@Wolgwang OK, I've done it, but it's harder than it looks.
We need to find the path difference between Lā and Lā. I won't go through it because it is just geometry, but the path difference turns our to be Ī»/6 i.e. a phase difference of š/3.
So if we write the light from Sā and Sā as sin(Ļt) then the light from Sā is sin(Ļt + š/3) i.e. a phase difference of š/3.
So when all three slits are open the field (NB the field E not the intensity I ā E²) is: E = 2 sin(Ļt) + sin(Ļt + 𝜋/3)
And when only Sā and Sā are open the field is: E = sin(Ļt) + sin(Ļt + 𝜋/3)
Now let's find what E = 2 sin(Ļt) + sin(Ļt + š/3) is. To do this we draw a phasor diagram:
I've put the 2 sin(Ļt) horizontal and the sin(Ļt + š/3) at the angle of š/3 then we add them to get the resultant E.
And using Pythagoras we get: E² = (2¹āā)² + (ā3/2)² = 25/4 + 3/4 = 18/4 = 7
I won't draw it but for E = sin(Ļt) + sin(Ļt + 𝜋/3) the horizontal vector becomes 1 instead of 2 and we get: E² = (1¹āā)² + (ā3/2)² = 9/4 + 3/4 = 12/4 = 3
And we are there!
The intensity is proportional to E² so the ratio of the intensities is just 3/7.
But I have to say that is a haaaaaaaaard question!
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