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

[Nadi Yazbeck]

Homework Statement

An isotropic point source radiates electromagnetic energy, and its output is measured by a thin disc-like detector of radius R. Calculate the power measured by the detector at distance h from the source, assuming that the plane of the disc is orthogonal to the line of sight between source and disc centre.

Homework Equations

beer's law, and inverse square law

The Attempt at a Solution

i tried calculating the intensity over a radius, and integrating it over the whole circle.

 

[gneill]

The Attempt at a Solution

i tried calculating the intensity over a radius, and integrating it over the whole circle.

Please show details of what you did. We can't comment on what we don't see.

 

[Nadi Yazbeck]

sure, here it is.
thanks

 

[Nadi Yazbeck]

i think that's wrong, i tried using the intensity over every value of r(with an integral), accounting for attenuation with beer's law.
i will try including a picture as soon as I'm done with it, do you think it should work or is there some reasoning mistake i don't see?
thanks
nadi

 

[gneill]

sure, here it is.
thanks

I'm afraid that I can't make out the text in your picture. Can you provide a closeup?

The problem statement is not specific about how the quantity of electromagnetic energy being emitted by the point source is specified. Why not just assume that it's being emitted with an overall power $P_o$ isotropically?

 

[Nadi Yazbeck]

the paper with the answer isn't with me right now, but i can try to explain it here:
first, i did a derivative from 0 to tetamax of h/cos(teta) then i tried to integrate the result over the circumference of the circle(from 0 to 2pi.r) I'm not sure if that makes sense, every time i read it over it makes less sense to be honest.
the lecturer didn't say anything else, i copy pasted the question on here, i think he wants us to keep the variables (ie: r, R, h, ...), and precise any other variable we add.
what would i do with the overall power, is the a specific formula for attenuation of power depending on distance?
thanks a lo.
nadi.

 

[Nik_2213]

Perhaps you're overthinking the problem ? I can see a very, very simple way to solve this without calculus...

 

[Nadi Yazbeck]

Perhaps you're overthinking the problem ? I can see a very, very simple way to solve this without calculus...

what do you mean?
does your solution account for attenuation, and the angle at which it hits the detector(the bigger the angle the fewer particles hits the detector)?
thanks
nadi

 

[Nik_2213]

Yes.

 

[Nadi Yazbeck]

can i get a hint? hahahahaha

Yes.

 

[Nik_2213]

Find a physical analogue such as a football plus a coin of radius R, think about them for a while...

I'm sorry, I don't often play in this forum, so I'd prefer a mentor or moderator to give the nod before I say more.

 

[gneill]

If you assume that the electromagnetic energy is emitted isotropically, with overall power $P_o$, then at some distance r that power is spread out over some larger area (what sort of area surrounds the central point at a constant radius r?).

 

[Nadi Yazbeck]

but the detector is flat, so the power would be different for each radius...
is there a way to compute a general formula?
thanks
nadi

If you assume that the electromagnetic energy is emitted isotropically, with overall power $P_o$, then at some distance r that power is spread out over some larger area (what sort of area surrounds the central point at a constant radius r?).

 

[gneill]

but the detector is flat, so the power would be different for each radius...
is there a way to compute a general formula?
thanks
nadi

At any reasonable distance the area of the disk will be very close to its projection onto the surface of a sphere centered on the central point with that distance as its radius. If you want a solution for when the disk is close to the central point (so that the radial distance varies noticeably over the disk surface), then you'll need to do a bit of calculus, taking into account the power per square meter at each position of the disk. That power per square meter still varies inversely as the radius of the sphere that passes through that point on the disk.

 

[Nadi Yazbeck]

Find a physical analogue such as a football plus a coin of radius R, think about them for a while...

I'm sorry, I don't often play in this forum, so I'd prefer a mentor or moderator to give the nod before I say more.

i'm not really sure what you mean... i will try to think about it
thanks

 

[gneill]

Find a physical analogue such as a football plus a coin of radius R

A football brings to mind different objects depending upon where in the world you live

 

[Nadi Yazbeck]

A football brings to mind different objects depending upon where in the world you live

i was thinking about a soccer ball hahahahaah are you talking about an american football?

 

[Nik_2213]

{Cough...}
Soccer Ball. Volley ball. Basket ball. Beach ball.

