Φ21 autumn 2006 1HW26 SolutionsProblem K22.42White light (400-700 nm) is incident on a 600 line/mm diraction 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 diract come dierent 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 diraction 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 diraction pattern of 600 nm irradiate on a screen 1.0 m behind the diraction grating.(Notice the unlike the situations where the small angle approximation applies, below the dierent orders of the diraction 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 reection? Solution: The thinnest devastating interference comes with a phase dierence the ∆φ = π in between the wave reected o the rst surface and the one reected 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 dierence of π ). But, the1wave reected 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 diraction 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 dierent wavelengths room separated top top the screen by 3.20 mm. What is the dierence 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 diraction 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◦