How do you find the approximate interval that displays the solution?

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T = 350; %in K
epsi = 0.9; %emissitivity
c = 3e8; %speed of light
kb = 1.38e-23; %boltzman constant
h = 6.63e-34; %plancks constant
%full expression
Tb = (h*c)./(lambda*kb.*log(1+(1./epsi)*(exp((h*c)./(lambda*kb*T))-1)));
%Approximate expression
Tb_approx = epsi*T;
I need to find the approximate solution for both of these equations but I don't know how. Is there a way to plot both functions to find the interval results are displayed on?
  5 Comments
Walter Roberson
Walter Roberson on 12 Sep 2019
Tb = @(lambda) (h*c)./(lambda*kb.*log(1+(1./epsi)*(exp((h*c)./(lambda*kb*T))-1)));
fzero(@(lambda) Tb(lambda) - Tb_approx, rand())
The solution is about 3.20966703379108e+16

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Answers (1)

David Goodmanson
David Goodmanson on 14 Sep 2019
Edited: David Goodmanson on 14 Sep 2019
Hello Isabel,
I believe the idea here is that when lambda is large enough, the expression for Tb gets increasingly close to Tb_approx. So the question is, how large does lambda have to be so that Tb_approx is good? There is no hard and fast answer to that, but you can set up an array for lambda, plot Tb as a function of lambda, and compare to Tb_approx.
As far as the domain of lambda, there is a way to find a lambda0 that sets the scale of this problem. In the Tb expression, take a look at the Bose-Einstein factor in the denominator:
(exp((h*c)./(lambda*kb*T)) - 1)
A characteristic lambda0 occurs when the argument of the exponential is 1:
(h*c) / (lambda0*kb*T) = 1 --> lambda0 = (h*c) / (kb*T) = 4.12e-5.
You will have to go significantly above that to get close to Tb_approx. You can do something like
lambda = logspace(-7,-1,1000);
< calculate Tb here >
Tb_approxvec = Tb_approx*ones(size(lambda)); % horiz line
semilogx(lambda,Tb,lambda,Tb_approxvec); % log scale for lambda is appropriate
ylim([300 400])
It's interesting that for small lambda, the expession goes over to T.

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