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$\bullet$ Threshold of vision. Under controlled darkened conditions in the laboratory, a light receptor cell on the retina of a person's eye can detect a single photon (more on photons in Chapter 28 ) of light of wavelength 505 $\mathrm{nm}$ and having an energy of $3.94 \times 10^{-19} \mathrm{J} .$ We shall assume that this energy is absorbed by a single cell during one period of the wave. Cells of this kind are called rods and have a diameter of approximately 0.0020 $\mathrm{mm} .$ What is the intensity (in $\mathrm{W} / \mathrm{m}^{2} )$ delivered to a rod?

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$74.5 \times 10^{6} \mathrm{W} / \mathrm{m}^{2}$

Physics 102 Electricity and Magnetism

Physics 103

Chapter 23

Electromagnetic Waves and Propagationof Light

Electromagnetic Waves

Reflection and Refraction of Light

Cornell University

University of Michigan - Ann Arbor

Simon Fraser University

University of Winnipeg

Lectures

02:30

In optics, ray optics is a geometric optics method that uses ray tracing to model the propagation of light through an optical system. As in all geometric optics methods, the ray optics model assumes that light travels in straight lines and that the index of refraction of the optical material remains constant throughout the system.

10:00

In optics, reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. Common examples include the reflection of light, sound and water waves. The law of reflection says that for specular reflection the angle at which the wave is incident on the surface equals the angle at which it is reflected. Reflection may also be referred to as "mirror image" or "specular reflection". Refraction is the change in direction of a wave due to a change in its speed. The refractive index of a material is a measure of its ability to change the direction of a wave. A material with a higher refractive index will change the direction of a wave to a greater degree than a material with a lower refractive index. When a wave crosses the boundary between two materials with different refractive indices, part of the wave is refracted; that is, it changes direction. The ratio of the speeds of propagation of the two waves determines the angle of refraction, which is the angle between the direction of the incident and the refractive rays.

01:33

The retina of the eye cont…

04:56

Dark-adapted vision is pos…

03:21

Light detection by human e…

01:02

Go Given that the threshol…

03:48

Human vision sensitivity T…

02:56

Response of the eye. The h…

02:31

03:25

If the eye receives an ave…

05:10

The threshold of dark-adap…

02:20

A 0.75-mW laser emits a na…

01:48

Lasers are used extensivel…

02:27

02:36

Response of the Eye. The h…

04:57

02:13

Consult Interactive Learni…

in this problem, we're told to look at one whole cycle of the wave. And so first, we want the period of the way since this will give us the time for one full cycle, and then we can use this time to figure out the total power delivered to the surface area of the wrong. And so this is our first is art in him and so cold that the frequency is related to the speed in the wavelength with this relationship here and recall that the period is one over the frequency. And so plugging this in here we get that period is equal to the wavelength over the speed. It wishes the speed of light in this case. And so now I can play in the wavelength and the speed of light. And when we do that, we get that the period is 1.6 times 10 to the minus 15 seconds. And so it doesn't take much time at all to do one wave. And so now recall that the average power delivered is equal to the intensity times the area on the surface and this problem were given the power and we want to go out with the intensity is and so we divide by the area. Now they give us the size of the rock so we can get the area. But in order to figure out the average power, we have to take the energy. They give us the problem and divide by the time of that energy, which is the period which we found here. And so the power they give us is 3.94 times 10 to the nice 19. Jules, that's the energy they give us to get the power. We divide that by the time which is here. So this quantity, here's the power. And now we're going to divide by the area which is applying times. The radius squared in the radius is 0.1 times 10 to the minus 30 meters. And that's word. And so now we can do all of this and a calculator, and we get that the intensity is 7.46 times tend to the seven in the units are wants for me or swear, and that completes the problem.

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