{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: MIRAGESM.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 -1 227 "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 " }} {PARA 0 "" 0 "" {TEXT -1 114 "n(y) = a + by = a (1+e y), where a, b, a nd e = b/a are positive constants. Snell\222s law, n sin(theta) = con stant, " }}{PARA 0 "" 0 "" {TEXT -1 20 "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 -1 100 " y + 1/e=(y0 \+ + 1/e) cosh((x-x0)/(y0+1/e)), where (x0,y0) is the lowest point on the path." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 400 "We wish to launch several rays at different angles from \+ a fixed point (xp,yp) and plot their paths. From Snell\222s law we fi nd a different minimum height y0p for each ray: n(yp) sin(theta-p) = n (y0p), or (1+e yp) sin(theta-p) = 1+e y0p, or y0p=(1/e)((1+e yp) sin(t heta-p)-1). Next, we find x0p from yp+1/e = (y0p+1/e) cosh((xp-x0p)/( y0p+1/e)), or 1/sin(theta-p) = cosh((xp-x0p)/(yp+1/e)sin(theta-p))." } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 56 "Now we a re ready to plot the path starting from (xp,yp)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 126 "a) Let H = 20 m, a = 1.0 03, b = 1.14607*10^(-5), and A = 1.5 degrees. Plot the path of a ray \+ launched under these conditions." }}{PARA 0 "" 0 "" {TEXT -1 111 "b) F ind the location along the x-axis where the marage would be visible fo r a viewing height of 1.3 m to 1.4 m." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "Part A: Path of Ray" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 165 "Enter in the given parameters. Alph aP and ThetaP must be in radians. Don't put in the value for AlphaP f or 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 21 "AlphaP:=(1.5*Pi/180);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Thetap:=(Pi/2-AlphaP);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Now determine yop from the formula derive d in the problem statement." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "yop: =evalf((((1+e*yp)*sin(Thetap))-1)/e,20);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Now determine 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 "S olve the above equation for xop." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "xop2:=solve(xop1,xop);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "Now en ter in the formula for 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 "En ter in the initial conditions 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 "RayPathy:=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(RayPathy,x=0..3000,title=`Path of the \+ Ray`);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Part B: Where is the mi rage visible?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "Find when the ray is first visible." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "x1:=solve(RayPathy=1.4,x);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 38 "Find when the ra is no longer visible." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "x2:=solve(RayPathy=1.3,x);" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 91 "Thus, the mirage is visible when the observer is 2 768.5 m to 2786.4 m away from the object." }}}}{MARK "17 0 0" 3 } {VIEWOPTS 1 1 0 1 1 1803 }