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Discuss the diffraction at a grating and obtain the condition for the m^(th)maximum. |
Answer» Solution : (i) A plane transmission grating is represented by AB in Figure. Let a plane wavefront of monochromatic light with wave length `LAMDA` be incident normally on the grating.As the slits size is comparable to that of wavelength, the incident light diffracts at the grating.  (ii) A diffraction pattern is obtained on thescreen when the diffracted waves are focused on a screen using a convex lens. (III) Let US consider a point P at an angle `theta` withthe normal drawn from the center of the grating to the screen. The path difference `delta` between the diffracted waves from one pair of corresponding points is, ` delta = (a+b) sin theta` This path difference is the same for any pair ofcorresponding points. The point P will be bright,when `delta= m lamda `where m= 0, 1, 2, 3 . Combining the above two equations, we get, `(a+b) sin theta = m lamda ` Here, m is called ORDER of diffraction. Condition for zero order maximum, m=0 : For `(a + b) sin theta`= 0, the position, `theta = 0, sin theta = 0 ` and m=0. This is called zero order diffraction or central maximum. Condition for first order maximum, m=1: If `(a + b ) sin theta_1 = lamda` ,the diffracted light meet at an angle `theta_1` to the incident direction and the first order maximum is obtained. Condition for second order maximum, m=2 : Similarly, `(a + b) sin theta_2 = 2lamda`forms the second order maximum at the angular position `theta_2` Condition for higher order maximum : (i) On either side of central maxima different higher orders of diffraction maxima are formed at different angular positions.If we take, ` N = (1)/( a+ b)` (ii)Then, N gives the number of grating elements or rulings drawn per unit width of the grating. Normally, this number N is SPECIFIED on the grating itself. Now, the equation becomes ` 1/N sin theta =m lamda " or " sin theta = N m lamda` |
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