# Stuck in solving two variables from two expressions

After doing experiment with Michelson interferometer I need to calculate two coefficients: k and n. This involves solving two equations for k and n:

$$\frac{(8 \pi d)}{\lambda }{kn}=\frac{\text{\triangle I}}{I}$$

$$\text{\triangle l}_2=\left\{\frac{\lambda }{2 \pi }\right\} \tan ^{-1}\left(\frac{2 \left(n \sin \left(\frac{4 \pi d n}{\lambda }\right)+k\right)}{\frac{4 \pi d \left(n^2-1\right)}{\lambda }+\left(n^2+1\right) \cos \left(\frac{4 \pi d n}{\lambda }\right)}\right)$$

For the first one I have everything exept $\Delta$*r*. As for second one, I have all coefficients. But Mathematica won't solve it neither with Solve, neither with NSolve.

Solve[{k*n==0.00730621,
1.67/0.00539535==ArcTan[(2(k+n Sin[8.89655n]))/(8.89655(-1+n^2)+(1+n^2)Cos[8.89655 n])]},
{k,n}]


It just goes into ever increasing evaluation till computer starts to freeze or I stop evaluation.

Could anyone suggest how I tackle this problem? Thanks.

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• please put the complete code here. – Raymond Ghaffarian Shirazi Oct 7 '15 at 18:18
• Perhaps start by solving the first equation for, say, n, plugging that into the second equation, then plotting that equation as a function of n and seeing roughly where the roots are. Then use FindRoot. – march Oct 7 '15 at 19:12
• Thanks for response. @RaymondGhaffarianShirazi - it is all the code. I'm just trying to use Mathematica to calculate k,n. – JustasJ. Oct 7 '15 at 19:48
• Solve for k in terms of n and plug the result into the second equation. Then Plot[ f[n] , {n,-10,10}] to see that there is no zero. This suggests that there is something strange with the parameter numbers calculated before. – Kagaratsch Oct 7 '15 at 22:59

Following March's suggestion, if one takes the first expression we can determine that k is:

0.00730621/n


Create a function of n that represents the second equation. Replace k in the second equation with the above expression:

eq2[n_] := 1.67/0.00539535 - ArcTan[(2 (0.00730621/n + n Sin[8.89655 n]))/
(8.89655 (-1 + n^2) + (1 + n^2) Cos[8.89655 n])]


Plot it as a function of n. Since the equation has 8.89655 n as an argument to Sin and Cos it makes sense to limit n to be less than 8.89655/(2 π).

There is a discontinuity where the denominator goes to zero so it is excluded from the plot.

Plot[eq2[n], {n, 0, 8.89655/ (2 π)},
PlotRange -> {{0, 1.5}, {-1.8, 1.8}},
Exclusions -> {(8.89655 (-1 + n^2) + (1 + n^2) Cos[8.89655 n]) == 0}
] The plot shows that eq2 never approaches zero. At the discontinuity the ArcTan jumps from -π/2 to π/2.

• Thanks a lot! @JackLaVigne I just don't understand why you dont use LHS of second equation. – JustasJ. Oct 8 '15 at 7:16
• @JustasJ - Terribly sorry. 100% error on my part. The equation should be modified as shown by Willinski below. Following his and Kagaratsch's comment, it is clear that there is no solution. – Jack LaVigne Oct 8 '15 at 12:48
• I have edited my answer accordingly. – Jack LaVigne Oct 8 '15 at 13:08
• Thanks for going through all the trouble! @JackLaVigne – JustasJ. Oct 8 '15 at 15:10

There are no zeros (see @Kagaratsch comment).

n = 0.00730621/k;
eq = 1.67/0.00539535 - ArcTan[(2 (k +n Sin[8.89655 n]))/(8.89655 (-1 + n^2) + (1 + n^2) Cos[8.89655 n])] // FullSimplify Plot[eq, {k, -0.1, 0.1}, PlotRange -> All] 