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Text File  |  1993-08-31  |  2KB  |  14 lines

  1. "AST1CAL3 EQUATION VARIABLE","08-31-1993","18:35:44"
  2. "R┴=(COSD(Θ[I])-((N[T]/N[I])^2-SIND(Θ[I])^2)^0.5)/(COSD(Θ[I])+((N[T]/N[I])^2-SIND(Θ[I])^2)^0.5) R║=((N[T]/N[I])^2*COSD(Θ[I])-((N[T]/N[I])^2-SIND(Θ[I])^2)^0.5)/((N[T]/N[I])^2*COSD(Θ[I])+((N[T]/N[I])^2-SIND(Θ[I])^2)^0.5) R[N]=RND(100*0.5*(R┴^2+R║^2))"
  3. "OPTICS: REFLECTANCE OF NATURAL LIGHT AT AN INTERFACE.                                     incident      reflected                                                              \/        /\                                                                      \     /  air   n[i]                                                    angle Θ[i] \ /                                                               ╔════════════════════════════════╗                                              ╠     glass interface      n[t]  ╢                                              ╚════════════════════════════════╝                        R[N] is % reflectance of natural light which is the average of the magnitudes   of its perpendiclar R┴ and parallel R║ components.  Θ[I] is the angle of        incidence of the beam in degrees.  N[I] is the index of refraction of air (=1.0)and N[T] is the index of refraction of the glass substrate.                     *** Answer to Problems ***                                (c) PCSCC, Inc., 1993 Set N[I]=1, N[T]=1.61 and Θ[I]=75.  The reflectance is 27.1%.                                         Type any key to exit.                                                                                                                                    || (a) What percentage of incoming unpolarized light is reflectedat an air-glass interface (N[glass]=1.61) for a beam of natural light incident  at 75°?  Type comma key to see answers. Type (F2) to return to application file."
  4. 6
  5. -.6653652040734061,0,""
  6. -.3150533609387719,0,""
  7. 27.1,0,""
  8. 75,0,""
  9. 1.61,0,""
  10. 1,0,""
  11. 1
  12. 0
  13. 0
  14.