Φ21 autumn 2006 1HW26 SolutionsProblem K22.42White light (400-700 nm) is incident on a 600 line/mm diraction grating. What is the width of the rstorder rainbow on a screen 2.0 m behind the grating? provide your answer in cm. Execute not make little angle approximations. Solution:The two excessive wavelengths the light will diract come dierent angles according come mλ = d sin θThe distance between the present of the grating is d = (1/600) mm = 1.667 × 10−6 m = 1667 nm. Due to the fact that this is the rst-order rainbow, m = 1 because that both waves. So, the two angles space θred= sin−1θviolet= sin−1λ 700 nm = sin−1 = 24.834◦ d 1667 nm 400 nm = 13.887◦ 1667 nmThe actual distance y indigenous the center of the screen to the diraction line originates from tan θ = Ly , wherein L is the distance from the lattice to the screen. So, the 2 distances and the broad of the rainbow are: yred yviolet ∆y2= together tan θred = (2.0 m) tan (24.834◦ ) = 0.925 m = (2.0 m) tan (13.887◦ ) = 0.494 m = 0.431 m = 43.1 cmProblem K22.46How countless lines per millimeter walk the grating have? (Do no make tiny angle approximations). Solution:θ = tan−1The rst-order heat is hitting in ~ y = 0.436 m, for this reason the edge is = 22.557◦ . Utilizing mλ = d sin θ,0.436 1.0d=mλ (1) (600 nm) = = 1501 nm = 1.501 × 10−3 mm sin θ sin (22.557◦ )Therefore, the thickness of the grating is 666 lines/mm.Figure 1: The diraction pattern of 600 nm irradiate on a screen 1.0 m behind the diraction grating.(Notice the unlike the situations where the small angle approximation applies, below the dierent orders of the diraction pattern gain further and further apart.) 3Problem YF 35.29What is the thinnest lm that a coating with an index of refraction of n = 1.35 ~ above glass (with an table of contents of refraction ng > n) for which terrible interference the the red component (a wavelength the 630 nm) of an incident white irradiate beam in air deserve to take location by reection? Solution: The thinnest devastating interference comes with a phase dierence the ∆φ = π in between the wave reected o the rst surface and the one reected o the ago surface of the lm. Since n > 1 (as is true for every materials), the wave from the front surface ar is reverse (given a basic phase dierence of π ). But, the1wave reected indigenous the ago surface is additionally inverted due to the fact that n 4∆` λn) = 2π2y =π λ/nλ = 116.7 nm 4n=Problem YF 36.29Visible irradiate passes v a diraction lattice that has 900 slits/cm and also the interference sample is it was observed on a screen that is 2.36 m native the grating. In the rst-order spectrum, maxima for two dierent wavelengths room separated top top the screen by 3.20 mm. What is the dierence in this wavelengths (in meters)? (Here you need to make small angle approximations.) Solution:thatWe"ll do the little angle approximation the sin θ ≈ tan θ ≈ θ (with θ in radians). That means y = tan θ ≈ θ L∆θ ≈and∆y 0.0032 m = = 0.00136 rad together 2.36 mThe lattice spacing is d = (90000 lines/m)−1 = 1.11 × 10−5 m. In the diraction formula, the small-angle approx. V m = 1 gives: mλ = d sin θ ≈ dθ5→∆λ ≈ d∆θ = 15.1 nmProblem K36.33Plane monochromatic waves through wavelength 540 nm are incident normally on a airplane transmission grating having 350 slits/mm. Component A. Uncover the angle of deviation in the rst order. The cleft separation is d = (350000 slits/m)−1 = 2.857×10−6 = 2857 nm. An initial order means m = 1 in the formula mλ = d sin θ. Solution:θ1 = sin−1 part B.mλ dFind the edge of the 2nd order.
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θ2 = sin−1Part C.((Find the edge of the 3rd order. θ3 = sin−1()= sin−1Solution:mλ d)(540 2857)= 10.89◦Second order method m = 2, so= sin−1(2 · 540 2857)= 22.21◦Solution:mλ d)= sin−12(3 · 540 2857)= 34.54◦