In a Young's double slit experiment, the ratio of intensity between a dark spot and bright spot is 4:81. Using the equations that relate intensity to the amplitudes of the waves passing through each slit, the ratio of the widths of the two slits can be calculated. Solving the equations determines that the amplitudes through each slit are 5.5 and 3.5, and the ratio of the widths is 2.47.

1. 2D Interference
In a Young’s double slit experiment setup, the ratio of intensity of dark spot to
bright spot is a ratio of 4:81, what is the ratio of the width of the slits?
We know that I (intensity) of light at any point on a screen due to interference in
this experiment can be represented by the equation:
The a1 and a2 are the amplitudes of the waves with constant phase difference of ϕ.
The intensity of a wave is proportional to the square of its amplitude (for
simplicity’s sake in this LO). We know that a cosine graph has a range of -1 < y < 1.
Therefore:
Imax = a12 + a22 + 2a1a2(1)
Imin = a12 + a22 + 2a1a2(-1)
Simplifying these two equations, we get:
Imax = (a1 + a2)2
Imin =(a1 - a2)2
Where light intensity is max is where the light spots occur and the minimum is
where the dark spots occur, therefore we have a ratio where:
(Imin / Imax) = (a1 - a2)2 / (a1 + a2)2 = 4 / 81
a1 - a2 = 2 ; a1 + a2 = 9
a1 = 5.5 and a2 = 3.5
Intensity of the source of light is directly proportional to the width of the slits and
we established that intensity was proportional to the square of the amplitude,
therefore:
w1 / w2 = I1 / I2 = a12 / a22 = 30.25 / 12.25 = 2.47