Calculate the power measured by the detector at distance h from the source

[Nadi Yazbeck]

Typing out is always better. It makes quoting and commenting much easier. Many helpers will just abandon ship if they have to do too much work to respond.

But, making an effort to clip, edit, and post a piece of your image, can you explain the exponent of 4 in your equation:
View attachment 233381

thanks, i will type from now on,it's from the inverse square law(S/(4*pi*d^4)=ir)

 

[gneill]

Hmm. I'd say that the power is spread over the surface area of a sphere of a given radius, so the flux would go as

$\Phi = \frac{P_o}{4 \pi r^2}$

Where r is the distance from the source.

The same should apply to intensity.

 

[Nadi Yazbeck]

Hmm. I'd say that the power is spread over the surface area of a sphere of a given radius, so the flux would go as

$\Phi = \frac{P_o}{4 \pi r^2}$

Where r is the distance from the source.

The same should apply to intensity.

oh okay, so if i replace the 4 with a 2, the final result is i0(ln(h^2+R^2)-2ln(h))/4, does that make more sense?
thanks

 

[gneill]

Yes, it does to me

I think you could do some manipulation of the logs to get a more concise version, but otherwise I think you're doing fine.

 

[Nadi Yazbeck]

Yes, it does to me

I think you could do some manipulation of the logs to get a more concise version, but otherwise I think you're doing fine.

that's great! i have been working on this for almost a week and couldn't figure it out hahahaha
thanks a lot!
nadi