At night approximately 500 photons per second must enter an unaided human eye for an object to be seen. A small light bulb emits about $5.00 \times 10^{18}$ photons per second uniformly in all directions. The radius of the pupil of the eye is about $4 \times 10^{- 3}$ meters. What is the approximate maximum distance from which the bulb could be seen ?

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$2.0 \times 10^{4} m$

$5.0 \times 10^{3} m$

$2.0 \times 10^{5} m$

20 m

answer is B.

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Detailed Solution

Photons per area per second at a distance r are $5.00 \times 10^{18} / 4 {πr}^{2}$ . Photons per second entering the eye, radius R is then this times ${πR}^{2}$ . Set this product equal to 500 per second and solve for r. The result is [B].

$\frac{5.00 \times 10^{18}}{4 {πr}^{2}} \times {πR}^{2} = 500$

$\Rightarrow \frac{5.00 \times 10^{18}}{4 {πr}^{2}} \times π \times 16 \times 10^{- 6} = 500$

$\Rightarrow r = 2.0 \times 10^{5} m$

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