QJEEM24Ph9975: JEE Main 2024 Physics - Class 12 JEEM24Ph9975 Solution: Average force exerted on a non-reflecting surface at normal…

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 ( $3 \times 10^8 \mathrm{~m/s}$ ).

Given that the average force $F$ is $2.4 \times 10^{-4} \mathrm{~N}$ and the light energy flux $I$ is $360 \mathrm{~W/cm}^2$ , we first convert the flux to $\mathrm{W/m}^2$ :

$360 \mathrm{~W/cm}^2 = 360 \times 10^4 \mathrm{~W/m}^2$

Now we can use the relation between pressure and energy flux:

$P = \frac{I}{c} = \frac{360 \times 10^4}{3 \times 10^8} = 1.2 \mathrm{~N/m}^2$

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:

$A = \frac{2.4 \times 10^{-4} \mathrm{~N}}{1.2 \mathrm{~N/m}^2} = 2 \times 10^{-4} \mathrm{~m}^2$

The area of the surface is:

$A = 2 \times 10^{-4} \mathrm{~m}^2 = 0.02 \mathrm{~m}^2$

Hence, the correct option is:

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