Argh - De Broglie and the wave equation

[B1ueguy1]

This isn't a homework question, but something which has been bugging me. I can't figure it out. Maybe it's late, but it's probably a very stupid question

If light shines onto a denser medium, the light slows down in this medium right? If the light slows down, then by the wave equation, v = fλ, since f is constant, the wavelength must decrease.

So why is it that de broglie's equation λ = h/mv implies a decrease in velocity would increase the wavelength?

I'm so confused...

 

[qwerty2x]

v is in the denominator

 

[B1ueguy1]

v is in the denominator

eh?

de broglie: λ = h/mv
wave equation: λ = v/f

 

[B1ueguy1]

if anyone has any insight, could someone please help me? It's sort of urgent. I'm taking an exam tomorrow and this uncertainty is really making me nervous.

 

[bowma166]

Well... the de broglie equation is $$\lambda=\frac{h}{p}$$. I don't really think that $$p=mv$$ works for light, because photons have no mass.

I don't really know what I'm talking about though.

 

[americanforest]

This isn't a homework question, but something which has been bugging me. I can't figure it out. Maybe it's late, but it's probably a very stupid question

If light shines onto a denser medium, the light slows down in this medium right? If the light slows down, then by the wave equation, v = fλ, since f is constant, the wavelength must decrease.

So why is it that de broglie's equation λ = h/mv implies a decrease in velocity would increase the wavelength?

I'm so confused...

The equation $$p=mv$$ does not hold for photons (light quanta) since they have no rest mass. For a photon, its energy, $$E=h \nu$$ (Planck's constant multiplied by frequency) is equal to its momentum multiplied by the speed of light in vacuum (Generally, for relativistic motion, $$E^{2}=(pc)^{2}+(mc^{2})^{2}$$; Since m is zero in the case of the photon $$E=pc$$. In this case, de Broglie's equation becomes

$$\lambda = \frac{h}{p}=\frac{hc}{E}=\frac{hc}{h \nu}=\frac{c}{\nu} \Rightarrow \lambda \nu =c$$

Which is your original equation. Contradiction resolved.

 

[B1ueguy1]

The equation $$p=mv$$ does not hold for photons (light quanta) since they have no rest mass. For a photon, its energy, $$E=h \nu$$ (Planck's constant multiplied by frequency) is equal to its momentum multiplied by the speed of light in vacuum (Generally, for relativistic motion, $$E^{2}=(pc)^{2}+(mc^{2})^{2}$$; Since m is zero in the case of the photon $$E=pc$$. In this case, de Broglie's equation becomes

$$\lambda = \frac{h}{p}=\frac{hc}{E}=\frac{hc}{h \nu}=\frac{c}{\nu} \Rightarrow \lambda \nu =c$$

Which is your original equation. Contradiction resolved.

Thank you! So for electrons use debroglie, for light, use the other..