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

9.7 x 10¹¹ .

Step-by-step explanation:

2.5 % of 110 W = 2.75 J/s

energy of one photon

= hc / λ

=
`(6.6*10^(-34)*3*10^8)/(550*10^(-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

=
`(76.38*10^(17))/(4\pi*(2.8)^2) *\pi(2*10^(-3))^2`

= 9.7 x 10¹¹ .