Unformatted text preview: 39. The angular positions of the ﬁrstorder diﬀraction lines are given by d sin θ = λ. Let λ1 be the shorter wavelength (430 nm) and θ be the angular position of the line associated with it. Let λ2 be the longer wavelength (680 nm), and let θ +∆θ be the angular position of the line associated with it. Here ∆θ = 20◦ . Then, d sin θ = λ1 and d sin(θ + ∆θ) = λ2 . We write sin(θ + ∆θ) as sin θ cos ∆θ + cos θ sin ∆θ, then use the equation for the ﬁrst line to replace sin θ with λ1 /d, and cos θ with 1 − λ2 /d2 . After multiplying 1 by d, we obtain λ1 cos ∆θ + Solving for d, we ﬁnd d = = (λ2 − λ1 cos ∆θ)2 + (λ1 sin ∆θ)2 sin2 ∆θ
2 2 d2 − λ2 sin ∆θ = λ2 . 1 [(680 nm) − (430 nm) cos 20◦ ] + [(430 nm) sin 20◦ ] sin2 20◦ = 914 nm = 9.14 × 10−4 mm . There are 1/d = 1/(9.14 × 10−4 mm) = 1090 rulings per mm. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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