Question

How many photons per second enter one eye if you look directly at a 100 W light bulb 2.00 m away? Assume a pupil diameter of 4.00 mm and a wavelength of 600 nm. How many photons per second enter your eye if a 1.00 m W laser beam is directed into your eye? $\lambda =633nm)$

Answer 1

Given,
Wavelength of light $\lambda =600nm=600\times {10}^{-9}m$

Power $P=100W$

Diameter of pupil $D_p=4mm=4\times {10}^{-3}m$

Energy of one photon $E_{\lambda }=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{600\times {10}^{-9}}=3.3\times {10}^{-19}J$

Area of sphere (2m radius) $A_{2m}=4\pi r^2=4\pi \times 2^2=16\pi m^2$

Area of pupil $A_p=\frac{\pi {D_p}^2}{4}=\frac{\pi (4\times {10}^{-3})}{4}=\pi {10}^{-3}{m}^2$

Intensity at 2m from source $I_{2m}=\frac{P}{A_{2m}}=\frac{100}{16\pi }$

Power fall on pupil $P_p=I_{2m}\times A_p=\frac{100}{16\pi }\times \pi {10}^{-3}=6.25\times {10}^{-3}W$

Number of photons $N=\frac{P_p}{E_{\lambda }}=\frac{6.25\times {10}^{-3}}{3.3\times {10}^{-19}}=1.89\times {10}^{16}photons$

(b)

Given,

Laser wavelength $\lambda =633nm$

Energy fall on eye $P=1mW={10}^{-3}W$

Energy in one photon $E_{\lambda }=\frac{hc}{\lambda }=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{633\times {10}^{-9}}=3.14\times {10}^{-19}J$

Number of photons $N=\frac{P}{E_{\lambda }}=\frac{{10}^{-3}}{3.14\times {10}^{-19}}=3.14\times {10}^{16}photons$