Figure shows a narrow slit S illuminated b a monochromatic light of wavelength `lambda` in a double-slit experiment. In the path of the rays reaching the upper slit `S_(1)`, a tube to length L is interposed in which the index of reflection of the medium varies linearly as shown in figure . The position of the central maximum in the interference pattern on the screen was displaceed by N fringes. Find the value of N in terms of `mu_(0)`, L and `lambda`.

Answer

`[(1.5) mu_(0) - 1] L//lambda`

Explanation

Optical path difference,
`Delta x = 2 int_(0)^(L//2) mu d x - L = 2 mu_(0) int_(0)^(L//2) (1 + (2x)/(L)) dx - L`
`= (3)/(2) mu_(0) L - L = [(1.5) mu_(0) - 1] L`
Now, `Delta x = N lambda`
`implies N = Delta x// lambda = [(1.5) mu_(0) - 1 ] L// lambda`