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 {
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	   "source": [
       "# Chapter 4: Diffraction"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.10: Diffraction_of_Xrays.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
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"clc();\n",
"clear;\n",
"//Given:\n",
"N = 15000;//Number of lines per inch\n",
"a_plus_b = (2.54/N)*10^8 ;//Grating period in A\n",
"lambda = 1 ; //Wavelength in A\n",
"//Grating equation :(a+b)*sin(theta_n) = n*lambda\n",
"//First order maximum \n",
"theta1 = asind(lambda/a_plus_b); // angle in degrees\n",
"printf('The first order maximum will be obtained at : %.4f degrees .\n\n',theta1);\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.11: Resolution_of_human_eye.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
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"source": [
"clc();\n",
"clear;\n",
"//Given:\n",
"lambda = 6000; //Wwavelength in A\n",
"mu = 1.33; //Refractive index for cornea\n",
"D = 2; //Diameter of pupil in mm\n",
"//Yellow light wavelength in eye:\n",
"lambda1 = lambda/mu ; //Wavelength in A\n",
"//The angular resolution \n",
"//1 A = 1.0*10^-7 mm\n",
"theta_c = (1.22*lambda1*10^-7)/D; // angle in rad\n",
"//Maximum value for L\n",
"L = 1/tan(theta_c); // in mm\n",
"printf('Maximum value for L should be : %.1f mm',L);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.4: Interference_minima.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
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"clc();\n",
"clear;\n",
"//Given :\n",
"d = 8.8*10^-2 ; // slit width in mm\n",
"b = 0.7;// seperation between slits in mm\n",
"lambda = 6328 ; //Wavelength in A\n",
"//First diffraction minima is possible, when  d*sin(theta) = lambda\n",
"// 1 A = 1.0*10^-7 mm\n",
"theta = asind((lambda*10^-7)/d); // angle in degrees\n",
"printf('theta = %.3f degrees .\n\n',theta);\n",
"//interference minima is possible , when sin(theta) = ((p + 1/2)*lambda)/b\n",
"for p = 0 : 10\n",
"    //1 A = 1.0*10^-7 mm\n",
"    theta1 = asind((p + 1/2)*(lambda*10^-7/b)); // angle in degrees \n",
"    printf('When p = %d \n',p);\n",
"    printf('theta = %.3f degrees . \n\n',theta1);\n",
" if(theta1 > theta)\n",
"     printf(' When p >= %d ,  theta > %.3f degrees .\n\nBetween the first two diffraction minima , %d interference minima are possible.',p,theta,2*p);\n",
"     break;\n",
" end\n",
"end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.6: Angles_of_Diffraction.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
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	   "outputs": [],
"source": [
"clc();\n",
"clear;\n",
"//Given :\n",
"// a+b = (2.54/N)cm\n",
"N = 15000;//grating has 15000 lines\n",
"a_plus_b = 2.54/N ; // grating element in cm\n",
"//Grating equation, (a+b)*sin(theta_n) = n*lambda, we get :  theta_n = asind((n*lamba)/(a+b))\n",
"printf('For line D1  and Wavelength 5890 A:\n\n');\n",
"printf(' Angles at which first order and second order maxima will be observed are :\n');\n",
"lambda1 = 5890; //Wavelength in A\n",
"for n = 1:2 // First and second order maxima\n",
"// 1 A = 1.0*10^-7 mm\n",
"theta1_n = asind((n*lambda1*10^-8)/a_plus_b);// angle in degrees\n",
"printf(' Order :%d ,%.3f degrees \n',n,theta1_n);\n",
"end\n",
"printf('For line D2 and Wavelength 5895.9 A :\n\n');\n",
"printf(' Angles at which first order and second order maxima will be observed are :\n');\n",
"lambda2 = 5895.9 ; //Wavelength in A\n",
"for n1 = 1:2 //First and second order maxima\n",
"// 1 A = 1.0*10^-7 mm\n",
" theta2_n = asind((n1*lambda2*10^-8)/a_plus_b);// angle in degrees\n",
" printf('Order : %d, %.3f degrees \n',n1,theta2_n); \n",
"end\n",
