the flux to $$\mathrm{W/m}^2$$:
$$ 360 \mathrm{~W/cm}^2 = 360 	imes 10^4 \mathrm{~W/m}^2 $$
Now we can use the relation between pressure and energy flux:
$$ P = \frac{I}{c} = \frac{360 	imes 10^4}{3 	imes 10^8} = 1.2 \mathrm{~N/m}^2 $$
Next, we use the pressure formula to find the area of the surface:
$$ P = \frac{F}{A} \implies A = \frac{F}{P} $$
Substituting the values we have:
$$ A = \frac{2.4 	imes 10^{-4} \mathrm{~N}}{1.2 \mathrm{~N/m}^2} = 2 	imes 10^{-4} \mathrm{~m}^2 $$
The area of the surface is:
$$ A = 2 	imes 10^{-4} \mathrm{~m}^2 = 0.02 \mathrm{~m}^2 $$
Hence, the correct option is:
Option D $$0.02 \mathrm{~m}^2$$

Answer

$$20 \mathrm{~m}^2$$

Answer

$$0.2 \mathrm{~m}^2$$

Answer

$$0.1 \mathrm{~m}^2$$

Answer

$$0.02 \mathrm{~m}^2$$

EM waves

Average force exerted on a non-reflecting surface at normal incidence is $2.4 \times 10^{-4} \mathrm{~N}. If 360 \mathrm{~W} / \mathrm{cm}^2$ is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:

Choose the correct answer: