Equation to find the intensity of diffraction patterns: explanations please

• DottZakapa
In summary, angles in mathematical formulas are almost always expressed in radians. 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
DottZakapa
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 anyone so kind to explain to me why is like that?

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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?

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

DottZakapa said:
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.

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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1. What is the equation used to find the intensity of diffraction patterns?

The equation to find the intensity of diffraction patterns is given by I = I0sin2(θ)/ (θ)^2, where I0 is the incident intensity and θ is the angle of diffraction.

2. How is the intensity of diffraction patterns related to the angle of diffraction?

The intensity of diffraction patterns is inversely proportional to the square of the angle of diffraction. This means that as the angle of diffraction increases, the intensity of the diffraction pattern decreases.

3. What is the significance of the incident intensity in the equation?

The incident intensity, represented by I0, is the initial intensity of the light beam before it passes through the diffracting object. It affects the overall intensity of the diffraction pattern and can be adjusted to change the visibility of the pattern.

4. Can the equation be used for all types of diffraction patterns?

Yes, the equation can be used for all types of diffraction patterns as long as the diffraction is caused by a single slit, double slit, or diffraction grating. However, the value of θ may need to be adjusted depending on the type of diffraction pattern being studied.

5. Are there any limitations to using this equation?

Yes, there are some limitations to using this equation. It assumes that the diffracting object is infinitely long and thin, and that the incident light is monochromatic (single wavelength). It also does not take into account any other factors that may affect the intensity of the diffraction pattern, such as the size of the diffracting object or the wavelength of the incident light.

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