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+//example-22.11

+//page no-674

+//given

+//depth of penetration at 3K is 39.8 nm and at 7.1 K is 1730 A

+T1=3  //K

+T2=7.1  //K

+dp1=39.6*10^-9  //m

+dp2=1730*10^-10  //m

+//as we know that depth of penetration and temp are related as

+//(dp(T)/dp(T0))=1/(1-t^4/Tc^4)

+//so we get

+//at 3K

+//let dp(T0)=dp0

+//dp0=sqrt(dp1^2*(1-T1^4/Tc^4))  -(1)

+//also

+//dp0=sqrt(dp2^2*(1-T2^4/Tc^4))  .----(2)

+//solving 1 and 2 we get

+//((Tc^4-81)/(Tc^4-(7.1)^4))=(173)^2/(39.6)^2

+//so we get

+Tc=(48417.9/18.085)^(1/4)  //K\

+//depth of penetration at absolute zero willbe

+dp0=sqrt(dp1^2*(1-T1^4/Tc^4))*10^9  //nm

+printf ("critica temp is %f K\n and depth of penetration at critica zero is %f nm",Tc,dp0)