{VERSION 2 3 "IBM INTEL NT" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Fo nt 0" -1 256 1 {CSTYLE "" -1 -1 "Helvetica" 1 12 255 0 0 1 2 1 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Font 2" -1 257 1 {CSTYLE "" -1 -1 "Courier" 1 12 0 0 0 1 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "PHYSICS RESOURCE PACKETS P ROJECT File: MIRAGECM.MWS" }}{PARA 0 "" 0 "" {TEXT -1 110 "Rose-Hulman Institute of Technology \+ Authors: Perry Peters & Greg Williby" }}{PARA 0 "" 0 "" {TEXT -1 97 "http://www.rose-hulman.edu/~moloney \+ Software: Maple V Release 4" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 361 "A mirage will occur when surface air is hotter than the air above it. Then a flat road surfac e will appear to give glassy reflections of objects near the horizon. \+ The index of refraction increases with height, so we will say n(y) = \+ a + by = a (1+e y), where a, b, and e = b/a are positive constants. S nell\222s law, n sin(theta) = constant, applies to all rays." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 326 "Call the point (x0,y0) the point where a ray is traveling parallel to the ground, an d the angle theta in Snell\222s law is 90 degrees. In Snell\222s law \+ anywhere on the path, sin(theta) = dx (dx^2 + dy^2)^(-1/2) = (1+(dy/dx )^2)^(-1/2). y0 is the lowest point on the path; all other y values w ill be greater. Snell\222s law now reads" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 " n sin(theta) = a (1+ey)(1+(dy /dx)^2)^(-1/2) = constant = n(y0) sin(Pi/2) = n(y0) = a(1+ey0)" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 275 "Rearrang ing this gives (y + 1/e)/(y0 + 1/e) = (1+(dy/dx)^2)^(1/2). This equat ion has a straightforward solution, when we make the substitution (y + 1/e)/(y0 + 1/e) = cosh(z), which leads to dy/dx = sinh(z), and finall y to x=(y0 + 1/e) z. The full equation of a ray is then:" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 " y + 1/e=(y0 + 1/e) cosh((x-x0)/(y0+1/e))" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 400 "We wish to launch several rays at different an gles from a fixed point (xp,yp) and plot their paths. From Snell\222s law we find a different minimum height y0p for each ray: n(yp) sin(th eta-p) = n(y0p), or (1+e yp) sin(theta-p) = 1+e y0p, or y0p=(1/e)((1+e yp) sin(theta-p)-1). Next, we find xop from yp+1/e = (yop+1/e) cosh( (xp-xop)/(yop+1/e)), or 1/sin(theta-p) = cosh((xp-xop)/(yp+1/e)sin(the ta-p))." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 56 "Now we are ready to plot the path starting from (xp,yp)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 186 "a) Let H = 20 \+ m, a = 1.003, b = 1.14607*10^(-5), and A = 1.5 degrees. Plot the path of a ray launched under these conditions. Repeat for rays launched a t 1.48 degrees and 1.49 degrees." }}{PARA 0 "" 0 "" {TEXT -1 171 "b) E stimate the location of the image from you plot. This will involve dr awing lines by hand on the plot, or using Maple or Mathematica to find tangent lines to the rays." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Digits:=20;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "P art A: Path of Ray" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 165 "Enter in the given parameters. AlphaP and ThetaP must be in radians. Don't put in the value for AlphaP for now so that it \+ remains a variable in the final equation." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "xp:=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "yp :=30;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "a:=1.003;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "b:=1.14607*10^(-5);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "e:=b/a;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "AlphaP:=1.5*Pi/180;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "Thetap:=Pi/2-AlphaP;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Now determine yop from the formula derived in the problem statement." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "yop:=evalf((((1+e*yp )*sin(Thetap))-1)/e);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Now dete rmine xop from the formula derived in the problem statement." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "xop1:=yp+1/e=(yop+1/e)*cosh((xp-xop)/(yop +1/e));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "Solve the above equati on for xop." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "xop2:=solve(xop1,xop );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "Now enter in the formula fo r the path of the ray." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "RayPath1: =y+1/e=(y0+1/e)*cosh((x-x0)/(y0+1/e));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 "Enter in the initial cond itions for x0 and y0." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "RayPath2:= subs(\{y0=yop,x0=xop2\},RayPath1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "Solve RayPath2 for y." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "Ray Pathy:=solve(RayPath2,y);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Now \+ plot the path of the ray." