Radiation Pressure (Modern Physics)
im getting 2IRH/3c
im on PC so i cant take a pic but heres what i did :
divide the cone into disks of height dh. take a disk of radius r. consider a small area on it at angle theta as rd(theta) * dh. consider over a time diff dt on this momentum change = IAcos^2(theta)/c * dt
dP/dt = F = IA cos^2(theta)/c
now take component along direction of radiation cuz vectors will cancel and only the one in the direction of Incident ray remain.
dF(net) = IA cos^3(theta)/c
A = rdtheta * dh
integrate wrt theta from -pi/2 to pi/2
F (for small height dh at specific r) = 4Ir/3c * dh
r/h = R/H => r = Rh/H
F = 4IR/3Hc * h dh
integrate wrt h from 0 to H
F = 4IR/3Hc * H^2/2 = 2IRH/3c
what step is wrong and why
55 Replies
@Gyro Gearloose
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Integrate kyu kara ha?
how
cuz area is changing all the time and the angle between beam and area vector too
Aise na kar
Pehli baat to aadhe cone par koo radiation nahi pad rahi hogi
Ab ek kaam kr cone ko upar se dekh
So area = 1/2 * 2r*h
F=irh/c
Sorry D hoga answer
2r nahi lia tha kal
@AY
yes thats why limit is from -pi/2 to pi/2
where did u pull this out from?
of what
ok
and yes i have taken component of force
thats why cos^3(theta) comes
still didnt understand whats wrong
i might be overcomplicating but this is how we were taught
i wish there was something like gauss law for this
ek component bacha ha radiation ka, and that component ke liye area cone ke beech vale triangle ka
uska area 1/22rh
ok i think i kinda get it. u took projection of > along a plane
ok tell me how you would solve this :
wait ur ans makes sense
this is the ans to this
so for q like this
simply
projected area by radiation
=A
and do IA/c
the area shaded in yellow
but whats wrong with my approach ðŸ˜
that ans is correct
Haan vo dekhna padega thoda never thought it that way
Yesss
well so this works for any shape? (symmetrical like cone/sphere obv)
Yes
Same thing here also
ok and for not symmetrical ones too?
Haan usme bhi lag jayega
yes actually u get the same ans
but they used my approach
to get it
Can u send that
their approach?
Yep
yes just a sec
part 1
Is this cengage?
no my coaching module
this is a illustration
part 2
Both method should work i think
Aapke vale par kal dekhta hu firse integration vala part
ok
also one last
q
about ur approach
if surface is reflecting
or absorbing
does it affect it?
or the force is still the same
Then 2I/c
i thought the force should double for reflection cuz of momentum
yeah
but
last part of the solution
this confuses me
Wow
Gotta check this also
should not have missed the lecture ðŸ˜
Mera bhi modern phy chl raha ha abhi vese to revise krke pura fir dekhta iska answer
my phy teach is busy with other batches and wont take doubts till thursday
so cant even clear it
and i have full syl test on sunday
Oh shi
Hamare bhi phy teacher kehre the full test krvadein
1-2hfte mein kehdia humne.
allen?
Kehte agar bekar number aye aage se nahi dogeðŸ˜
Na na pw
oh vidyapeeth
Yess
nice
U allen?
my nearest centre is trash
nah
bakliwal tutorials
I also think pw is trash, but kya karein ab
Oh nice
nah pw isnt. vidyapeeths are good (some are better than allen in their area). but some centres have bad management
and thus faculty is changed alot in my nearest centre
@Prasan any news?
Lol folks yhou should really close this out. and stay on point. this one went wayyy offtrack eh
Yeah i got the answer for this
But aapke integration vale ka nahi pata chala mereko flaw
I think ki reflecting mei nahi kr skte use projection vala method because har jagha alag alag angle par padh rahi light, to force bhi alag alag lg rahi hogi iss karan se
@AY
+solved @Prasan
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