Tag Info

New answers tagged

3

You're perfectly correct. Referring to Classical Electrodynamics by Jackson, we see that the index of refraction $n$ is given by: $$n=\sqrt{\frac{\mu}{\mu_{0}}\frac{\epsilon}{\epsilon_{0}}} = \sqrt{\mu_{r}\epsilon_{r}}.$$ But Jackson notes that for most optical frequencies (and non-meta-material media), $\frac{\mu}{\mu_{0}}=\mu_{r}\approx 1$, which is why ...


0

The idea behind this exercise is, to show you that everything CAN act like a wave in quantum mechanics, see slit experiments. You will have to use the same formula for all three entities. As stated before, the de Broglie wavelength is the way to go when confronted with such tasks. For the non-relativistic case, the de Broglie formula states: $$ \lambda = ...


1

Here is a hint - de Broglie wavelength for the neutron and electron - does that help?


2

According to Maxwell's theory of electromagnetism, a light pulse (or generic electromagnetic wave) carries momentum, which can be transferred to an absorbing surface hit by the pulse. This momentum transfer is known under the name 'radiation pressure'. Despite carrying momentum, light carries no mass. Yet a light pulse does carry energy. For a light pulse ...


-3

The total energy of a photon, the carrier of the electromagnetic force, is given by $ E=hf $ where $ h $ is Planck's constant and $ f $ is the frequency of the light. So yes, if two EM waves have the same energy, they will have the same frequency and wavelength, meaning they have the same colour. Photons have no rest mass, but they do have a relativistic ...


0

The very definition of a frequency is the rate at which something occurs over time. Therefore by shortening the time interval in which something occurs you are increasing its frequency, and by increasing the time interval you decrease the frequency. Hence, Frequency is inversely proportional to time.


1

First, the context is a function of time that is periodic which means that it is repetitive with repetition period $T$. $$g(t) = g(t + T)$$ So, if one sampled the function every $T$ seconds, one would get the same value each time. Now, we have the period of time $T$ which tells how long it takes for the signal to go through one cycle. The inverse ...


3

To answer the question "same wavelength, different frequencies, which arrives first?": naturally, the one with biggest speed, which is proportional to frequency AND wavelength according to the formula: $$v = \lambda f$$ So, for the same wavelength $\lambda$, the one with bigger frequency $f$ will have bigger speed $v$, thus arriving earlier. For the ...


2

The size or path-length that the wave has nothing to do with it. There are two explanations on offer for why the frequency of the radiation remains unchanged as it crosses the boundary between different media. (a) If a medium is considered to be lots of driven oscillators, then they will oscillate at the frequency of the driving force - the electric field ...


1

I often try to comprehend exactly why light wants to keep it's frequency, yet alter it's wavelength as it travels through a medium Have you not heard the standard (macroscopic, continuum approximation) explanation? To whit: the electric and magnetic fields must both be continuous at the boundary, and must therefore have the same time variation on ...


3

$\lambda = c/nf$ While in the medium, $n$ is greater than one, in vacuum $n=1$. The medium responds to whatever is driving it. The molecules of the medium oscillate at whatever frequency shakes them. Each molecule is then a tiny radiator, generating light at that same frequency. The light that the polarized, oscillating, molecule produces interferes ...



Top 50 recent answers are included