What is the equation for making theta the subject in this scenario?

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Homework Help Overview

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.

Discussion Character

  • Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore methods to manipulate the equation to isolate θ, discussing steps such as dividing by constants and applying inverse functions. There are questions about the typesetting of the equation and clarification on variable notation.

Discussion Status

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.

Contextual Notes

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.

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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
 
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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

What are you trying to express? It=Iosin2(4[tex]\theta[/tex]sin2([tex]\pi[/tex]dnp/[tex]\lambda[/tex])

Maybe some of that needs to be amended as
[tex]\frac{\pi dnp}{\lambda}[/tex]
The typesetting tools available in the Compose for replying are new to me.

I know that did not work, I still have trouble with these typesetting features.
 
Last edited by a moderator:
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{It}{Iosin^2(\pi dn p/\lambda)}[/tex]
Get rid of the "2" by doing the opposite: square root
[tex]sin(4\theta)= \sqrt{\frac{It}{Iosin^2(\pi dn p/\lambda)}[/tex]
and get rid of the sin by using arcsin:
[tex]4\theta= arcsin\left(\sqrt{\frac{It}{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!
 
Last edited by a moderator:
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!

hi, thanks for the reply. where did the In bit come from?
 
Sorry, that was supposed to be your It.

I have gone back and editted to change In to It.
 
Thank you for your help.
 

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