"printf(' When n = 3,  sin(theta)= ((n*lambda*10^-8)/a_plus_b)>1 , which falls outside the sine range, hence third order maximum is not visible');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.8: Dispersion_and_resolving_power.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
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"clc();\n",
"clear;\n",
"// Given :\n",
"//(a) 15000 lines per inch \n",
"N1 = 15000; //15000 lines per inch\n",
"a1_plus_b1 = (2.54/N1)*10^8 ; //grating element in A\n",
"lambda1 = 5890; //Wavelength in A\n",
"lambda2 = 5895.9 ; // Wavelength in A\n",
"deltalambda1 = lambda2-lambda1; //in A\n",
"//For first order\n",
"n =1;\n",
"theta1 = 20.355; // in degrees\n",
"deltatheta1 = ((n*deltalambda1)/((a1_plus_b1)*cosd(theta1)));// dispersion in degrees/A\n",
"rp1 = n*N1; // resolving power\n",
"\n",
"\n",
"//(b)15000 lines per cm\n",
"// 1 cm = 0.393701 inches, so We have 15000 lines per 0.393701 inches. \n",
"//Therefore, For 1 inch we have 15000/0.393701 = 38099.979 or 38100 lines\n",
"N2 = 38100 ; //38100 lines per inch\n",
"a2_plus_b2 = (2.54/N2)*10^8 ; //grating element in A\n",
"//For first order\n",
"theta_1 = asind((n*lambda1)/(a2_plus_b2));// in degrees\n",
"deltatheta_1 = ((n*deltalambda1)/((a2_plus_b2)*cosd(theta_1)));//  dispersion in degrees/A\n",
"rp2 = n*15000; // resolving power\n",
"\n",
"\n",
"//(c)5906 lines per cm\n",
"// 1 cm = 0.393701 inches, so We have 5906 lines per 0.393701 inches. \n",
"//Therefore, For 1 inch we have 5906/0.393701 =  15001.232  or 15001 lines\n",
"N3 = 15001; //15001 lines per inch\n",
"a3_plus_b3 = (2.54/N3)*10^8; //grating element in A\n",
"//For first order\n",
"theta__1 = asind((n*lambda1)/(a3_plus_b3)); // in degrees\n",
"deltatheta__1 = ((n*deltalambda1)/((a3_plus_b3)*cosd(theta__1))); //  dispersion in degrees/A\n",
"rp3 = n*5906; // resolving power\n",
"\n",
"printf(' Number of lines \tGrating element (in A)\t Angle of diffraction(degrees)\t Dispersion (degrees/A) \t Resolving Power\n');\n",
"printf('%d /inch\t\t\t %.0f\t\t %.2f \t\t\t\t %.2f x 10^-3\t\t\t %d\n',N1,a1_plus_b1,theta1,deltatheta1*10^3,rp1);\n",
"printf('%d /cm\t\t\t %.0f\t\t %.2f \t\t\t\t %.2f x 10^-3\t\t\t %d\n',15000,a2_plus_b2,theta_1,deltatheta_1*10^3,rp2);\n",
"printf('%d /cm\t\t\t %.0f\t\t %.2f \t\t\t\t %.2f x 10^-3\t\t\t %d\n',5906,a3_plus_b3,theta__1,deltatheta__1*10^3,rp3);\n",
"// Error in textbook for dispersion values . Error in decimal point placement ."
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.9: Determination_of_separation_of_lines.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
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"source": [
"clc();\n",
"clear;\n",
"//Given:\n",
"//Wavelength\n",
"n=1; // first order diffraction\n",
"lambda1 = 4680 ;// Wavelength in A\n",
"lambda2 = 4800; //Wavelength in A\n",
"lambda3 = 5770 ; // Wave;ength in A\n",
"// First order diffraction angle\n",
"theta1 = 28.0; // angle in degrees\n",
"theta2 = 28.7; // angle in degrees\n",
"theta3 = 35.5; //angle in degrees\n",
"//Grating equation : (a+b) = n*lambda/sin(theta) \n",
"a1_plus_b1 = (n*lambda1)/sind(theta1); //spacing in A\n",
"a2_plus_b2 = (n*lambda2)/sind(theta2); //spacing in A\n",
"a3_plus_b3 = (n*lambda3)/sind(theta3); //spacing  in A\n",
"mean_spacing = (a1_plus_b1 + a2_plus_b2 + a3_plus_b3)/3; // mean spacing in A \n",
"printf('(a)Wavelength :%d  A \n Angle of 1st order Diffraction : %.1f degrees \n Spacing = %.1f A\n\n',lambda1,theta1,a1_plus_b1);\n",
"printf('(b)Wavelength :%d  A \n Angle of 1st order Diffraction : %.1f degrees \n Spacing = %.1f A\n\n',lambda2,theta2,a2_plus_b2);\n",
"printf('(c)Wavelength :%d  A \n Angle of 1st order Diffraction : %.1f degrees \n Spacing = %.1f A\n\n',lambda3,theta3,a3_plus_b3);\n",
"printf('Mean Spacing = %.1f A',mean_spacing);"
   ]
   }
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