Analysis of N-Slit Fraunhofer Diffraction Patterns: Optics Lecture

1. Chapter III OPTICS Lecture 3.6 Books:(1) Optics, 3rd edition: AjoyGhatak, McGraw-Hill Companies

2. N-slit Fraunhofer Diffraction Pattern (Plane diffraction grating): Here we shall consider the diffraction pattern produced by N parallel slits, each of width b. The distance between two consecutive slits is assumed to be d. f

3. To calculate the diffraction pattern, we use a method similar to that used for the case of a single slit. We assume that the slits consist of a large number of equally spaced point sources and that each point on the slit is a source of Huygens’ secondary wavelets. Before discussing the general case of N slits, we first assume the case of two slits as shown in fig 2. Let the point sources be at A1, A2, A3, . . . (in the first slit) and at B1, B2, B3, . . . (in the second slit). We assume that the distance between two consecutive points in either of the slits is ∆.

4. If the diffracted rays make an angle ɵ with the normal to the plane of the slits, then the path difference between the disturbances reaching point P from two consecutive points in a slit will be ∆ sin ɵ. The field produced by the first slit at point P will, therefore, be given by ---------(1) And due to 2nd slit it is, ---------(2) Where, --------------(3) represents the phase difference between the disturbances (reaching point P) from two corresponding points on the Slits.

5. The resultant filed of two slits at P is given by --------------(4) In the same sense, we can write the field at point P due to N slits. It consists of N terms, ---------------(5)

6. Above eqn can be written as, --------(6) where, ------------(7)

7. The intensity distribution is given by, --------(8) where represents the intensity distribution produced by a single slit. As can be seen, the intensity distribution is a product of two terms. The first term represents the diffraction pattern produced by a single slit. The second term represents the interference pattern produced by N equally spaced point sources.

8. For N = 1, Eq. (8) reduces to the single-slit diffraction pattern and for N = 2, to the double-slit diffraction pattern . In figures we plotted function as a function of ᵞ for N = 5 and N = 11. Note that as N increases the function becomes sharply Peaked at . Between two peaks the function vanishes when --------------(9)

10. Position of maxima and minima: For Maxima: When the value of N becomes very large, then one obtains maxima at . --------(10) (corresponding to sin γ =0) We also know, -------------(11) From eqn (10) and (11) we get ---------(12) Note that if sin Nγ =0, then becomes indeterminate so we use hospital rule -----------(13) Using properties of limits, hospital rule, Differentiate numerator and denominator

11. The resultant amplitude (eqn (6)) and corresponding intensities (eqn (8)) , using eqn (13), can be are written as -----(14) And ------------(15) Where, --------------(16) Using eqn (12), above eqn becomes, --------(17)

12. It may be noted that the above discussed maxima are known as principal maxima. m = 0 corresponds to zero order maxima. m = 1,2 , 3...corresponds to 1st order, 2nd order etc. principal maxima Physically, at these maxima the fields produced by each of the slits are in phase, and therefore, they add and the resultant field is N times the field produced by each of the slits. Consequently, the intensity has a large value unless itself is very small. Since , m cannot be greater than d/λ. Thus, there will only be a finite number of principal maxima.

13. For minima: From eqn (8) we observe that the intensity is zero when -----(18) (from sin Nγ = 0) ---(19) The angles of diffraction corresponding to Eq. (19) (using value of γ from (11) ) are -----(20) Where p can have all values except 0, N, 2N,.......nN, because for These values sinγ becomes zero and we get principal maxima. Thus p has values ±(1,2,3,.....(N-1)).

14. Note that between two principal maxima we have N – 1 minima. Between two such consecutive minima the intensity has to have a maximum; these maxima are known as secondary maxima. Exercise: Find the position of secondary maxima in N slits Diffraction pattern and also find the ratio of intensity of Secondary maxima to principal maxima. Hint: Differentiate intensity I w.r.t. γ (Eq. 8) and equate to zero.

15. Note the following two points: 1. A particular principal maximum may be absent if it corresponds to the angle which also determines the minimum of the single-slit diffraction pattern. This will happen when --------(21) ------------(22) are satisfied simultaneously, and it is usually referred to as a missing order. Even when Eq. (22) does not hold exactly (i.e., if is close to an integral multiple of λ ), the intensity of the corresponding principal maximum will be very weak .

16. 2. In addition to the minima predicted by Eq. (19), we will have the diffraction minima [see Eq. (18)]. However, when N is very large, the number of such minima is very small. Exercise: Find the angular half width or width of Principal maximum in N-slit Fraunhofer diffraction Pattern. Exercise: Discuss the case of secondary maxima in case of N-slit Fraunhofer Diffraction pattern.