Question

The filament of a light bulb has a surface area 64 $m{m^2}$. The filament can be considered as a black body at temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius 3 mm. Then
(Take Stefan-Boltzmann constant = \[5.67\text{ }\times \text{ }{{10}^{-8}}~W{{m}^{-2}}{{K}^{-4}}\],
Wien’s displacement constant =\[2.90\text{ }\times \text{ }{{10}^{-3}}m-K\],
Planck’s constant =\[6.63\text{ }\times \text{ }{{10}^{-34}}Js\],
speed of light in vacuum =\[3.00\text{ }\times {{10}^{8}}~m{{s}^{-1}}\])
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 \[3.15\text{ }\times \text{ }{{10}^{-8}}~W\]to \[3.25\text{ }\times \text{ }{{10}^{-8}}~W\]
C. the wavelength corresponding to the maximum intensity of light is \[1160\text{ }nm\]
D. taking the average wavelength of emitted radiation to be\[1740\text{ }nm\], the total number of photons entering per second into one eye of the observer is in the range \[2.75\times {{10}^{11}}~to\text{ }2.85\times {{10}^{11}}\]

Answer 1

Hint: It is crucial to remember all the black body radiation ideas when answering questions like the one above, including what creates it, what influences it, if it can be raised or lessened, and if so, how. Remember that a black body radiation's total power can be computed.

Complete answer:
Given:
$A=64m{{m}^{2}}$, $T=2500K$ (A = surface area of filament, T= temperature of filament, d is distance of bulb from observer,${{R}_{e}}$= radius of pupil of eye).
Point source d = 100m
${{R}_{e}}=3mm$

(a)$P=\sigma Ae{{T}^{4}}$
$=5.67\times {{10}^{-8}}\times 64\times {{10}^{-6}}\times 1\times {{(2500)}^{4}}$ (e = a black body)
$=141.75w$
Option A is wrong.

(b) Power reaching to the eye
$=\dfrac{P}{4\pi {{d}^{2}}}\times (\pi R_{e}^{2})$
$=\dfrac{141.75}{4\pi \times {{(100)}^{2}}}\times \pi {{(3\times {{10}^{-3}})}^{2}}$
$=3.189375\times {{10}^{-8}}W$
Option (b) is correct.

(c) ${{\lambda }_{m}}T=b$
${{\lambda }_{m}}\times 2500=2.9\times {{10}^{-3}}$
${{\lambda }_{m}}=1.16\times {{10}^{-6}}$
 $=1160nm$
Option C is correct.

(d) Power received by one eye of observer $\left( \dfrac{hc}{\lambda } \right)\times N$
N = number of photons entering into eye per second
$\Rightarrow 3.189375\times {{10}^{-8}}$
$=\dfrac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1740\times {{10}^{-9}}}\times N$
$\Rightarrow N=2.79\times {{10}^{11}}$
Option D is correct.

Hence, options B, C and D are correct.

Note: The temperature at the top of the flame is higher than the temperature at the sides because the airflow carries heat upwards. Radiation can be detected by differential air thermometers, bolometers, thermopiles, etc.