Question

If a 110-W lightbulb emits 2.5 % of the input energy as visible light (average wavelength 550 nm) uniformly in all directions. Part A How many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 2.8 m away? Express your answer using two significant figures.

Answer 1

Answer:

9.7 x 10¹¹ .

Explanation:

2.5 % of 110 W = 2.75 J/s

energy of one photon

= hc / λ

=$\frac{6.6\times10^{-34}\times3\times10^8}{550\times10^{-9}}$

= .036 x 10⁻¹⁷ J

No of photons emitted

= 2.75 / .036 x 10⁻¹⁷

= 76.38 x 10¹⁷

Now photons are uniformly distributed in all directions so they will pass through a spherical surface of radius 2.8 m at this distance

photons passing per unit area of this sphere

= 76.38 x 10¹⁷  / 4π ( 2.8)²

Through eye which has surface area of π x ( 2 x 10⁻² )² m² , no of photons passing

= $\frac{76.38\times10^{17}}{4\pi\times(2.8)^2} \times\pi(2\times10^{-3})^2$

= 9.7 x 10¹¹ .