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.

1.67/0.00539535==ArcTan[(2(k+n Sin[8.89655n]))/(8.89655(-1+n^2)+(1+n^2)Cos[8.89655 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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    – user9660
    Oct 7, 2015 at 17:59
  • $\begingroup$ please put the complete code here. $\endgroup$ Oct 7, 2015 at 18:18
  • 1
    $\begingroup$ 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. $\endgroup$
    – march
    Oct 7, 2015 at 19:12
  • $\begingroup$ Thanks for response. @RaymondGhaffarianShirazi - it is all the code. I'm just trying to use Mathematica to calculate k,n. $\endgroup$ Oct 7, 2015 at 19:48
  • 1
    $\begingroup$ 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. $\endgroup$
    – Kagaratsch
    Oct 7, 2015 at 22:59

2 Answers 2


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


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}

Mathematica graphics

The plot shows that eq2 never approaches zero. At the discontinuity the ArcTan jumps from -π/2 to π/2.

  • $\begingroup$ Thanks a lot! @JackLaVigne I just don't understand why you dont use LHS of second equation. $\endgroup$ Oct 8, 2015 at 7:16
  • $\begingroup$ @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. $\endgroup$ Oct 8, 2015 at 12:48
  • $\begingroup$ I have edited my answer accordingly. $\endgroup$ Oct 8, 2015 at 13:08
  • $\begingroup$ Thanks for going through all the trouble! @JackLaVigne $\endgroup$ Oct 8, 2015 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

enter image description here

Plot[eq, {k, -0.1, 0.1}, PlotRange -> All]

enter image description here


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