Sorry, I had the misfortune to attend a sports-mad school.
In Winter, we played through ankle-deep mud or over frozen corrugations with a brown, prolate spheroid.
In Summer, we played 'Russian Roulette' with red, fist-sized cannon balls...
;-))

 

[gneill]

i was thinking about a soccer ball hahahahaah are you talking about an american football?

'fraid so. Just sayin'

 

[Nadi Yazbeck]

At any reasonable distance the area of the disk will be very close to its projection onto the surface of a sphere centered on the central point with that distance as its radius. If you want a solution for when the disk is close to the central point (so that the radial distance varies noticeably over the disk surface), then you'll need to do a bit of calculus, taking into account the power per square meter at each position of the disk. That power per square meter still varies inversely as the radius of the sphere that passes through that point on the disk.

i need to account for for the change in distance, but even if i find a way to calculate it, i can't figure out how to find the air to apply the intensity over(i thought about doing an integral with the circumference, what do you think?)

 

[gneill]

i can't figure out how to find the air to apply the intensity over

Not sure what that means. What air is involved? Ah, do you mean that there's attenuation by the air through which the radiation passes?

 

[Nadi Yazbeck]

Not sure what that means. What air is involved? Ah, do you mean that there's attenuation by the air through which the radiation passes?

i meant area, sorry I'm used to french hahahahaha

 

[Nadi Yazbeck]

That's my work, do you think it's right?
https://s.amsu.ng/6HHsQQx466QN

 

[gneill]

That's my work, do you think it's right?
https://s.amsu.ng/6HHsQQx466QN

I'm sorry, I tried to read your image but had to give up trying to make best guesses at what I was seeing versus what I thought I should be seeing

Can you describe what your approach was and maybe type out a few equations?

If I were to approach this problem I'd assume an isotropic power emission $P_o$ and then define the flux at a given radial distance from the source to be $\Phi = \frac{P_o}{4 \pi r^2}$, basically spreading the power evenly over the surface of a sphere of radius r. Then I'd think about how various spheres of some radius r intersect the disk. Then integrate over the disk to sum the intercepted flux.

 

[Nadi Yazbeck]

I'm sorry, I tried to read your image but had to give up trying to make best guesses at what I was seeing versus what I thought I should be seeing

Can you describe what your approach was and maybe type out a few equations?

If I were to approach this problem I'd assume an isotropic power emission $P_o$ and then define the flux at a given radial distance from the source to be $\Phi = \frac{P_o}{4 \pi r^2}$, basically spreading the power evenly over the surface of a sphere of radius r. Then I'd think about how various spheres of some radius r intersect the disk. Then integrate over the disk to sum the intercepted flux.

hahahaha sure, i will try to explain the best i can:
first i used Pythagoras to find the distance (sqrt(r^2+h^2))
then i used the inverse square law to find the intensity when the radius is r (ir=i0/(4.pi.d^4))
then i integrated ir*Ar (area) from 0 to R (max radius)
Ar changes depending on the radius, = 2*pi*r
so P= integral(0,R) of ir*Ar=((i0/(4.pi.sqrt(r^2+h^2))*(2*pi*r))
then i just plugged the integral in an online integral calculator.i will try doing it like you said, could you check if what i did makes sense, and as soon as I'm done, i will post the result i got working like you said?
thanks a lot
nadi

 

[gneill]

i will try doing it like you said, could you check if what i did makes sense, and as soon as I'm done, i will post the result i got working like you said?

I will if I'm online and can read your work. You might want to investigate using LaTeX syntax to type out your equations. Then helpers can easily read, quote, and comment on them.

 

[Nadi Yazbeck]

o

I will if I'm online and can read your work. You might want to investigate using LaTeX syntax to type out your equations. Then helpers can easily read, quote, and comment on them.

okay, let me know what you think about my work please, my final result is: (-R^2 * i0)/(4h^2 (h^2 + R^2))
with: R: radius of detector, h: distance between the detector and the source, i0: intensity at the source.
thanks

 

[gneill]

o

okay, let me know what you think about my work please, my final result is: (-R^2 * i0)/(4h^2 (h^2 + R^2))
with: R: radius of detector, h: distance between the detector and the source, i0: intensity at the source.
thanks

I find it hard to comment instructively without seeing the details of your work, but I can say that I'm rather dubious about the minus sign.

Can you type out the integration you performed?

 

[Nadi Yazbeck]

Is a picture fine?
https://s.amsu.ng/5zxMdp9eKk6N
If not let me know and i will type it out
Thanks
Nadi

 

[gneill]

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:

 

[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