1.

Two identical beam A and B of wavelength lambda fall cylindeical screen. The angle between the directions a point P on the screen at angular position phi from the beam A as shown in the figure. Find the distance between adjacent interference fringes on the screen near the point P. Asume that the distance beta between adjacent fringes is much less than the radius of the cylinder.

Answer»

`(lambda)/(2sin ((phi)/(2))cos(theta-(phi)/(2)))`
`(lambda)/(2sin((phi)/(2))cos(theta+(phi)/(2)))`
`(lambda)/(2sin((theta)/(2))cos(theta+(phi)/(2)))`
`(lambda)/(2sin((theta)/(2))cos(phi+(theta)/(2)))`

Solution :The two parallel BEAMS of rays INTERFERING at P MAKE an angle `phi` and `theta + phi` with the normal at P. considering a tangential place, the FRINGE width is
`(lambda)/(sin(theta+phir)-sin theta)` (same as in the previous PROBLEM)
or`beta = (lambda)/(2sin((phi)/(2))cos(theta+(phi)/(2))`


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