help Monochromatic light of wavelength lambda is incident on a pair of slits sep
ID: 2244614 • Letter: H
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Monochromatic light of wavelength lambda is incident on a pair of slits separated by 2.05 times 10-4 m and forms an interference pattern on a screen placed 2.00 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. 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. 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. Using the result of part (b) and dsin theta bright = m lambda, calculate the wavelength of the light. Compute the angle for the 50th-order bright fringe from d sin theta bright = m lambda Find the position of the 50th-order bright fringe on the screen from y bright =L tan theta bright. Comment on the agreement between the answers to parts (a) and (e).Explanation / Answer
d = 2.05*10^-4 m
D = 2 m
y1 = 4.51*10^-3 m
a) y50 = 50*y1 = 50*4.51*10^-3 = 0.2275 m = 22.55 cm
b) theta = tan^-1(y1/D)
= tan^-1(4.51*10^-3/2)
= 0.1292 degrees
c)
d*sin(theta) = m*lamda
lamda = d*sin(theta)
= 2.05*10^-4*sin(0.1292)
= 4.62*10^-7 m
= 462 nm
d)
d*sin(theta) = 50*lamda
theta = sin^-1(50*462*10^-9/2.1*10^-4)
= 6.32 degrees
e) y_50 = L*tan(6.32)
= 2*tan(6.32)
= 0.2213 m
= 22.13 cm
f)
a and f are same. due to calculation errors we are getting small difference.
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