Saved Bookmarks
| 1. |
Making use of the result obtained in the foregoing problem, find the conditions defining the angular position of maxima of the first, the second, and the third order. |
|
Answer» Solution :SINCE `I(alpha)` is `+ve` and vanishes for `b sin varphi = k LAMBDA` i.e for `alpha = k lambda`, we expect maxima of `I(alpha)` between `alpha =+ pi & alpha =+ 2pi`, etc. We can get these values by. `(d)/(d alpha) (I(alpha)) = I_(0)2(sin alpha)/(alpha)(d sin alpha)/(alpha) = 0` `(alpha cos alpha- sin alpha)/(alpha^(2)) = 0` or `tan alpha = alpha` SOLUTIONS of this transcendental equation can be obtained graphically. The first three solutions are `alpha_(1) = 1.43pi, alpha_(2) = 2.46, alpha_(3), = 3.47 pi` on the `+ve` side. (On the negative side the solution are `-alpha_(1), -alpha_(2),- alpha_(3), .......)` THUS `b sin varphi_(1) = 1.43 lambda` `b sin varphi_(2) = 2.46 lambda` `b sin varphi_(3) = 3.47 lambda` Asymptotocally the solutions are `b sin varphi_(m) ~= (M+(1)/(2)) lambda` |
|