Question

133 frequeneies other than the carrier and two sideband of a tuned cireit UDE UNEAR MODULATION reuit make use ristics of diode e levels becase eteristic of a yoltage 6. Consider a plane wave incident normally on a rectangular aperture of width b (along the axis and width a (along the n axis) placed on the aperture plane i.e. x and y following 1Eb/2 and Inlk a/2 UCEn 0)A elsewhere for all values of n. Calculate the corresponding Fraunhoffer diffraction. 7. Analyze using Fresnel integrals the diffraction of a plane wave incident normally on a long narrow slit. Further analyze the transition of Fresnel diffraction to Fraunhoffer diffraction. [5] 8. In a straight edge diffraction pattern, if the wavelength of the light used is 6000 A° and the distance between the screen and the straight edge is 100cm, calculate the distance between the most intense maximum and the next minimum. Find approximately the distance in centimetres inside the geometrical shadow where I/I=0.1. b 1+1] t are [5] at a Table for Fresnel integrals is given below first FRESNEL INTEGRALS Sis) d on ? cone l to ocal Cs) 0.000 0.004 0.033 0.111 0.249 0.438 0.623 0.714 0.000 0.200 0.398 0.0 0.2 0.4 0.6 0.561 0,723 0.780 0.715 0.8 1.0 1.2 the 0,543 0.366 0.334 1.4 1.6 0.638 0.451 0.343 1.8 2.0 2.5 re 0.488 0.457 0.606 0.533 0.498 0.500 0.619 0.496 0.415 0.420 0.500 3.0

Can anyone solve Q8
Fresnel Analysis

Answer 1

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For a given experimental set up, the determination of maxima and minina positions is very easy Let the posihon bey's be wavelength of light) de distance between screen and straight edge Now, the above cuwe is a universal awwe. For a quien value of & andd, one can calculate the value for vo, as the obsewection pont moves along y axis Therefore, first three maximas ofcula at, VOE-122 with I -1.37 ILA VO = -2-34 with I z1-2016 Vy = -3.08 with I. 11151. Similarly first three minmas ocau at, VOP-1.87 urth 1.0.778 I, VO2-274 urth I-0-8431 Vo = -3.39 with I, 0.872 I. The formula to be used for this problem is y=-fad vo .} Let the most intense maxoma be g = "max and next minimum occurs at y = ymin . { x= 6000 Å - 6x15cm then f Volymax) = -1.22 l d=100cm. Volymin) - -|-87 from 0, 00, ymax = -122 E ymun = -1187 :: Sy = Ymın - Ymax = (1-87-1-27) Ad 65) 6x105x6m) x 100 cm) 0.0356 m

Now, the intensity vamation is given by the formula, I = Jo [{ i-com} + { [ -5(vo);?) Vo = y. Cknown equation') Using table of Fresnel integrals, we can forid TlIo by the above equaliön, This answer is Obtained by making cutain approximations At Vo=0, ccvo) = 0.00 SCVO )= 0.00 >Fat Io { 6) +(+)?) .: 141,- 0-25 but we necd 1/16 = 0.1, hence we keep on doing trial and allor method. > At v=0.2 ccvo) -0.2 Skvo) - 0.004 1,4 to [(4-02)? ++ -.004) ?] 1/1,20.168 > At Vo=04, CCvo) = 0-398 SCUo) = 0-033 Inţ Io ( 6 -0382 + (-0-03)] 157 0.114 > At Vo = 0.6 ,c(vo-0.581 s(v) = 0111 Iotto ( E -o-su)? +( -01073) 77, 30.074 So Vo has to be in between 0.4 and o.s. On some research, I was able to find that at Vo-0.5, c(vo) = 0.49234 and slvo) 0.064732, which gives it a 0.095 -0.1. Hence, we can assume VO 2015 Sonly approximate 3 on substtäting it on yo voy de 6cco x10$ xloo 0.027 cm 4-6-5) , 6.9) Como aloo ~ 0027 cm