1.

The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the distance between two fringe from the centre is

Answer» Total intensity at any point is given by
`I = kR^(2) = k (a^(2) + b^(2) + 2 ab cos phi)`
when `b = a, I = k (a^(2) + a^(2) + 2 a^(2) cos phi)`
`= 2 ka^(2)(1 + cos phi)`
At the centre of a bright fringe, `phi = 0^(@)`
`:. I_(1) = 2 ka^(2)(1 + cos 0^(@)) = 4 ka^(2)`
Distance between two fringes `= beta`, which is proportional to wavelength `(lambda)`
Now `(lambda)/(4)` corresponds to a phase diff.
`= (2pi)/(4) = (pi)/(2)`
`:. I_(2) = 2 ka^(2) (1 + cos pi//2) = 2 ka^(2)`
`:. (I_(1))/(I_(2)) = (4 ka^(2))/(2 ka^(2)) = (2)/(1)`


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