Monochromatic light of wavelength A is incident on a pair of slits separated by

Question

Monochromatic light of wavelength A is incident on a pair of slits separated by 2.00 times 10^-4 m and forms an interference pattern on a screen placed 2.10 m from the slits. The first-order bright fringe is at a position y_bright = 4.51 mm measured from the center of the central maximum. From this information, we wish to predict where the fringe for n = 50 would be located. (a) Assuming the fringes are laid out linearly along the screen, find the position of the n = 50 fringe by multiplying the position of the n = 1 fringe by 50.0. m (b) Find the tangent of the angle the first-order bright fringe makes with respect to the line extending from the point midway between the slits to the center of the central maximum. (c) Using the result of part (b) and d sin theta_bright = m lambda, calculate the wavelength of the light. nm (d) Compute the angle for the 50th-order bright fringe from d sin theta_bright = m lambda. degree (e) Find the position of the 50th-order bright fringe on the screen from y_bright = L tan theta_bright. m

Explanation / Answer

(a)

the first order bright fringe is located at ybright = 4.51 mm, so this is the position of the n = 1

So, the position of n = 50 fringe is at y = (4.51 mm)(50) = 0.225 m

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(b)

tan = ybright/(2.10 m) = 4.51 mm/2.10 m = (4.51X10-3 m)/(2.1 m)

or, tan = 2.15X10-3 is the required tangent.

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(c)

From part (b) we can calculate sin = tan/[1+tan2] = 2.15X10-3

dsinbright = m, where d is the separation between the slits.

or (2X10-4 m)(2.15X10-3 ) = 1. (m =1 as was calculated for first order fringe)

or = 4.3X10-7 m =

or, = 430 nm

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(d)

dsinbright = m, here m = 50, so we have,

(2X10-4 m)sinbright = 50X4.3X10-7 m

or bright = 6.170

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(e)

ybright = Ltanbright = (2.1 m)tan6.170

or ybright = 0.227 m

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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