Question

It is not good for you to stare at the Sun, or to shine laser pointers into the eyes; Compare the two.

(a) The intensity (power per area) of the Sun at the surface of the Earth is 1.367 kW/m². Estimate the number of photons that enter your eye if you look for a tenth of a second at the Sun.

(b) What energy is absorbed by your eye during that time, assuming that all the photons are absorbed?

(c) How long would you have to have a 1mW laser pointer shine in your eye, to absorb the same energy? Clearly, state any assumptions/estimates you are making for values you need to solve this problem.  

Answer 1

total output power of Sun = 4.2 *10^26 Watts

Intensity = power/area

area = 4pi r^2

r = dstance between the sun and the earth = 1.5*10^11 m

I = (4.2 *10^26) /(4pi * 1.5*10^11)^2

I = power/area

Power = energy/time = E/t

I = E/(A*t)

E = I A per second

E = (4.2*10^26)(3.14 * 1*10^-3)^2/(4pi * 1.5 *10^9)^2

E = 4.7 milli JOules

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Eenrgy absorbed per one secind = E = 0.1* 4.7 mJ

E = 0.47 mJ

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En for single photon E = hc/L

En = (6.626*10^-34 * 3*10^8)/(550 nm)

En = 3.6 *10^-19 J

No. of photons = 0.47 *10^-3/(3.6 *10^-19)

N = 1.3 *10^15

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Power = Eenery/time

time   = 0.47 *10^-3/(1*10^-3)

t = 0.47 secs