A 100-W lightbulb radiates visible light at a rate of about 10 W ; the rest of the EM radiation

A 100-W lightbulb radiates visible light at a rate of about...

A 100-W lightbulb radiates visible light at a rate of about $10 \mathrm{~W} ;$ the rest of the $\mathrm{EM}$ radiation is mostly infrared. Assume that the lightbulb radiates uniformly in all directions. Under ideal conditions, the eye can see the lightbulb if at least 20 visible photons per second enter a dark-adapted eye with a pupil diameter of $7 \mathrm{~mm}$. (a) Estimate how far from the source the lightbulb can be seen under these rather extreme conditions. Assume an average wavelength of $600 \mathrm{~nm}$. (b) Why do we not normally see lightbulbs at anywhere near this distance?

A 100-W lightbulb radiates visible light at a rate of about $10 \mathrm{~W} ;$ the rest of the $\mathrm{EM}$ radiation is mostly infrared. Assume that the lightbulb radiates uniformly in all directions. Under ideal conditions, the eye can see the lightbulb if at least 20 visible photons per second enter a dark-adapted eye with a pupil diameter of $7 \mathrm{~mm}$.
(a) Estimate how far from the source the lightbulb can be seen under these rather extreme conditions. Assume an average wavelength of $600 \mathrm{~nm}$. (b) Why do we not normally see lightbulbs at anywhere near this distance?

Key Concepts

Photon Energy

Photon energy is determined by the relation E = hc/?, where h is Planck’s constant, c is the speed of light, and ? is the wavelength. This concept is essential for converting the energy output in watts into the number of photons emitted per second when dealing with light at a particular wavelength.

Inverse Square Law

The inverse square law describes how the intensity of radiation from a point source decreases with distance. Specifically, the power per unit area is inversely proportional to the square of the distance from the source. This concept is critical when calculating how many photons from a uniformly radiating lightbulb reach a small receiver, such as the human eye, at a given distance.

Detector (Pupil) Area and Photon Flux

The area of the detector, in this case the pupil of the eye, determines how many photons are intercepted from the total radiated flux. By calculating the pupil’s area, one can use the inverse square law to estimate the photon flux entering the eye at a certain distance. This is key in estimating detectability based on a minimum photon threshold.

Visual Threshold of the Human Eye

The human eye, especially under dark-adapted conditions, has a minimum photon detection threshold, meaning a certain number of photons per second must enter for the light source to be seen. This concept involves understanding the sensitivity of the visual system and is used to set the criteria—such as 20 photons per second—in problems involving visual detection.

Practical Observational Limitations

Although theoretical calculations using idealized conditions may suggest that a small number of photons is sufficient for detection at a large distance, practical factors—such as atmospheric absorption, ambient light interference, and inefficiencies in the eye’s photon capture and processing—often prevent us from actually observing such sources at these extreme distances. This highlights the difference between idealized models and real-world conditions.

Transcript

00:01 In this exercise, we have a light bulb that radiates at 100 watts, and the power at the visible spectrum is 10 watts.

00:14 We also have the information that the number of photons, the minimum number of photons that the human eye can detect is 20 photons per second, and the diameter of a human pupil is 7 millimeters on average.

00:32 The wavelength of the lights that we're going to have to consider in this exercise is 600 nanometers.

00:42 And for question a, we're asked to calculate what is the maximum distance, r, at which a human eye can still see a light bulb that radiates at 10 watts.

00:56 So the first step here is going to be to calculate the intensity of the light, and that's going to be p over 4 p .r.

01:06 Squared, okay, so the intensity of light source is given by the power of the light source divided by the area of the sphere that has a radius r.

01:18 Then i'm going to calculate what is the power received by the pupil.

01:24 So the pupil is going to receive a power that equals the intensity of the light times the area of the pupil.

01:37 Then i will calculate how many photons per second the pupil receives.

01:47 And that's going to be the power received by the pupil divided by the energy of a single photon.

01:55 And that's i a pupil over the energy of a single photon.

02:03 That's h .c.

02:08 Times lambda.

02:12 I'm sorry, it's hc over lambda.

02:17 And so we can substitute i in here, in this equation.

02:23 So we're going to have n equals p over 4 pi r squared the area of the pupil times lambda over hc.

02:37 Okay...