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A coherent parallel beam of microwaves of wavelength A = 0.5 mm fails on a Young's double slit apparatus. The separation between maxima is measured on a screen placed paraiiei to the plane, of the slits at a distance of 1.0 m from it as shown in figtird. The separation between the slits is 2d - 1mm. (a) If the incident beam fails normally on the double slit apparatus, find the y-coordinates of all the interface minima of the screen. (b) If the incident beam makes an angle of 30° with the x-axis (as shown in fig. find the y-coordinates of the first minima on either side of the central maximum. |
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Answer» Solution :(a) Givenseperation between two slits = 2d 1.0mm, D=1.0m, Path diffenrece = `2dsin theta` For minima `2 d SINTHETA = (2n -1 ) lambda/2` `therafore 1 xx 10^(-3) = ((2n-1)xx(0.5xx10^(-3)))/2` as `theta=90^@` for the HIGHEST possible order of minimum. or `(2n-1) = (2xx10^(-3))/(0.5xx10^(-3)) implies n=2.5` Thus, only two minima are possible on either side of central maxima. Thus total number of minima = 2+2 =4 For minima `y= pm ((2n-1)lambdaD)/(4d) = (2n-1)0.25` `y_(1)=0.25m , y_(2)=0.75m` and `y_(4)= -0.75m` (b) Initial path DIFFERENCE = `2d sin 30^@` For the central maxima `therefore y/D xx2d= 2d sin30^@` `y=Dsin30^@=0.5m` For first minima `y/D xx2d -2dsin 30^@ = om lambda/2` or `y= pm 0.25 + 0.5` or `y_(1)0.75m ` and `y_(2) = 0.25 m`. |
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