The minimum energy of the visible lights of wavelength 500to 600 nm that can just be detected by the eye is 10e-17Watts. Only 1% of power input to a 60W light bulb is emitted between these wavelengths. Estimate the maximum distance that it might be possible to see the lit bulb. The pupil of the eye will be at its maximum diameter say 8mm. At its maximum sensitivity the eye only detect 8% of the light incident on it.(adsbygoogle = window.adsbygoogle || []).push({});

My answer

Minimum power the eye can detect= 10e-17w

1% of 60w=0.6w

Therefore, the maximum distance to see the lit bulb,

Power output ÷ surface area of a sphere = power per meter square

0.6÷(4πr^2)=10e-17

0.6÷10e-17=(4πr^2)

(6X10^16)/4π=r^2

so r=sqrt(6X10^16)/4π)

r= 217080km however in answer is 140km.

Please help me with this. Thanks.

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# Homework Help: Finding maximum distance of a lit power source

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