//chapter-1,Example1_4_6,pg 1-34

//For nth dark ring       Dn^2 =4*R*n*wavelength

D_5=0.42                  //Diameter of 5th dark ring

D_10=sqrt(2*D_5^2)               //as number of ring double, the diameter is sqrt(2) times the diameter of original ring

printf('\nThe diameter of 10th dark ring is  D10 = %.3f cm',D_10)