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//Chapter-1,Example 1_14,Page 1-41
clc()
//Given Data:
n=4 //4th dark ring
m=12 //m=n+p
D4=0.4*10^-2 //Diameter of 4th ring
D12=0.7*10^-2 //Diameter of 12th ring
//Calculations:
//(Dn+p)^2-Dn^2=4*p*lam*R
//Solving, (D12^2-D4^2)/(D20^2-D4^2)
//We get above value =1/2. Hence
D20=sqrt(2*D12^2-D4^2) //Diameter of 20th ring
printf('Diameter of 20th ring is =%.5f m \n',D20)
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