B Equation to find the intensity of diffraction patterns: explanations please

  • Thread starter DottZakapa
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can anyone explain why i divide degrees by radians
I = I0[ sin (φ/2)/(φ/2)]2

Given such equation I don't understand why in sin(φ/2) are used degrees and in (φ/2) at the denominator radians. If there is any reason for it i forgot about it 😓.
Is there any one so kind to explain to me why is like that?

Thanks in advance
 

Charles Link

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Angles in mathematical formulas are almost always expressed in radians=the one exception is in trigonmetric functions like sine, cosine, and tangent, where calculators are made to give results when the angles are expressed in degrees. If you have seen Taylor series for sin(x): ## \sin{x}=x-x^3/3!+... ##, it is necessary that the ## x ## on the right side be measured in radians. The reason is the derivation uses that the derivative of ## f(x)= \sin{ x} ##, ## f'(x)= \cos{x} ##, and that is the case only if you measure in radians. In the derivation, the limit as ## \Delta x \rightarrow 0 ## of ## \frac{\sin(\Delta x)}{\Delta x}=1 ## only when measuring in radians. ## \\ ## The diffraction formula results from an integration of ## \cos{x} ##, and that's how it gets the factor in the denominator that it does. The integral of ## \cos{x} ## is equal to ## \sin{x} ## only if measuring in radians. If we mesure in degrees, there is an additional multiplicative factor. It's much, much easier to use radians to avoid the multiplicative factors that result if we were to take derivatives or to integrate ## \cos{x} ## and ## \sin{x} ## when ## x ## is measured in degrees. ## \\ ## Notice in the diffraction formula, for ## \phi /2 \rightarrow 0 ##, the intensity ## I=I_o ##. The angle ## \phi ## needs to be measured in radians to get this correct limit.## \\ ## e.g. if you take ## \frac{\sin{x}}{x} ## where ## x ## is a small angle such as ##x=1^{\circ} ##, you need to convert to radians to get the calculation of ## \frac{\sin{x}}{x} ## to give the answer of approximately "1" as it should.
 
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So in a formula, anytime i have a sine or cos or tan i solve it in degrees and what is in the denominator must be in radiant or also if trigonometric at denominator it stays always in degrees and what is at the numerator is in radiant.
For example if x=90º, in the trig function i plug in 90º while for x not in a trig function must be converted into pi/2.
Did i get it right?
 

Charles Link

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Yes. If your calculator is set on "degrees",(as opposed to "radians"), then you need to use ##90^{\circ} ## for the angle in the trig functions, but anywhere else, where it appears simply as ## x ## in the formula, you need to call it ## \frac{\pi}{2} ##.
 
Yes. If your calculator is set on "degrees",(as opposed to "radians"), then you need to use ##90^{\circ} ## for the angle in the trig functions, but anywhere else, where it appears simply as ## x ## in the formula, you need to call it ## \frac{\pi}{2} ##.
ok, many thanks
 

jtbell

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anytime i have a sine or cos or tan i solve it in degrees
Numerically, you can use either degrees or radians, depending on which mode your calculator is set to, but it's often easier to use degrees. For example:
If your calculator is set on "degrees",(as opposed to "radians"), then you need to use ##90^{\circ}## for the angle in the trig functions, but anywhere else, where it appears simply as ##x## in the formula, you need to call it ##\frac \pi 2##
If your calculator is set to "radians" mode, and your calculator has a ##\pi## button, you can enter ##\pi/2## (or ##\pi~2~/## if you use an H-P calculator like mine) but that's one more keystroke than ##90^{\circ}##. On the other hand, if you need to enter ##180^\circ##, you can simply press ##\pi## in radians mode. :cool:

On the third hand, if you need an "odd" angle such as 35° in radians mode, you need to enter something like 35 x π / 180. ?:)

On the fourth hand (or second foot), your calculator might have a button to convert degrees to radians, allowing you to enter something like 35 [RAD] [SIN].
 
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