Understanding Interference Patterns and the Wave/Particle Duality of Light

[Gewitter_05]

Homework Statement

Helium-neon lasers like those we've used in lab typically have power ratings of 1mW. They produce monochromatic light with a wavelength of 632.8 nm. Recall that c=λf

I solved a and b, correctly I assume. I am having problems with c and d.

c) At the power rating determined in part b the interference pattern behind the slits forms slowly. It needs to be observed with a long exposure time photographic plate or a bank of detectors. Assume the detector bank is used and consider a detector placed so that maxima forms at its position. How much energy is absorbed each time the detector makes a reading? Explain how you know.

d) A crafty physicist decides he can determine which slit the single photon passes through. He places a detector over one of the slits (it blocks the slit). He feels satisfied in that he does get a series of readings (clicks) with his detector. How much energy is associated with each click?
He becomes frustrated though, in that the pattern that forms behind the slits is very different from the double slit pattern. What does this have to do with the wave/particle duality? And, which slit (or slits) does the single photon go through when the detector is removed and the pattern begins to form again? Discuss

Homework Equations

f=c/λ, Δt=d/c, P=nhf, E= PΔt

The Attempt at a Solution

For letter b I got the answer of 1.88x10^-10 W. Letter a was to find the rate (#/s) and I used the equations P=nhf and used the 1mW=nhf, then solved for n. I think I solved it right and got a number of 3.18x10^15. For letter b I used Δt=d/c and found the time, 1.67x10^-9s, E=PΔt ⇔ hf=PΔt = 1.88x10^-10 W.

I'm not sure how to find the energy absorbed or the energy in letter d. I remember my teacher talking about the double slit experiment and how when one lens or something was blocked, light acted more like a wave, but in the beginning a particle which is how the wave/particle duality came about.

 

[mfb]

(c) is just the absorption of a single photon, so E=hf=hc/λ.
(d) same energy as ín (c). If one slit is blocked, the light detected behind the splits can only come from the unblocked slit, and a single-slit pattern forms.