Figure. Illustrates the interference axperiment with Fresnel mirrors. The angle between the mirrors is `alpha = 12'` the distance from the mirrors intersection line to the narrow slit `S` and the screen `Sc` are equal to `r= 10.0 cm` and `b = 130 cm` resoectively. the wavelength of light is `lambda = 0.55 mu m`. Find:
(a) the width of a fringe on the screen and the number of possible maxima,
(b) the shift of the interference pattern on the screen when the slit is displaced by `del l = 1.0 mm` along the are of radius `r` with centre at the point `O`,
(c) at what maximum width `del_(max)` of the slit the interference fringes on the screen are still observed sufficiently sharp.

Explanation

Here `S' S'' = d = 2r alpha`
Then `Deltax = ((b + r)lambda)/(2 alpha)`
Putting `b = 1.3` metre, `r =1` metre
`lambda = 0.55 mu m, alpha = 12 = (1)/(5 xx 57)` radian
we get `Delta x = 1.1 mm`
Number of possible maxima `= (2b alpha)/(Delta x) + 1~~8.3 + 1~~9`
(`2b alpha` is the length of the spot on the screen which gets light after refection from both mirror. We add `1` above to take account of the fact that in a distance `Delta x` there are two maxima).
(b) What the silt moves by `del l` along the are of radius `r` the incident ray on the mirror rotates by `(del l)/(r )`, this is the also the roation of the reflected ray. there is then a shift of the fringe of magnitude.
`b(del l)/(r ) = 13 mm`.
(c) If the width of the slit is `del` then we can imagine the slit to consist of two narrow slits with separation `del`. The fringe pattem due to the wide slit is the susperposition of the patten due to these two narrow slits. The full patten will not be sharp at all if the patten due to the two narrow slits are `(1)/(2) Deltax` apart because then the maxima due to one will fill the minima due to the other . Thus we demand thet
`(b del_(max))/(r ) = (1)/(2) Deltax = ((b + r)lambda)/(4r alpha)`
or `del_(max) = (1-(r)/(b)) (lambda)/(4 alpha) = 42 mu m`.