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Maths question

  1. Oct 15, 2008 #1
    How can dn be made the subject in the following equation?

    It = Io sin2 x 4 x theta x sin2 (pi x dn x p / lambda)

  2. jcsd
  3. Oct 15, 2008 #2


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    [itex]I_t= I_0 sin(2) 4\theta sin^2(\pi d_n p/\lambda)[/itex]?
    I am assuming that the first sin2 is sin(2) rather than [itex]sin^2[/itex] because sin (and so [itex]sin^2[/itex] is a <b>function</b> not a number and does not make sense if not applied to some number.

    You start just the way I'm sure you have learned before: since the [itex]d_n[/itex] you want to solve for is inside the [itex]sin^2[/itex] you first divide both sides by every thing multiplying that: [itex]\frac{I_t}{4 sin(2)\theta} = sin^2(\pi dn p/\lambda)

    Now get rid of the square by doing the opposite of that: take the square root of each side to get [itex]\sqrt{\frac{I_t}{4 sin(2)\theta}}= sin(\pi dn p/\lamba)[/itex]

    Now get rid of the sin by doing <b>its</b> opposite: arcsine or [itex]sin^{-1}[/itex] (which is NOT sin to the negative one power!):
    [itex]sin^{-1}(\sqrt{\frac{I_t}{4 sin(2)\theta}})= \pi dn p/\lambda[/itex]

    Finally, multiply both sides by [itex]\lambda[/itex] and divide both sides by [itex]\pi p[/itex].
  4. Oct 15, 2008 #3
    Thanks for your reply.

    Both sins are sin ^ 2. Would this make a difference?
  5. Oct 15, 2008 #4


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    In that case, we don't understand the problem. The following string of symbols:
    [tex]I_t= I_0 sin^2()\cdot 4\theta sin^2(\pi d_n p/\lambda)[/tex]
    does not form a well-defined equation because there's nothing in the (). What did you actually mean? Perhaps
    [tex]I_t= I_0 sin^2\left(4\theta sin^2(\pi d_n p/\lambda)\right)[/tex]?
  6. Oct 15, 2008 #5

    thanks for the reply. it is (4 x theta) sin ^ 2
  7. Oct 16, 2008 #6


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    You mean it is really It= Iosin^2(4 theta) sin^2(pi dn p/lambda)? Even easier. First divide both sides by Iosin^2(4 theta) to get
    It/(Io sin^2(4 theta))= sin^2(pi dn p/lambda)

    Take square roots:
    sqrt(It/(Io sin^2(4 theta)))= sin(pi dn p/lambda)
    Take the inverse sine of each side
    arcsin(sqrt(It/(Io sin^2(4 theta)))= pi dn p/lambda
    and finally, divide both sides by pi d/lambda

    lambda arcsin(sqrt(It/Io sin^2(4 theta)))/(pi p)= dn.
  8. Nov 1, 2008 #7
    Hi. Thanks for the previous input. how would theta be made the subject? any input is appreciated.
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