00:01
Hello students in this question we have to obtain the diffraction pattern differaction pattern of annular aperture bounded by the circle of radii a1 and a2 okay a1 and a2 and where a2 is greater than a1 okay so we can say here that for the annular aperture the field of the field of at a point p is given by we can write the field value that up is equals to a divided by iota lambda z to the power iota k z exponential of i iota k r squared divided by 2 z and this double integration from 0 to a and from 0 to 2 pi e to the power minus i k k r, sine theta, and this cos -fi here, cos -5 and row, d -row, and this d -sy.
01:15
Okay, so now, but here the limits of the row, limits of the row varies from a1 to a2.
01:23
Okay, so we can rewrite the given equation as up, this is equals to the a divided by i, lambda, z, the power iota k z exponential of i k k r squared divided by two z and one by k sine theta and this whole square okay and integration uh sorry this these are the limits here so uh limits we can substitute the value of row that is from a 1 to a 2 and the next integration from 0 to 2 pi e to the minus i k row sine theta cos phi row d row and d psi okay now we can use the basal function basal function to integrate the above and we can rewrite that this will be equals to the a divided by i lambda z and e to the power iota k z exponential of i k r squared divided by 2 z and solving for the integration of this function so we will get one by k sine theta and this whole square okay and integration from k a1 sine theta from 2 k a 2 sine theta and this j -not function of jeta and this d jeta okay here we have considered that jeta is equal to k row sign theta okay now we can use this special function and we can integrate this function so we will get that u p is equals to the a divided by iota lambda z e to the power i k z exponential of i k r square divided by two z and this two pi devere by k sine theta and this whole square and this jeta and function of first derivative of g1 zeta and this values okay and the limits from k a1 sine theta to the k a2 sine theta okay now we can substitute this value of function so we can rewrite this equation as u p equals to a divided by i lambda z to the power i k z exponential of i k r square divided by 2 z and this 2 pi divided by k sine theta and this whole square and multiplied by k a2 sine theta and this j1 of k a2 sine theta theta and multiplied by this j1 k a1 sine theta okay now the above equation can be written in the next another form so we can write here that u p is equals to the i not divided by 1 minus alpha and this whole square okay and bracket 2 j1 of new 2 divided by new 2 minus alpha square multiplied by this 2 j 1, new 1 divided by this new 1, okay, and this whole square where i not is intensity intensity and alpha is equal to a1 divided by a 2 and this v1 or new 1 is equal to k a 1 sine theta and this new 2 is equals to k a 2 sine theta okay now the the diffraction function of the annular aperture will be given by finally up is equals to i0 divided by 1 minus alpha and this whole square and bracket 2 j1, new 2 divided by new 2 minus alpha square of this 2 j1, new 1, divide by this new 1 and this whole square...