questions_uk
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hi. with It = Io sin^2 x (4 x theta) x sin^2 (pi x dn x p / lambda) what would the equation be for making theta the subject? thanks for your input
The discussion revolves around an equation involving intensity, specifically It = Io sin²(4θ) sin²(πdn p / λ), and the goal is to isolate θ as the subject of the equation.
Some participants have provided guidance on how to approach the problem, suggesting steps to rearrange the equation. There is an ongoing clarification regarding variable notation, and the discussion reflects a collaborative effort to understand the equation better.
Participants note challenges with typesetting and the need for clarity in variable representation. There is an acknowledgment of potential ambiguities in the functions involved, particularly regarding the one-to-one nature of sine and squaring operations.
questions_uk said:hi. with It = Io sin^2 x (4 x theta) x sin^2 (pi x dn x p /lambda) what would the equation be for making theta the subject? thanks for your input
HallsofIvy said:questions_uk, start by "reversing" what is there, just like solving any equation. Everything except [itex]sin^2(4\theta)[/itex] is a constanat so just divide both sides by [itex]Io sin^2(\pi dn p/\lambda[/itex] to get
[tex]sin^2(4\theta)= \frac{In}{Iosin^2(\pi dn p/\lambda)}[/tex]
Get rid of the "2" by doing the opposite: square root
[tex]sin(4\theta)= \sqrt{\frac{In}{Iosin^2(\pi dn p/\lambda)}[/tex]
and get rid of the sin by using arcsin:
[tex]4\theta= arcsin\left(\sqrt{\frac{In}{Iosin^2(\pi dn p/\lambda}\right)[/tex]
Finally, of course, divide both sides by 4.
Since squaring and sine are not "one-to-one" functions, you might need to think about other possible values.
Symbolipoint, use "\", not "/" inside LaTex. And I recommend that you put entire equations in LaTex, not just individual symbols. In the second one here, you had a "tex", "/tex" pair inside another!