(a) Assuming the fringes are laid out linearly along the screen, find the positi
ID: 2244592 • Letter: #
Question
(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.
(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 dsin?bright = m?, calculate the wavelength of the light.
(d) Compute the angle for the 50th-order bright fringe from dsin?bright = m?.
(f) Comment on the agreement between the answers to parts (a) and (e).
Monochromatic light of wavelength ? is incident on a pair of slits separated by 2.10 times 10-4 m and forms an interference pattern on a screen placed 1.50 m from the slits. The first-order bright fringe is at a position ybright = 4.55 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 d sin? bright = m?, calculate the wavelength of the light. Compute the angle for the 50th-order bright fringe from d sin? bright = m?. Find the position of the 50th-order bright fringe on the screen from ybright = L tan? bright. Comment on the agreement between the answers to parts (a) and (e).Explanation / Answer
d = 2.1*10^-4 m
D = 1.5 m
y1 = 4.55*10^-3 m
a) y50 = 50*y1 = 50*4.55*10^-3 = 0.2275 m = 22.75 cm
b) theta = tan^-1(y1/D)
= tan^-1(4.55*10^-3/1.5)
= 0.1738 degrees
c)
d*sin(theta) = m*lamda
lamda = d*sin(theta)
= 2.1*10^-4*sin(0.1738)
= 6.37*10^-7 m
= 637 nm
d)
d*sin(theta) = 50*lamda
theta = sin^-1(50*637*10^-9/2.1*10^-4)
= 8.724 degrees
e) y_50 = L*tan(8.724)
= 1.5*tan(8.724)
= 0.23 m
= 23 cm
f)
a and f are same. due to calculation mistakes we are getting small difference.
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