Saved Bookmarks
| 1. |
A plane light wave with wavelength lambda = 0.50mu m falls normally on the face of a glass wedge with an angle Theta = 30^(@). On the opposite face of the wedge a transparent diffraction grating with period, d = 2.00 mu m is inscibed, whose lines are parallel to the wedge's edge. Find the angles that the direction of incident light froms with the directions to the principle Fraunhofer maxima of the zero and the first order. waht is the highest order of the spectrum? At what angle to the direction of incident light of the observed? |
|
Answer» Solution :The diffraction formula is easily obtaind on TAKING account of the fact that the optical PATH in the glass wedge acquires a factor `n`(refractive index). We get `d(n sin THETA -sin(Theta theta_(k))) = k lambda` Isnce `ngt0, Theta - theta_(0) gt Theta` and so `theta_(0)` must be nagative. we get, USING `Theta = 30^(@)` `(3)/(2) xx (1)/(2)=sin (30^(@) - theta_(0)) = sin48.6^(@)` Thus `theta_(0) =-18.6^(@)` Also for `k =1` `(3)/(4)-sin (30^(@) - theta_(+1)) = (lambda)/(d) = (0.5)/(2.0) = (1)/(4)` Thus `theta_(+1) = 0^(@)` We calaculate `theta_(k)` for varoius `k` by the above fromula. For `k =6`. `sin(theta_(k) - 30^(@)) = (3)/(4) rArr theta_(k) = 78.6^(@)` For `k = 7` `sin(theta_(k) - 30^(@)) =+ 1 rArr theta_(k) = 120^(@)` This is in ADMISSIBLE. Thus the highest order that can be observed is `k = 6` corresponding to`theta_(k) = 78.6^(@)` (for `k =7` the diffracted ray will be grazing the wedge). 
|
|