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A plane light wave with wavelength lambda = 0.60mum falls normally on the face of a glass wedge with refracting angle Theta = 15^(@). The opposite face of the wedge is opaque and has a slit of width b = 10mum parallel to the edge. Final: (a) the angle Delta theta between the direction to the Fqaunhofer maximum of zeroth order and that of incident light, (b) the angular width of the frqunhofer maximum of the zeroth order. |
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Answer» Solution :This case is analogous to the previous one except that the incident wave moves in GLASS of `R.I. n`. Thus the expression for the PATH DIFFERENCE for light diffracted at angle `theta` from the NORMAL to the hypotenuse of the wedge is `b(sin theta - n sin Theta)` we write `theta = Theta + Delta theta` Then for the direction of principle Fraunhofer maximum `b(sin(Theta + Delta theta) - n sin Theta) = 0` or `Delta theta = sin^(-1) (n sin Theta) - Theta` Using `Theta = 15^(@), n = 1.5` we get `Delta theta = 7.84^(@)` (b) The width of the central maximum is obtained from `(lambda = 0.60 MU m, b = 10mu m)` `b(sin theta_(1) - n sin Theta) = +- lambda` Thus `theta_(+1) sin^(-1) (n sin Theta +(lambda)/(b)) = 26.63^(@)` `theta_(-1) = sin^(-1) (n sin Theta(lambda)/(b)) = 19.16^(@)` `:. delta theta = theta_(+1) - theta _(-1)= 7.47^(@)` 
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