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Physics: Post your doubts here!

mehria

mehria

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Thought blocker said:
:/ Its a sad thing, tum logoko jo hard lagta hain, meko wo easy lagta hain, meko jo hard lagta hain, appko easy lagta hain :/
Click to expand...
ok listen... as the light is passing through the polariser then it's amplitude is chnged by cos60... lets suppose the initial amplitude is 1... and when light passes through the polariser then the amplitude will chnge to 1cos60= 0.5
when the amplitude is halved then the intensity decreases by 4...
so I/4=(1/2)^2
and our ans will be 0.25I
 
S

sagar65265

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Thought blocker said:
Again, now hope I am helpful to you :(
9)use the formula v^2 = u^2 + 2as
where
positive direction = direction of train
u = original velocity (speed of train)
v = final velocity (zero)
a = acceleration of train (in this case, acceleration is negative)
s = distance (from point where velocity = u to where velocity = v)

The problem uses x for distance, not s, so the equation we'll use is v^2 = u^2 + 2ax

because v = 0, we can write
u^2 + 2ax = 0, rearranging as:
x = -(u^2) / 2a . . . . . . . . Note - a has a negative value, thus making x positive
the deceleration doesn't change, hence, x is directly proportional to u^2.....
x = Ku^2 where K = -1 / (2a)
ie. x varies as the square of u
so if u increases by 20% (ie changes by a factor of 1.2), then x will change by a factor of (1.2)^2
which is a factor of 1.44
thus the minimum distance between yellow and red must now be 1.44x


25)Not sure ._. Suchal Riaz
1I ∝ 1A^(2) ---> (we can write it as 2I ∝ 2A^(2) means any value of I will be same as A^(2))
So [a cos (60)]^(2) = 0.25A^(2)
Hence we can write it as 0.25I ∝ 0.25A^(2)

I am not sure about 25 I think this is only the way... if anyone knows any other way, do tell me...
Click to expand...

This is the formula that relates Intensity I to Amplitude x:

I = 2(π)^2 * ρ(ƒ)^2v(Δxmax)^2

Where I = Intensity, ρ = density of medium where wave is travelling, ƒ = frequency of vibration, v = speed of propagation of wave and Δxmax= maximum amplitude.

(This is from an advanced website, paper me nahi pucchenge).

Since the wave is traveling from air into air, the density of medium is constant. The frequency of the wave doesn't change (orange light pass kare to orange dim ho jata hai, par orange hi rehta hai) so f is constant. The speed of the wave depends on medium, since medium doesn't change speed doesn't change. Amplitude changes, that is why Intensity changes.

So I = kA^2 before light passes through polarizing filter.

After it passes through filter,

I(final) = kA(final)^2
I(final) = kA2(cos 60)^2

Dividing both equations, k cancels out because it doesn't change. A^2 cancels because it is the same variable. So,

I(final)/I = (cos 60)^2

So, I(final) = (0.5)(0.5) * I

I(final) = 0.25I = A.

Hope this helped!

Good Luck for all your exams!
 
Thought blocker

Thought blocker

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mehria said:
ok listen... as the light is passing through the polariser then it's amplitude is chnged by cos60... lets suppose the initial amplitude is 1... and when light passes through the polariser then the amplitude will chnge to 1cos60= 0.5
when the amplitude is halved then the intensity decreases by 4...
so I/4=(1/2)^2
and our ans will be 0.25I
Click to expand...
I m getting more confused ._.
y u assumed 1 :/
y u halved, hw u got 4
??? :/
 
Thought blocker

Thought blocker

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sagar65265 said:
This is the formula that relates Intensity I to Amplitude x:

I = 2(π)^2 * ρ(ƒ)^2v(Δxmax)^2

Where I = Intensity, ρ = density of medium where wave is travelling, ƒ = frequency of vibration, v = speed of propagation of wave and Δxmax= maximum amplitude.

(This is from an advanced website, paper me nahi pucchenge).

Since the wave is traveling from air into air, the density of medium is constant. The frequency of the wave doesn't change (orange light pass kare to orange dim ho jata hai, par orange hi rehta hai) so f is constant. The speed of the wave depends on medium, since medium doesn't change speed doesn't change. Amplitude changes, that is why Intensity changes.

So I = kA^2 before light passes through polarizing filter.

After it passes through filter,

I(final) = kA(final)^2
I(final) = kA2(cos 60)^2

Dividing both equations, k cancels out because it doesn't change. A^2 cancels because it is the same variable. So,

I(final)/I = (cos 60)^2

So, I(final) = (0.5)(0.5) * I

I(final) = 0.25I = A.

Hope this helped!

Good Luck for all your exams!
Click to expand...
Got it, <3 ty ;)
 
mehria

mehria

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Thought blocker said:
I m getting more confused ._.
y u assumed 1 :/
y u halved, hw u got 4
??? :/
Click to expand...
luk we knw that I is directly proportional to A^2
this means that when the A is doubled then the intensity increases by 4 ... it increases by 4 because only A is squared not the I
so when A is halved then I will decrease by 4
i assumed 1 for a bcuz we can see that a is not chnged... the only extra thing we r gettng is cos60 after the light passes through the polariser...n ths thing is the one that is chngng the A... and as cos60=0.5 so we can see that it halves the value of the amplitude... that's the reason y the intensity is decreased by 4...
 
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kitkat <3 :p said:
great -_- i solve it in detail and the problem in solved
Click to expand...
nyways ty ;)
mehria said:
luk we knw that I is directly proportional to A^2
this means that when the A is doubled then the intensity increases by 4 ... it increases by 4 because only A is squared not the I
so when A is halved then I will decrease by 4
i assumed 1 for a bcuz we can see that a is not chnged... the only extra thing we r gettng is cos60 after the light passes through the polariser...n ths thing is the one that is chngng the A... and as cos60=0.5 so we can see that it halves the value of the amplitude... that's the reason y the intensity is decreased by 4...
Click to expand...
can we assume anything to be 1 if it is unchanged ?
well I got it ty ;)
 
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