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A plane light wave wityh wavelength 0.60 mu m falls normally on a long opaque strip 0.70 mm wide. Behind it a screen is placed at a distance 100cm. Using Fig. find the ratio of intensities of light in the middle of the diffraction pattern and at the edge of the geometrical shadow. |
Answer» Solution :We shall use the equation written down in `5.103`, the Fresnel-Huyghens FORMULA.  Suppose we want to find the intensity at `P` which is such that the corrdinates of the edges (x-coordinates) with respect to `P` are `x_(2)` and `-x_(1)`. Then, the amplitude at `P` is `E = intK(varphi) (a_(0))/(r )e^(-ikr) dS` We write `dS = dxdy, y` is to integrated from `-oo + 0 + oo`. We write `r = b + (x^(2) + y^(2))/(2b) .......(1)` `(r` is the disatnce of the element of surface on `I` from `P`. It is `sqrt(b^(2) +x^(2) + y^(2))` and hence appoximately (1)). Wer then get `E = A_(0)(b) [int_(0)^(oo) e^(-ikx^(2)//2b) dx + int_(-oo)^(-x_(1))e^(-ikx^(2)//2b)dx]` `= A'_(0)(b) [int_(v_(2))^(oo) e^(-i)(piu^(2))/(2)du+ int_(-oo)^(-v_(1)) e^(-ipiu^(2)//2 du)]` where `v_(2) = sqrt((2)/(b lambda))x_(2), v_(1) sqrt((2)/(b lambda))x_(1)` The intensity is the SQUARE of the amplitude. In our case, at the centre `v_(1) = v_(2) = sqrt((2)/(b lambda)).(a)/(2) = sqrt((a^(2))/(2b lambda)) = 0.64` `(a =` width of the STRIP `= 0.7 mm, b = 100CM, lambda = 0.60mu m)` At, say, the lower edge `v_(1) = 0, v_(2) = 1.28` Thus `(I_(centre))/(I_("edge")) =|(int_(0.64)^(oo) e^(-IPI u^(2)//2)du + int _(-oo)^(0.64)e^(-ipiu^(2)//2) du)/(int_(1.28)^(oo) e^(-ipi u^(2)//2)du + int_(-oo)^(0)e^(-ipiu^(2)//2) du)|^(2) = 4(((1)/(2)-C(0.74))^(2) + ((1)/(2)-S(0.64))^(2))/(1-C(1.28)^(2)+(1-S(1.28))^(2))` where `C(v) = int_(0)^(v) cos .(piu^(2))/(2)du` `S(v) = int_(0)^(v) sin. (piu^(2))/(2)du` Rough evaluation of the intergrals using corn's spiral gives `(I_(centre))/(I_("edge")) ~~2.4` (We have used `int_(0)^(oo) cos. (piu^(2))/(2)du =int_(0)^(oo) sin. (piu^(2))/(2)du = (1)/(2)` `C(0.64) = 0.62, S(0.64) = 0.15` `C(1.28) = 0.65 S(1.28) = 0.67` |
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