intensity of ultrasound!! phy app

[WellWIshER]

guys i went through the booklet and and da book but i still didnot understand the part where we use da equation of attenuation and da reflection absorption co-efficient 2gether???

any one help !!

also to be specific i didnot get question on pg 418 of da endorsed cie book

 

[zeebujha]

guys i went through the booklet and and da book but i still didnot understand the part where we use da equation of attenuation and da reflection absorption co-efficient 2gether???

any one help !!

also to be specific i didnot get question on pg 418 of da endorsed cie book

It may be helpful to think of the equation in terms of the radioactive decay. You may notice that there is a striking similaritiy between the two. You can take the intensity to be the no. of undecayed nuclei and k(absorption coefficient) to be the decay constant.

Thinking in those terms, all the equation states is that the intensity of the ultrasound (the same applies for X-ray or any other wave) decreases exponentially.

 

[WellWIshER]

yes yes but check out da question

i dont get it when it says reflecting..bl bl bla........

 

[zeebujha]

Oh, sorry . Should have read the question more carefully!
The question basically involves absorption on two occasions and reflection on one occasion.

First , when the wave passes through the 3.5 cm of muscle, use the absorption coeff. formula
I1/I = exp(-kx)= exp (- 3.5 x 0.23) =0.4471 I1= transmitted energy I= Initial total energy
What this means is that the fraction of initial energy that gets transmitted through the muscle layer is 0.4471.

now, that fraction of intensity gets reflected off a bone layer
IR/I= (Z2-Z1)^2/ (Z2+Z1)^2. When you plug in the values you will get the answer 0.34
Here is the TRICK!! This 0.34 is the fraction of the energy of the wave energy incident on the bone. The wave energy which is incident is already attenuated . It is 0.4471 of the original intensity. So, the wave that gets reflected off the bone is 0.34 of 0.4471 of original intensity i.e. 0.4471x0.34=0.152001 (not emphasising too much on significant figures here)
Finally, this fraction of intensity goes through the muscle layer as well. We alredy figured out that the transmitted energy is 0.4471 of incident energy
i.e 0.4471 of 0.152001 = 0.4471x0.152001=0.0068
So, the wave intensity that finally returns in 0.0068 of the original intensity i.e 0.68%

Hope it was clear enough

 

[WellWIshER]

thanks

but i am still confused. using dis way i tried da following q which cums aftr dis but i couldnt do it

can u do da other q

i dont get y dont we sue da fact tht"when some energy is transmitted, some is also reflected."?
i was trying to do in terms of diagrams but didnt really work out 4 me

 

[princesszahra]

there's a question abt this topic in the pastpapers , chk that out !if u understand the way markscheme has solved it, well and good!
otherwise try solving form anyother phyics book
or try this
http://en.wikipedia.org/wiki/Attenuation_coefficient

 

[WellWIshER]

thanks

but i am still confused. using dis way i tried da following q which cums aftr dis but i couldnt do it

can u do da other q

i dont get y dont we sue da fact tht"when some energy is transmitted, some is also reflected."?
i was trying to do in terms of diagrams but didnt really work out 4 me

anyone!!!!!!!!!

 

[WellWIshER]

a parallel beam of ultrasound passes through a thickness of 4 cm of muscle. it is then incident normally on a bone having a specific acoustic impedance of 6.4x10^6. the bone is 1.5 cm thick . CALCULATE THE FRACTION OF THE INCIDENT INTENSITY THAT IS TRANSMITTED THROUGH THE MUSCLE AND BONE.

LINEAR ABSORPTION COEFFICIENT

bone--------------------0.13
muscle-----------------0.23


anyone!!


thanx

 

[WellWIshER]

like anyone helped me!!!!