Recent content by says

So to answer the question I would say the distance would be approximately 45 cm, however, the Fraunhofer condition under these circumstances are unreliable, because if let the distance be shorter the bright blob will be smaller than the aperture, which is impossible

I still get an answer of 45.45cm when I solve for L...

so Δx= Lλ / b will give us the distance between m=0 (central maximum) and m=±1

What is Δx in your working?

The only other equation I have found is: df = 2λz / W where df=width of central band in diffraction pattern λ=wavelength z=viewed at a distance of... W=slit width z = (0.5mm)(0.5mm)2 / 2(550nm) = 22.72 cm This equation is for a slit of infinite depth. I'm not totally sure how to convert it...

Homework Statement A square aperture with a side of length 0.5 mm is illuminated with light of wavelength 550 nm. At what distance from the aperture would the Fraunhofer diffraction pattern have a central maximum with a width also equal to 0.5 mm? What can you say about the Fraunhofer condition...

So boundary conditions cause quantisation of energy levels. In solids, it causes quantised energy bands. As we decrease the dimensions and size of something it approaches discrete energy levels (atomistic)

Confinement of particles in a potential field

What causes the quantization though?

I've been reading a bit about the quantum confinement effect on nanowires, particularly how it changes the band structure. I'm trying to find an explanation on why the density of states splits into sub-bands. At the moment all I'm running into is 'because of the quantum confinement effect' which...

Question is asked in regards to LEDs

I'm guessing greater energy = better emitter = lower wavelength

Homework Statement Calculate the expected peak wavelength and spectral bandwidth (in units of wavelength) of the emission for both a GaAs and silicon LED at liquid nitrogen temperature (77 K) and room temperature (300 K). Which of these cases would you expect to result in the best emitter and...

Will do! I've attached my notes to this comment. Pages 12-16 discuss Fraunhofer diffraction and power densities in more detail, but it doesn't have the exact equation we were using.

ahhh ok. So the final answer will just have a 3 in the denominator of the sine squared term?