Light of wavelength 588.2 nm illuminates a slit of width 0.63 mm. (a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.86 mm from the central maximum? (b) Calculate the width of the central maximum. Step 1 (a) As shown in the figure, dark bands or minima occur where sin 0 = m(2/a). For the first minimum, m = 1 and the distance from the center of the central maximum to the first minimum is y₁ = L tan 8, where L is the distance of the viewing screen from the slit. 32 sin dark = 22/a 31 sin dark = λ/a HE 0 -1 sin dark = -λ/a -2 sin dark = -22/a Viewing screen a Because is very small, we can use the approximation tan sin 0 = m(2/a). Substituting the approximation and solving for the distance to the screen, we have 6.3 x 10 m ³ m ) (₁ L = = y ₁ ( ² ) = x 10-3 m x 10-⁹ m m.
Light of wavelength 588.2 nm illuminates a slit of width 0.63 mm. (a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.86 mm from the central maximum? (b) Calculate the width of the central maximum. Step 1 (a) As shown in the figure, dark bands or minima occur where sin 0 = m(2/a). For the first minimum, m = 1 and the distance from the center of the central maximum to the first minimum is y₁ = L tan 8, where L is the distance of the viewing screen from the slit. 32 sin dark = 22/a 31 sin dark = λ/a HE 0 -1 sin dark = -λ/a -2 sin dark = -22/a Viewing screen a Because is very small, we can use the approximation tan sin 0 = m(2/a). Substituting the approximation and solving for the distance to the screen, we have 6.3 x 10 m ³ m ) (₁ L = = y ₁ ( ² ) = x 10-3 m x 10-⁹ m m.
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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
Transcribed Image Text:Light of wavelength 588.2 nm illuminates a slit of width 0.63 mm.
(a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.86 mm from the central maximum?
(b) Calculate the width of the central maximum.
Step 1
(a) As shown in the figure, dark bands or minima occur where sin 0 = m(2/a). For the first minimum, m = 1 and the distance from the center of the central
maximum to the first minimum is y₁ = L tan 8, where L is the distance of the viewing screen from the slit.
32 sin dark = 22/a
31 sin dark = λ/a
HE
0
-1 sin dark = -λ/a
-2 sin dark = -22/a
Viewing screen
a
Because is very small, we can use the approximation tan sin 0 = m(2/a). Substituting the approximation and solving for the distance to the screen, we have
6.3 x 10 m
³ m ) (₁
L =
= y ₁ ( ² ) =
x 10-3 m
x 10-⁹ m
m.
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