The filament of a light bulb has surface area 64 mm^2. The filament can be considered as a black bod

Q: Why do we divide the total power by $4\pi d^2$ to find the intensity?

Because the light bulb acts as a point source emitting radiation equally in all directions. By the time the light travels a distance $d$, the total power $P$ is spread uniformly over the surface of an imaginary sphere of radius $d$, which has an area of $4\pi d^2$.

Q: Why did we take emissivity $e = 1$?

The problem explicitly states that the filament 'can be considered as a black body'. By definition, a perfect black body is an ideal emitter and absorber of radiation, which corresponds to an emissivity $e = 1$.

Q: How do I convert $mm^2$ to $m^2$ without making a mistake?

Remember that $1 \text{ mm} = 10^{-3} \text{ m}$. Therefore, when you square it, $(1 \text{ mm})^2 = (10^{-3} \text{ m})^2 = 10^{-6} \text{ m}^2$. So, $64 \text{ mm}^2$ becomes $64 \times 10^{-6} \text{ m}^2$.

Q: In option D, why do we use the average wavelength instead of the peak wavelength $\lambda_m$?

A black body emits radiation over a continuous spectrum of wavelengths, not just at the peak wavelength. To accurately estimate the total number of photons, we must use the average energy per photon, which corresponds to the average wavelength of the entire emitted spectrum.

Q: Is the area of the pupil taken as a flat circle or a curved surface?

Since the distance $d$ ($100 \text{ m}$) is vastly larger than the radius of the pupil ($3 \text{ mm}$), the wavefront of the light hitting the eye is essentially flat (a plane wave). Therefore, we can safely treat the pupil as a flat circular area $\pi R_e^2$ intercepting the light.

Question

The filament of a light bulb has surface area $64 	ext{ mm}^2$. The filament can be considered as a black body at temperature $2500 	ext{ K}$ emitting radiation like a point source when viewed from far. At night the light bulb is observed from a distance of $100 	ext{ m}$. Assume the pupil of the eyes of the observer to be circular with radius $3 	ext{ mm}$. Then
(Take Stefan-Boltzmann constant $= 5.67 	imes 10^{-8} 	ext{ Wm}^{-2} 	ext{ K}^{-4}$, Wien's displacement constant $= 2.90 	imes 10^{-3} 	ext{ m-K}$, Planck's constant $= 6.63 	imes 10^{-34} 	ext{ Js}$, speed of light in vacuum $= 3.00 	imes 10^8 	ext{ ms}^{-1}$)

Options:

(A) power radiated by the filament is in the range $642 	ext{ W}$ to $645 	ext{ W}$

(B) radiated power entering into one eye of the observer is in the range $3.15 	imes 10^{-8} 	ext{ W}$ to $3.25 	imes 10^{-8} 	ext{ W}$

(C) the wavelength corresponding to the maximum intensity of light is $1160 	ext{ nm}$

(D) taking the average wavelength of emitted radiation to be $1740 	ext{ nm}$, the total number of photons entering per second into one eye of the observer is in the range $2.75 	imes 10^{11}$ to $2.85 	imes 10^{11}$

Sigma Percentile · Accepted Answer

Imagine you are standing in the dark, looking at a light bulb glowing a hundred meters away. This seemingly simple scenario is a beautiful playground for the laws of thermal radiation and quantum physics. Let's break down the physics of what is actually happening and how much energy and how many photons are reaching your eye.

### Analyzing the Setup

We are given a light bulb filament with a surface area $A = 64 	ext{ mm}^2$. It acts as a black body at a temperature $T = 2500 	ext{ K}$. We are observing it from a distance $d = 100 	ext{ m}$, and our eye pupil has a radius $R_e = 3 	ext{ mm}$.

Before we jump into calculations, we must ensure all our units are in the standard SI system. The area $A = 64 	imes 10^{-6} 	ext{ m}^2$ and the pupil radius $R_e = 3 	imes 10^{-3} 	ext{ m}$.

### The Total Power Radiated

First, let's determine the total power radiated by the filament. According to the **Stefan-Boltzmann Law**, the power $P$ radiated by a black body is given by:

$$P = \sigma A e T^4$$

Since it's considered a black body, the emissivity $e = 1$. Substituting the values:

$$P = (5.67 	imes 10^{-8}) 	imes (64 	imes 10^{-6}) 	imes 1 	imes (2500)^4$$

$$P = 141.75 	ext{ W}$$

This tells us that option (A) is incorrect, as the power is exactly $141.75 	ext{ W}$, not in the $642-645 	ext{ W}$ range.

### Power Reaching the Eye

Now, this $141.75 	ext{ W}$ of power doesn't all go into our eye. It spreads out isotropically (equally in all directions), forming a giant sphere of radius $d = 100 	ext{ m}$. The intensity $I$ of the radiation at this distance is the power divided by the surface area of this giant sphere:

$$I = \frac{P}{4\pi d^2}$$

Our eye pupil only intercepts a tiny fraction of this spherical wavefront. The power entering the eye, $P_{	ext{eye}}$, is the intensity multiplied by the area of the pupil:

$$P_{	ext{eye}} = I 	imes (\pi R_e^2) = \frac{P}{4\pi d^2} 	imes (\pi R_e^2)$$

Substituting the values:

$$P_{	ext{eye}} = \frac{141.75}{4\pi (100)^2} 	imes \pi (3 	imes 10^{-3})^2$$

$$P_{	ext{eye}} = 3.189 	imes 10^{-8} 	ext{ W}$$

This perfectly matches the range given in option (B), making it a correct choice.

### The Peak Wavelength

Next, let's find the wavelength at which the filament emits the maximum intensity of light. This is governed by **Wien's Displacement Law**, which states:

$$\lambda_m T = b$$

Where $b$ is Wien's constant ($2.90 	imes 10^{-3} 	ext{ m-K}$). Solving for $\lambda_m$:

$$\lambda_m = \frac{2.90 	imes 10^{-3}}{2500} = 1.16 	imes 10^{-6} 	ext{ m}$$

Converting this to nanometers, we get **$\lambda_m = 1160 	ext{ nm}$**. Thus, option (C) is also correct.

### Counting the Photons

Finally, let's look at the quantum nature of this light. The radiation is composed of photons. If we take the average wavelength of the emitted radiation to be $\lambda_{	ext{avg}} = 1740 	ext{ nm}$, we can find the energy of a single average photon using Planck's equation:

$$E = \frac{hc}{\lambda_{	ext{avg}}}$$

The total number of photons entering the eye per second, $\dot{N}$, is simply the total power entering the eye divided by the energy of a single photon:

$$\dot{N} = \frac{P_{	ext{eye}}}{E} = \frac{P_{	ext{eye}} \lambda_{	ext{avg}}}{hc}$$

Plugging in our numbers:

$$\dot{N} = \frac{3.189 	imes 10^{-8} 	imes 1740 	imes 10^{-9}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}$$

$$\dot{N} \approx 2.79 	imes 10^{11}$$

This falls right into the range specified in option (D), making it our final correct choice.

In conclusion, by systematically applying the laws of thermodynamics and quantum mechanics, we've successfully decoded the light from a distant bulb!

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Total Power Radiated}

\text{Calculating Total Power}

\text{Power Reaching the Eye}

\text{Calculating Power to Eye}

\text{Wavelength of Maximum Intensity}

\text{Calculating } \lambda_m

\text{Photon Count per Second}

\text{Calculating } \dot{N}

\text{Conclusion}

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Total Power Radiated

Power Reaching the Eye

The Peak Wavelength

Counting the Photons

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