X-ray Flux density and a differential equation for photon scattering

[razidan]

Homework Statement

Consider interactions of a X-ray beam at a depth, x, within a material. The flux density is:
density flux = $$\frac{I}{A}$$

where I is the intensity of the beam that cross a unit area A at right angles to the beam. Let dx be a small slice at the depth x and let dI(x) be the decrease in the intensity of the X-ray beam due to the interaction or scattering of the photons by the atoms of the material contained in the slice between x and (x + dx).
Determine the ability of a beam of X-ray photons to form a sharp image by calculating the loss of photons scattered out of the beam by the slice of material of thickness dx at a depth x in the material.
(Hint: Find the expression for the flux density rate)

Homework Equations

The Attempt at a Solution

so i can almost "read" off the question that I need to derive $$\frac{dI}{dx}=-I$$ but I'm not sure how to set it up.
I tried to say there is "conservation of flux" and write $$\phi (x+dx) - \phi (x) = d\phi (x)$$ but that's always true. it doesn't lead me anywhere...
I am also unsure what "the ability to form a sharp image" means.
Thanks.
R.

 

[Dr Dr news]

I think the intensity is the flux/area. The higher the intensity, the sharper the image. I would guess that the intensity falls off as a negative exponential of the depth of penetration.

 

[haruspex]

not sure how to set it up.

What will be the relationship between the intensity reaching depth x, the thickness dx, and the loss through that thickness? You will need to create a constant.

 

[razidan]

What will be the relationship between the intensity reaching depth x, the thickness dx, and the loss through that thickness? You will need to create a constant.

I wrote that the incoming flux (at x) will br equal to the outgoing (at x+dx) plus the loss.
From there I got my equation (missing a minus on the RHS).
But that's not saying much. How can I construct the differential equation based on what I'm given?
Also, I inderatund there should be some attenuation factor, but do I "artificially" set it in my equation?

 

[haruspex]

I wrote that the incoming flux (at x) will br equal to the outgoing (at x+dx) plus the loss.
From there I got my equation (missing a minus on the RHS).
But that's not saying much. How can I construct the differential equation based on what I'm given?
Also, I inderatund there should be some attenuation factor, but do I "artificially" set it in my equation?

Consider a slice of thickness dx. Let the intensity reaching it be I and the loss through it ΔI.
What if we were to double the intensity reaching it to 2I? What loss would you expect then?

 

[razidan]

Consider a slice of thickness dx. Let the intensity reaching it be I and the loss through it ΔI.
What if we were to double the intensity reaching it to 2I? What loss would you expect then?

Twice the photons will have twice the probability to scatter. So twice the loss.
I think...

 

[haruspex]

Twice the photons will have twice the probability to scatter. So twice the loss.
I think...

Right, so can you write that as a proportionality?
Next, how about doubling dx (but for very small dx)?

 

[razidan]

Right, so can you write that as a proportionality?
Next, how about doubling dx (but for very small dx)?

The lossvwill be squared...?

 

[haruspex]

The lossvwill be squared...?

No, not squared. I think you meant multiplied by four if both intensity and dx are doubled.
I meant, keeping the incoming intensity constant but doubling dx, but no matter.

So as an algebraic equation, how does the loss of intensity depend on I and dx?

 

[razidan]

No, not squared. I think you meant multiplied by four if both intensity and dx are doubled.
I meant, keeping the incoming intensity constant but doubling dx, but no matter.

So as an algebraic equation, how does the loss of intensity depend on I and dx?

Why is the loss bot squared (without doubling the flux).
For each dx , there is some loss, so wouldn't for 2dx the loss would be proportional to dx²?

 

[haruspex]

Why is the loss bot squared (without doubling the flux).
For each dx , there is some loss, so wouldn't for 2dx the loss would be proportional to dx²?

So if dx<1, there is less loss through 2dx than through dx?

Suppose the loss through the first dx is 1%. What would you expect the loss to be through the next dx?

 

[razidan]

So if dx<1, there is less loss through 2dx than through dx?

Suppose the loss through the first dx is 1%. What would you expect the loss to be through the next dx?

Hmm...
Ok, I hear you.

 

[Mahnaz]

Hmm...
Ok, I hear you.

Do you mind sharing your method as I am stuck on this exact same question. Thanks

 

[haruspex]

Do you mind sharing your method as I am stuck on this exact same question. Thanks

It is not evident that razidan has solved it. Anyway, you should show your own attempt, so maybe best if you create a new thread.