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "plot(R ayPathy,x=0..3000,title=`Path of the Ray`);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 18 "Part B: Vary Alpha" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 127 "Now find the path of the ray when AlphaP is 1.49 degrees. All commands are the same, so only the final path l ength is printed." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "AlphaP:=evalf( 1.49*Pi/180):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Thetap:=ev alf(Pi/2-AlphaP):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "yop:=( ((1+e*yp)*sin(Thetap))-1)/e:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "xop1:=yp+1/e=(yop+1/e)*cosh((xp-xop)/(yop+1/e));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "xop2:=solve(xop1,xop);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "RayPath1:=y+1/e=(y0+1/e)*cosh((x-x0 )/(y0+1/e)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "RayPath2:=s ubs(\{y0=yop,x0=xop2\},RayPath1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "RayPathya:=solve(RayPath2,y);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "Now find the path of the ray for when AlphaP is 1.48 degrees." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "AlphaP:=evalf(1.48*Pi/ 180):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Thetap:=evalf(Pi/2 -AlphaP):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "yop:=(((1+e*yp )*sin(Thetap))-1)/e:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "xop 1:=yp+1/e=(yop+1/e)*cosh((xp-xop)/(yop+1/e));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "xop2:=solve(xop1,xop);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "RayPath1:=y+1/e=(y0+1/e)*cosh((x-x0)/(y0+1/e)): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "RayPath2:=subs(\{y0=yop ,x0=xop2\},RayPath1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Ra yPathyb:=solve(RayPath2,y);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "No w plot the path of the three rays for AlphaP=1.5,1.49,1.48" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "plot(\{RayPathy,RayPathya,RayPathyb\},x=200 0..2800,title=`Rays for AlphaP=1.5,1.49,1.48`);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Part C: Determination of image location" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 148 "Assume that the o bserver is at x=2770 and that the observer can see all rays between 1. 3 and 2.3 m high. Where does the image appear to be located?" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 131 "In order to determine the image's location, the tangent lines to the three pat hs must be calculated and then simultaneously solved." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "At x=2770, the corresp onding y-values are:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "yvalue:=eva lf(subs(x=2770,RayPathy));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "yvaluea:=evalf(subs(x=2770,RayPathya));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "yvalueb:=evalf(subs(x=2770,RayPathyb));" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "The slope at the points calculated is the derivative of the paths at the calculated point." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "slope:=evalf(subs(x=2770,diff(RayPathy,x))); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "slopea:=evalf(subs(x=27 70,diff(RayPathya,x)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 " slopeb:=evalf(subs(x=2770,diff(RayPathyb,x)));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "The tangent lines at the points, in point-slope form , are:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "line:=(y-yvalue)=slope*(x -2770);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "linea:=(y-yvalue a)=slopea*(x-2770);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "line b:=(y-yvalueb)=slopeb*(x-2770);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "Solving each equation for y gives:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "y1:=solve(line,y);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "y2:=solve(linea,y);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 " y3:=solve(lineb,y);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "Find the i ntersection of lines y1 and y2." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 " sol1:=solve(\{line,linea\},\{x,y\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "Find the intersection of lines \+ y1 and y3" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "sol2:=solve(\{line,lin eb\},\{x,y\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 189 "The three rays do not converge on the same point which indicates the the fuzzy appea rance most mirages. Howerer, the rays are relatively close together, \+ which allows the object to be seen." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 70 "Thus the image appears to be comming fro m approximately (-1.4,-13.846)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "plot(\{y1,y2,y3,RayPathy,RayPathya ,RayPathyb\},x=-10..2800,y=-14..10,color=black,title=`Path of rays`); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "48 0 0" 0 } {VIEWOPTS 1 1 0 1 1 1803 }