Two sources `S_(1)` and `S_(2)` emiting light of wavelenght 600 nm placed a distance `1.0 xx 10^(-2)` cm apart.
A detector can be moved on line `S_(1) P` which is perpendicular to `S_(1) S_(2)`
(a) What would be the minimum and maximum path difference at the detector as it moves along line `S_(1) P` ?
(b) Locate the psoition of the farthest minima detected
.

Explanation

a. Path difference at P,`Delta x = S_(2) P - S_(1) P`
Minimum path difference will occur when the observer will be at a large distance from the sources. The Path difference will be maximum when the observer is near source `S_(1)`. As observer moves from `S_(1)` toward `+ x-` direction, the path difference decreases from d to zero.
`(Delta x)_(max)` : when `S_(1) P`, i.e., detector is at position`S_(1)`.
`:. (Delta x)_(max) = S_(1) S_(2) = d = 1.0 xx 10^(-2) m`
`(Delta x)_(min)` : when detector is at large distance and
`(Delta x)_(min)` will be apprximately zero.
b. Let the farthest minimum occur at P. As we move away increases as we move toward `S_(2)`. At the position of the farthest minimum path difference is
`S_(2) P - S_(1) P = lambda//2`
`(D^(2) + d^(2))^(1//2) - D = lambda//2`
`[D(1 + (d^(2))/(D^(2)))^(1//2)] - D = lambda//2`
`(d^(2))/(2D) = (lambda)/(2) implies D = (d^(2))/(lambda) D = (d^(2))/(lambda) = (lambda)/(4) = 1.7 cm`
.