InterviewSolution
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InterviewSolution
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A single slit of width b is illuminated by a coherent monochromatic light of wavelength lambda. If the second and fourth minima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum? (i.e., distancebetween first minimum on either side of the central maximum) |
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Answer» `1.5 cm` Wavelength of the light = `lambda` The condition for secondary minima = `b sin theta_n ~~ b theta_n = n theta` `theta_n = (n lambda)/b` For angular position of the second minima, n = 2 `theta_2 = (2 lambda)/b` For angular position of the fourth minima , n = 4 `theta_2 = (4lambda)/b` Distance of the screen from the slit, D = 1m Distance of the `n^(th)` secondary minima from the central maxima is GIVEN by `x_n = D theta_n` THUS, distance of `2^(nd)` secondary minima form the central maxima is given by : `x_2 = (2 lambda)/b xx D = (2 lambda)/b "" [ :. D = 1m]` Similarly, distance of `4^(th)` secondary minima from the central maxima is given by: `x_4 = (4 lambda)/b` Thus, the distance between fourth and second secondary minima is given by: `x_4 - x_2 = (4 lambda)/b - (2 lambda)/b = (2 lambda)/b` Given that, `x_4 - x_2= 3 cm` Thus, `(2 lambda)/b = 3 cm " or " b = (2 lambda)/((3 cm))` Width of central maxima is given by: `beta = (2 D lambda)/b = (2 lambda)/(2 lambda//(3cm)) = 3cm`. |
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