Question

Consider an incident electron beam on a system of two slits A
and B of unequal length, A being wider than B. A screen is arranged
behind the slits and we observe where the electrons arrive on the screen. Call
N_A the number of electron incidents on the screen per second when only the slit
A is open and N_B this number when only slit B is open. It happens that
N_A = 25 N_B. When both slits are open, we can observe an interference pattern on the screen. We ask to find the ratio of the probability that an electron is detected at a maximum of the pattern on the probability that it is
detected at an adjacent minimum. (Hint: Admit that the number of electrons
detected at a position of the screen is proportional to the density of the
probability.)

Answer 1

The number of electrons represents the amplitude of the light, so when both beams are incident together with the amplitude at maxima will be

A_max = N_A + N_B = 25N_B + N_B = 26N_B

The amplitude of resultant beam at minima will be the difference of two amplitudes

A_min = N_A - N_B = 25N_B - N_B = 24N_B

the probability of finding an electron is proportional to the intensity. The intensity is proportional to the square of the amplitude

the ratio of the probability that an electron is detected at a maximum of the pattern on the probability that it is
detected at an adjacent minimum is 1.17