How many photons per second, N, would enter an aperture of area A = 3cm2 located a distance D = 5 m from the light bulb?

N = 6.4886 *10¹²

Explanation:

Assumption:

2.5% of electrical energy is converted into light.

Power = 100W

So, to calculate energy of light per scoend, wee simply multiply 100 to 2.5%

E = 100 * (2.5/100) = 2.5J

To calculate number of photons, we have formula

N = (2.5 * λ)/hc

Now, using distance of 5m, we find area of sphere using formula 4λR²

A=4*3.14*5² = 314m²

Now, to determine number of photons passing through area of 3cm² or 0.0003m²

N = (3*10⁻⁴/314)*[(2.5*λ)/hc]

Now by putting the values we gett,

N = (3*10⁻⁴/314)*[(2.5*540*10⁻⁹)/((6.626*10⁻³⁴J.s)(3*10⁸))]

N = 6.4886 *10¹²

that would be because of both gravity and friction. so     c