Question

If a 120-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 1.6 m away?

Express your answer using two significant figures.

Part B

How many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 1.5 km away?

Express your answer using two significant figures.

If a 120-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 1.6 m away? Express your answer using two significant figures. TO ADD A 0 2 ? photons/s Submit Request Answer Part B How many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 1.5 km away? Express your answer using two significant figures. ALQ 0 2 ? photons/s Submit Request Answer

Answer 1

Solution:

Given :

P = 120 W ; % = 2.5 ;

Wavelength ($\lambda$) = 550 nm

Part (A) Solution:

Here, d = 4 mm ; D = 1.6 m

P = 0.025 x 120 = 3 W

We know that the energy carried by a single photon is:

E = h c / $\lambda$ = (6.626 x 10^-34)(3 x 10⁸) / (550 x 10^-9) = 3.61 x 10^-19 J

We know that, intensity is:

I = P/A = 3 / { 4 x 3.14 x (1.6)² } = 0.09325 W/m²

Therefore: at eye ; P = I x A

P = (0.09325 W/m²)(3.14) x (2 x 10^-3)² = 1.172 x 10^-6 J/s

Thus: N = ( 1.172 x 10^-6 J/s) / (3.61 x 10^-19) = 3.246 x 10¹² photons/s

Hence, N = 3.246 x 10¹² photons/s

Part (B) Solution:

Here, I = P/A = (3 W) / {4 x 3.14 x (1500 m)²} = 1.1 x 10^-7 W/m²

P = I x A (at eye)

P = 1.1 x 10^-7 x 3.14 x (2 x 10^-3)² = 1.4 x 10^-12 J/s

N = (1.4 x 10^-12 J/s) /(3.61 x 10^-19) = 3.9 x 10⁶ photons/s

Hence, N = 3.9 x 10⁶ photons/s