Assuming that 10.0% of a 100 W light bulb's energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, and that the photons spread out uniformly and are not absorbed by the atmosphere, how far away (in km) would you be if 468 photons per second enter the 3.30 mm diameter pupil of your eye

Explanation:

10% of 100 W = 10 W .

energy of each photon = hν , h is plank's constant and ν is frequency

= hc / λ , c is velocity of light , λ is wavelength

wavelength given λ  = 580 x 10⁻⁹ m .

energy of each photon = 6.6 x 10⁻³⁴ x 3 x 10⁸ / 580 x 10⁻⁹

= .034137 x ⁻¹⁷ J

no of photons in 10 W

= 10 /  .034137 x ⁻¹⁷

= 292.9 x 10¹⁷

If d be the required distance

no of photons per unit area per second  at distance d

= 292.9 x 10¹⁷ / 4πd²

no of photons through the area of eye

= π x (1.65 x 10⁻³)² x 292.9 x 10¹⁷ / 4πd²

= 199.355 x 10¹¹ / d²

199.355 x 10¹¹ / d² = 468 ( given )

d² = 199.355 x 10¹¹ / 468

= .42597 x 10¹¹

= 42597 x 10⁶

d = 206.4 x 10³ m

= 206.3 km.

electrical to thermal