A narrow beam of electrons passes immediately over the surface of a metallic mirror with a diffraction grating with period `d = 2.0 mu m` inscribed on it. The electrons move with velocity `v,` comparable to `c`, at right angles to the lines of the grating. The trajectory of the electrons can be seen in the from of a strip, whose colouring depends on the observation angle `theta` (Fig.) Interpret this phenomenon. Find the wavelength of the radiations observed at an angle `theta = 45^(@)`.

Explanation

An electron moving in front of a metal mirror sees an image change of equal and opposite type. The two together constitute a dipole. Let us look at the problem in the rest frame of the electon. In this frame the grating period is Lorenz contracted to
`d' = d sqrt(1-v^(2)//c^(2))`
Because the metal has etchings the dipole moment of electron-image pair is periodically disturbed with a period `(d')/(v)`
The corresponding frequency is `(v)/(d')` which is also the proper frequency of radiation emitted. Due to Doppler effect the frequency observed at an angle `theta` is
`v=v'(sqrt(1-(v//c)^(2)))/(1-(v)/(c )cos theta) (v//d)/(1-(v)/(c ) co stheta)`
The corresponding wave length is `lambda = (c )/(v) = d ((c)/(v) - cos theta)`
putting `c~~v, theta = 45^(@), d = 2mu m` we get
`lambda = 0.586 mu m`