AAMC FL1 C/P #16

[I'm ugly and I'm proud]

So after reading over the passage again, I can see how D is a stronger answer than A, since it says that the oxygen levels affect the number of photons emitted. But just in general, can't the propagation speed/frequency of radiation affect the intensity? E = hf, so higher frequency means more energy per photon, which I would assume means higher intensity.

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[5words]

Nope, the concept being tested here is that constant do not affect proportionality relationships. The speed of an electromagnetic wave is the constant C = celerity, Thus doesnt affect the relationship between E and f or E and the wavelenght.

So, the frequency and the wavelenght both affect E but c (speed of a wave) doesn't because it's a constant.

 

[aldol16]

So after reading over the passage again, I can see how D is a stronger answer than A, since it says that the oxygen levels affect the number of photons emitted. But just in general, can't the propagation speed/frequency of radiation affect the intensity? E = hf, so higher frequency means more energy per photon, which I would assume means higher intensity.

What did Einstein teach us about the speed of light?

 

[I'm ugly and I'm proud]

What did Einstein teach us about the speed of light?

I was thinking of frequency when they were talking about propagation speed. I already know that the speed of light is constant lmao

 

[aldol16]

I was thinking of frequency when they were talking about propagation speed. I already know that the speed of light is constant lmao

Then I think you should make sure you understand the distinction between propagation speed (velocity) as opposed to frequency. They're not equivalent.

 

[5words]

I was thinking of frequency when they were talking about propagation speed. I already know that the speed of light is constant lmao

Light is an Electromagnetic wave, just like radiation. A radiation can be a UV or Infra-red and else. The propagation speed up there is coming from the Blue LED; hence it;s supposed to stay constant as it's depends on both C and n(index of refraction of the meduim). And what @aldol16 said as well.

 

[JKetlerP]

I am confused as to why it doesn't also depend on frequency.

If I understand what you are saying, then you are saying that because the light stays blue we know frequency isn't changing. But how do you know it isn't changing color at all? (I which case the change in frequency could be the cause of a change in intensity...)

 

[aldol16]

I am confused as to why it doesn't also depend on frequency.

If I understand what you are saying, then you are saying that because the light stays blue we know frequency isn't changing. But how do you know it isn't changing color at all? (I which case the change in frequency could be the cause of a change in intensity...)

It is a fundamental physical fact that intensity of radiation is directly related to number of photons emitted. Perhaps it would be helpful if you look up the photoelectric effect and how those experiments were designed. You might also think of intensity and energy in this way - say you have a row of birds on a wire and you're trying to shoot them with an automatic. Your gun shoots a bullet out that has some constant velocity (energy). We'll neglect drag and air resistance. You shoot one bullet. You hit a bird and it drops off. This is low intensity. Then you engage in full automatic mode and shoot 10 bullets. They're all coming out at the same velocity, mind you. There's just more of them. Ten birds drop. Your gun is emitting a high intensity of bullets - but they all have the same energy because they have identical velocities. This is changing intensity. Changing energy in this analogy would be actually changing the gun so that the bullet(s) are faster or slower.

 

[JKetlerP]

It is a fundamental physical fact that intensity of radiation is directly related to number of photons emitted. Perhaps it would be helpful if you look up the photoelectric effect and how those experiments were designed. You might also think of intensity and energy in this way - say you have a row of birds on a wire and you're trying to shoot them with an automatic. Your gun shoots a bullet out that has some constant velocity (energy). We'll neglect drag and air resistance. You shoot one bullet. You hit a bird and it drops off. This is low intensity. Then you engage in full automatic mode and shoot 10 bullets. They're all coming out at the same velocity, mind you. There's just more of them. Ten birds drop. Your gun is emitting a high intensity of bullets - but they all have the same energy because they have identical velocities. This is changing intensity. Changing energy in this analogy would be actually changing the gun so that the bullet(s) are faster or slower.

I understand what you are saying. I just mean that since intensity = energy/time, then if there are 2 ways to increase energy, either one would have the effect of increasing intensity. Increasing photons is one way to accomplish this, but increasing frequency should ALSO be able to (E=hf)...

But thanks for referencing the photoelectric effect--I will try to take a look later today!

 

[aldol16]

I understand what you are saying. I just mean that since intensity = energy/time, then if there are 2 ways to increase energy, either one would have the effect of increasing intensity. Increasing photons is one way to accomplish this, but increasing frequency should ALSO be able to (E=hf)...

No, you have the equation for intensity confused. Power is energy per unit time. Intensity is power per unit area.

 

[deleted647690]

No, you have the equation for intensity confused. Power is energy per unit time. Intensity is power per unit area.

If intensity is power per unit area, and E=hf, then E is proportional to f, and thus intensity is proportional to f.

Is B eliminated because the wavelength is constant due to it being a single color of light?
Wouldn't the speed of the light change as it leaves the medium of the vitreous humor and travels to the photodiode?

 

[aldol16]

If intensity is power per unit area, and E=hf, then E is proportional to f, and thus intensity is proportional to f.

Frequency is fixed. It's blue light. It's not changing. So frequency here can't be proportional to anything.

Is B eliminated because the wavelength is constant due to it being a single color of light?

Yes. If wavelength is constant, what do we know about frequency by the equation c = wavelength*frequency?

Wouldn't the speed of the light change as it leaves the medium of the vitreous humor and travels to the photodiode?

What did Einstein teach us about the speed of light?

 

[deleted647690]

Frequency is fixed. It's blue light. It's not changing. So frequency here can't be proportional to anything.



Yes. If wavelength is constant, what do we know about frequency by the equation c = wavelength*frequency?



What did Einstein teach us about the speed of light?

Yeah, but isn't the speed of light 'c' a constant for traveling in a vacuum? Wouldn't speed here be given by v, as in v = c/n, where the medium decreases the speed below c?

 

[aldol16]

Yeah, but isn't the speed of light 'c' a constant for traveling in a vacuum? Wouldn't speed here be given by v, as in v = c/n, where the medium decreases the speed below c?

Yes, you're talking about the index of refraction. It's a bit more nuanced than the light simply slowing down, but the overall observed effect is a slowing. Why does light slow down when entering a prism but speed up when it exits? Where does it get the energy to increase it's velocity?

But here, we're all in one phase so there's no slowing or speeding up. The speed of light is some constant in the vitreous humor, given by v = c/n.