Question

The filament of a light bulb has a surface area $64m{m}^2$. The filament can be considered as a black body at temperature $2500K$emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of$100m$. Assume the pupil of the eyes of the observer to be circular with radius $3mm$. Then

• A. Power radiated by the filament is in the range$642W$ to $645W$
• B. Radiated power entering into one eye of the observer is in the range $3.15\times {10}^{-8}W$ to $3.25\times {10}^{-8}W$
• C. The wavelength corresponding to the maximum intensity of light is $1160nm$
• D. Taking the average wavelength of emitted radiation to be $1740nm$, the total number of photons entering per second into one eye of the observer is in the range $2.75\times {10}^{11}$ to $2.85\times {10}^{11}$

Answer 1

Answer: (B), (C), (D)

Taking the average wavelength of emitted radiation to be $1740nm$, the total number of photons entering per second into one eye of the observer is in the range $2.75\times {10}^{11}$ to $2.85\times {10}^{11}$

Step 1: Given data

The surface area of filament$\left(A\right)$=$64m{m}^2=64\times {10}^{-6}{m}^2$

$1m{m}^2={10}^{-6}{m}^2$

The distance between bulb and observer$\left(d\right)$=$100m$

The radius of the pupil eyes$\left(R_e\right)$=$3mm=3\times {10}^{-3}m$

Temperature$\left(T\right)=2500K$

The filament can be considered a black body, so emissivity $e=1$being the perfect absorber

Step 2: Formula used

Use the Stefan-Boltzmann law to determine the power radiated by the filament. This law can be expressed as,

$P=\sigma Ae{T}^4,whereP=powerradiated,e=emmisitivity,A=radiatingarea,T=temperatureand\sigma =Stefen\text{'}scons\tan t=5.67\times {10}^{-8}W/m^2-K^4$

Wien's displacement law can be expressed as-

${\lambda }_m=\frac{b}{T},whereT=absolutetemperate,b=cons\tan tand\lambda =wavelength$

$Wien\text{'}sdisplacementcons\tan tvalueb=2898\mu m.K$

Step 3: Compute power radiated by filament

The power radiated by filament can be calculated using the formula of Stefan-Boltzmann law -

Substitute the known values in the formula,

$$\begin{gathered} P=\left(5·67\times {10}^{-8}\right)\times \left(64\times {10}^{-6}\right)\times 1\times {\left(2500\right)}^4 \\ P=141·75W \end{gathered}$$

Thus, the power radiated by filament will be $141·75W$.

Step 4: Compute wavelength corresponding to the maximum intensity of light

The power observed by the eyes can be determined using the Wien's displacement law formula,

Substitute known values in the formula,

$$\begin{gathered} {\lambda }_m=\frac{2898\times {10}^{-6}}{2500} \\ {\lambda }_m=1.1592\mu m\approx 1160nm \end{gathered}$$

Thus, the wavelength corresponding to the maximum intensity of light is $1160nm$

Step 5: Calculating the radiated power entering one’s eye

The power reaching to the eyes or wavelength can be determined using the following formula,

$\begin{array}{rcl}& =& \frac{P}{A}\times Areaofpupileyes\\ & =& \frac{P}{4{\mathrm{\pi d}}^2}\times {\mathrm{\pi R}}_{\mathrm{e}}^2\\ & =& \frac{141·75}{4\times {100}^2}\times {\left(3·0\times {10}^{-3}\right)}^2\\ & =& 3.189\times {10}^{-8}\mathrm{W}\end{array}$

Thus, the radiated power entering one’s eye is $\begin{array}{l}3.189\times {10}^{-8}\mathrm{W}\end{array}$.

Step 6: Compute the number of photons entering the eye of the observer

Taking the average wavelength of emitted radiation to be $1740nm=1740\times {10}^{-9}m$, the total number of photons entering per second into one eye of the observer is in the range

$$\begin{gathered} N\frac{hc}{\lambda }=PwhereN=numberofphotons,h=cons\tan t,\lambda =wavelength,P=radiatedpowerandc=energyoflight \\ h=6.63\times {10}^{-34}J.s \end{gathered}$$

$$\begin{gathered} P=N\times \frac{hc}{\lambda } \\ N=\frac{P\lambda }{hc} \\ N=\frac{3·189\times {10}^{-8}\times 1740\times {10}^{-9}}{6·63\times {10}^{-34}\times 3·0\times {10}^8} \\ N=2·77\times {10}^{11}photons \end{gathered}$$

Thus, the number of photons entering one's eyes will be $2·77\times {10}^{11}photons$.

So, the power radiated by filament will be $141·75W$ and radiated power entering into one eye of the observer is in the range $3.15\times {10}^{-8}W$ to $3.25\times {10}^{-8}W$ which we got as$\begin{array}{l}3.189\times {10}^{-8}\mathrm{W}\end{array}$, the wavelength corresponding to the maximum intensity of light is $1160nm$ and the number of photons entering one's eyes will be $2·77\times {10}^{11}photons$

Hence, options B, C, and D are the correct answers.