Kaplan Problem - Physics (Electromag Waves)

[hokiemon]

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I ran across a problem in a kraplan practice test that said the intensity at freq X is greater then the intensity at freq Y because freq X is greater then Y.

Umm. isn't that crap? Energy is equal to hf but im pretty sure we've been schooled on how intensity is the # of photons and not the energy of them.. basically if we take photoelectric effect and shine like with E=hf on it, we ll get some electrons.. if we crank up the INTENSITY of the light, we're still emitting E=hf, but just a bunch of more photons..

fellow physics enthusiasts.. can you all confirm this? THANKS

 

[scotties123]

intensity is directly related to w^2 and w = 2pi(frequency)

 

[BrokenGlass]

I ran across a problem in a kraplan practice test that said the intensity at freq X is greater then the intensity at freq Y because freq X is greater then Y.

Umm. isn't that crap? Energy is equal to hf but im pretty sure we've been schooled on how intensity is the # of photons and not the energy of them.. basically if we take photoelectric effect and shine like with E=hf on it, we ll get some electrons.. if we crank up the INTENSITY of the light, we're still emitting E=hf, but just a bunch of more photons..

fellow physics enthusiasts.. can you all confirm this? THANKS

Well, light is both a particle and a wave. You are thinking of something similar to photoelectric effect. But I guess in this problem the solution assumes the wave-like behavior and uses the equation that says intensity is proportional to the square of frequency and square of the amplitude for all waves.

 

[majik1213]

I ran across a problem in a kraplan practice test that said the intensity at freq X is greater then the intensity at freq Y because freq X is greater then Y.

Umm. isn't that crap? Energy is equal to hf but im pretty sure we've been schooled on how intensity is the # of photons and not the energy of them.. basically if we take photoelectric effect and shine like with E=hf on it, we ll get some electrons.. if we crank up the INTENSITY of the light, we're still emitting E=hf, but just a bunch of more photons..

fellow physics enthusiasts.. can you all confirm this? THANKS

i think I see what your saying, and I hope what I'm about to say clarifies. Intensity is defined as magnitude, as of energy or a force per unit of area, volume, time, etc. Energy per unit area can be increased, then, by increasing the frequency of incident particles on a certain area.

Now, studies on the photoelectric effect revealed that intensity did NOT increase by increasing the number of photons per unit area when the photons were of a uniform frequency, and this finding contradicted classical expectations. Classically we'd expect a summation effect; that is, increasing the intensity would come naturally by increasing the number of incident particles. Instead, we discovered that increasing the frequency of the incident photons led to an increase in intensity. (i.e. "[W]e'd been schooled" is correct .. you *were* schooled in intro physics classes, and by solely relying on that material you would invariably make that logical conclusion, that increasing the # of photons would increase intensity, but your intro-level schooling has mislead you; if you take an upper level physics course, you will find out that intensity does not increase with the number of incident photons on a material.)