When light from two sources (say slits `S_(1)` and `S_(2)`) interfere, they form alternate dark and bright fringes. Bright fringe is formed at all point where the path difference is an odd multiple of half wavelength. At the condition of equal amplitudes, `A_(1) = A_(2) = a`, the maximum intensity will be `4 a^(2)` and the visibility improves, The resultant intensity can also be indicated with phase factor as `I = 2 a^(2) cos^(2) (phi // 2)`.Using this passage, answer the following questions.
At point having a path difference of `(lambda)/(4)`, the intensity

Options

A. 0
B. `a^(2)`
C. `2 a^(2)`
D. `a^(2) // 2`

Answer

b

Explanation

For a path difference of `lambda // 4`, the phase difference will be
`phi = (2 pi)/(lambda) (lambda)/(4) = (pi)/(2)`
`:.` intensity `= 2a^(2) cos^(2) ((phi)/(4)) = a^(2)`
So, choice (b) and the other are incorrect.