I don't understand how I can use that to work out the central diffraction maximum. $$\Delta Py*\Delta y<h/4\pi$$ So is $$\Delta y$$ the central diffraction maximum? Where does the x component come into it?
Homework Statement
A horizontal beam of laser light of wavelength 474 nm passes through a narrow slit that has width 5.80×10−2 mm . The intensity of the light is measured on a vertical screen that is 2.40 m from the slit.
Use the result of part A to estimate the width of the central diffraction...
I did it for both and the difference is small so I guess either works. I got $$\lambda = 7.79*10^-11$$ and when I used this in the equation I can't get it to work. It worked out to be $$\cos\phi = -23.99$$, which doesn't work
I assumed it meant the scattered wavelength but is it the original wavelength and I need to find the scattered wavelength?
I can use it to get the wavelength I need because for the photons $$p=h/\lambda$$
Homework Statement
A photon with wavelength 0.1385 nm scatters from an electron that is initially at rest. What must be the angle between the direction of propagation of the incident and scattered photons if the speed of the electron immediately after the collision is 9.30×106 m/s?
Homework...
E^2 = (mc^2)^2 + (pc)^2
I used mc^2 = 139.6 MeV
I put p = mv so the pc = mvc but m = 139.6/c^2 and v = (1-Δ)c = (1-(3.04*10-5))c so pc = 139.6 MeV
So then E = sqrt(139.6^2 + 139.6^2) = 197.4 MeV