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// Scilab Code Ex2.55:: Page-2.46(2009)
clc; clear;
lambda1 = 6000e-008; // First visible wavelength, cm
lambda2 = 4500e-008; // Second visible wavelength, cm
R = 100; // Radius of curvature of the lens, cm
// As diameter of nth dark ring due to lambda1 is
// D_n^2 = 4*n*R*lambda1 and D_nplus1^ = 4*(n+1)*R*lambda2, so that D_n^2 = D_nplus1^2 gives
n = lambda2/(lambda1-lambda2); // Order of interference for dark fringes
D_n = sqrt(4*n*R*lambda1); // Diameter of nth dark ring due to lambda1
printf("\nThe diameter of nth dark ring due to wavelength of %4d angstrom = %4.2f cm", lambda1/1e-008, D_n);
// Result
// The diameter of nth dark ring due to wavelength of 6000 angstrom = 0.27 cm
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