Normalized intensity on value I(0)

[RA_10315344]

I have generated three plots for different z or distance 

now i would like to compare the width of axial maximum and the width of central maximum is the value r from 0 to the point where the cure achieves zero. 

I want to normalies and assume the value achive r=0 

and i want to divide all values of I of I(r=0) how its possible to do this in mathcad

please help!

[Werner_E]

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and i want to divide all values of I of I(r=0) how its possible to do this in mathcad

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Simply do it as you wrote.

In the plot at the placeholder for the function you simply type in I(r,100)/I(0,100) instead of just I(r,100).

You may also plot all three graphs in one plot region using different colors. Guess you know how to add a second and third trace by either using the menu or by just typing Shift-Enter.

For further questions I would suggest that you attach your worksheet.

[RA_10315344]

I tried this but still i cant see any difference they were all the same i will upload all the note , and please tell me how can i combine all three in one ? 

[Werner_E]

About adding further traces

Its also not wrong to consult the help

https://support.ptc.com/help/mathcad/r9.0/en/index.html#page/PTC_Mathcad_Help/about_traces_and_axis_expressions.html

When  I asked you you attach your worksheet, I did not mean you to send a full screenshot picture but rather meant that you should attach the Prime file itself. You can do it by drag and drop or by browsing:

EDIT: I just look at the definition of your function I(r,z).

Changing z does just mean a scaling in ordinate direction. So when you "normalize" by dividing by I(0,z) you get the exact same function/graph, basically you cancel al the factors and what you see is the square of the Bessel function J0(k.0*r*sin(beta).

So of course you could not see any difference if you plot all traces in one plot - they are exactly the same!

Just a notice:

Prime can do quite well with units - You should use them!

No need to define units like inches and no need to define the imaginary unit (you get it by typing 1i or 1j without a space or multiplication inbetween)

Admittedly it looks ugly because on contrary to real Mathcad Prime does not suppress the leading "1" when it displays non-real numbers.

[Werner_E · Accepted solution]

One additional remark because you initially wrote that you are looking for "the value r from 0 to the point where the cure achieves zero. ".

The zeros correspond with the zeros of the Bessel function, divided by the factor you used for r in J0. And of course they are independent from the value of z.

Here are the first two positive zeros of your function - they correspond to the first two negative zeros of J0 because your factor sin(beta) is negative: