# Determining radius at which 50% of energy is in the profile.

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1. Nov 7, 2014

### nabeel17

1. The problem statement, all variables and given/known data
I have to determine the radius at which 50% of energy is in a Gaussian profile.

2. Relevant equations
The intensity is given by I=Ioe^(-r/2c)^2. This is just a gaussian function ofcourse.

3. The attempt at a solution
I know c is the standard deviation. I searched through charts that 50% of the data set is contained within about 0.67 standard deviations. So can i set I/i0 to 0.5 and set c=0.67? I feel like this is not the correct method though. What I need is the area under the gaussian curve that is 50%. So I need to integrate between values of -r and r that will give me an answer of 50 % but I dont know what values those are.

2. Nov 7, 2014

### vela

Staff Emeritus
$c$ sets the length scale of the problem. Your value for $r$ will be expressed as some multiple of $c$.

You need to be a little more careful here. Your Gaussian is a two-dimensional Gaussian, not the one-dimensional Gaussian you've assumed. That is, you actually have
$$e^{-\frac{r^2}{2\sigma^2}} = e^{-\frac{x^2+y^2}{2\sigma^2}}.$$ You have to integrate with respect to polar coordinates $r$ and $\theta$, not just $r$ alone.

3. Nov 7, 2014

### nabeel17

right so I can just multiply by 2pi and integrate with respect to r. But I don't know how to integrate this. My value of r will be a multiple of c...so I can rewrite r as n*c? and try integrating that?

4. Nov 7, 2014

### vela

Staff Emeritus
Not exactly. What's the area element for polar coordinates?

5. Nov 8, 2014

### nabeel17

rdrdtheta but what are the limits?