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psa
5:43 AM
@JohnRennie hiya
 
@psa morning :-)
 
psa
boy am I ever having trouble with a couple SR problems : (
the final is on saturday morning (it's thursday night here) unfortunately lol
 
@JohnRennie Good Morning sir :-)
 
psa
 
@user8718165 morning :-)
 
psa
5:45 AM
@JohnRennie just part d) I find extremely confusing
we've worked on this one before as well but it eludes me once again
what I chose to do was denote the double primed frame to be the dragon's frame. I denoted the event "dragon passes Hermione" to be (t,x) = (t'',x'') = (0,0)
in the dragon's frame, the events "hermione casts spell" and "harry casts spell" are equidistant and simultaneous, so $x''_d = \frac{1}{2}(x''_{hermione} + x''_{harry})$
from there I did some other stuff with LT's and got the wrong answer
I hate these harry potter questions
 
If the spells reach the dragon at the same time in the dragon's frame they reach the dragon at the same time in all frames.
 
psa
oops, yes you're right
it's just one event
but the dragon's frame is probably the only one where they're equidistant
so I guess it's a nice frame to choose?
 
Yes, because we know they were cast at the same time in the dragon's frame. In Hermione's and Harry's frames the casting events wouldn't have been simultaneous.
I would probably work in the dragon's frame.
 
psa
right
do you want me to show you what I did in full?
or do you want to just work through it from scratch?
 
No, I'm a bit busy this morning and won't have time to go through it in detail.
 
psa
5:54 AM
oh OK no worries
can I ask you a quicker MC question then?
 
@psa yes
 
psa
I understand how altering the conditions in this way could lead to A or C, but what about B?
I mean, is it just possible that you could decrease the frequency (so lower the energy of the incoming photons) and yet increase the intensity (so increase the number of photons) such that the current would increase?
 
If you are above the threshold frequency then the number of photons per second is $N = I/(h\nu)$ where $I$ is the intensity and $\nu$ is the frequency. The number of photoelectrons per second is $N$ times some constant that represents the efficiency of PE production.
And the current is proportional to the number of photoelectrons.
So the current is proportional to $I$ and inversely proportional to $\nu$.
@psa OK so far?
 
psa
yes
 
In the question we are increasing $I$ and decreasing $\nu$, so as long as we stay above the threshold the current will increase.
 
psa
6:05 AM
wow well that's quite helpful
I never actually knew $N = I/h\nu$
 
Well $I$ is the number of joules per second, and $1/h\nu$ is the number of photons per joule. Yes?
 
psa
yes
ah, so that makes sense
 
:-)
 
@JohnRennie hello sir
 
@user8718165 hi :-)
@user8718165 that's quite a fun question. I'm just discussing Python in another room, but I'll have a look at that question when I'm finished.
 
psa
6:13 AM
@JohnRennie when you're done with user - just to clarify, the current could also decrease (how?) or stop (if you went below the threshold frequency) as well right?
 
@JohnRennie okay sir...please ping me.
 
psa
I actually don't see any way you could decrease the current at all by doing this... but the answer says you could.
as in, you could do all three.
unless they're being pedantic and saying "stop" is a form of decrease
 
@psa well I suppose A implies C i.e. if you reduce the frequency far enough you go below the threshold and the current stops because no photoelectrons are produced. And for the current to stop it must reduce ...
 
psa
that seems like trickery
 
Yes. I think C is rather confusing.
I can't think of a way for the current to decrease except by getting near the threshold. I assume the trun-off isn't instant so the current will decrease smoothly to zero as you pass through the cut-off frequency.
 
psa
6:21 AM
I thought it was just a linear relationship
oh no
that's energy I believe
or something else as a function of frequency
not current
that's true though
yeah, that is certainly what it looks like when you look up graphs
kind of like a diode...
 
@JohnRennie hi, sir, Alesha here.
 
@yuvrajsingh hi Alesha. I'm busy in another room, but I'll ping you when I'm free.
 
6:42 AM
@JohnRennie
I will wait.
 
7:13 AM
@JohnRennie hello sir
 
7:30 AM
@JohnRennie done sir.
 
@yuvrajsingh hi, yes I'm free for about half an hour before I need to work again.
 
@JohnRennie can I.
 
Yes, if you have a question ask now :-)
 
OK, sir it is about prism.
But before asking the question I have a request.
Sir, please answer my mistake, or what error I am making whole thinking it.
So prism has prism angle of 45°
Now I take a light ray, which has gazing incidence.
 
OK ...
 
7:37 AM
And refracted at critical angle, now if this refracted ray when strikes to second surface, it should also have gazing emergence am I right or not, if I am wrong is it because of prism angle that light does it gaze while emerging.
@JohnRennie
 
The angle the ray strikes the second surface will depend on the value of the critical angle i.e. the value of the refractive index. I can't see any reason why the light should hit the second surface at any special angle.
 
OK sir, when light got refracted from first surface, $r_1$=theta c.
And this be the same angle for the second surface, with normal.
Then it should gaze.
 
Let's draw a diagram ...
@yuvrajsingh I've just drawn some random value for $\theta_c$. There's no reason why it should strike the second surface at the same angle.
 
I have a question question in my book.
Same like this
A prism of prism angle 45°,it is found that angle of emergence is 45°,for grazing incidence calculate the refractive index of the prism @JohnRennie
 
@yuvrajsingh I think this is what the question means.
 
7:53 AM
Yes.
@JohnRennie
 
It's a tedious question. You just have to calculate all the angles and solve for $n$.
 
@JohnRennie hello sir...are you free for a while
 
@user8718165 I need to work now again I'm afraid.
 
@JohnRennie for how long sir? I just wanted to talk about that ball problem...
 
8:14 AM
@user8718165 suppose the radius of curvature of the wall and the ball are equal, then at the instant the ball hits the wall there is a normal force all along the quarter of the ball that is in contact with the wall. Yes?
 
@JohnRennie work over sir?
 
Yes
 
@JohnRennie yeah sir...
 
@JohnRennie yeah sir...got it
 
8:19 AM
So at the instant of the collision the net force looks like this.
 
@JohnRennie yeah sir...
 
So there is a net upwards force and the ball will accelerate upwards.
But ...
 
@JohnRennie gravity...
 
Suppose the ball moves up an infinitesimal distance $dx$. As soon as it moves up at all it loses contact with the curved part of the wall.
 
@JohnRennie yeah sir
 
8:22 AM
 
@JohnRennie okay sir
 
So as soon as it moves upwards at all the force becomes horizontal and there is no upwards force. This happens for even the smallest value of $dx$.
 
@JohnRennie yeah sir...
 
So the work done by the upwards component of the force is zero because the upwards part of the force doesn't act over any distance.
 
@JohnRennie yeah sir
 
8:25 AM
And that means the ball doesn't move up. It just bounces back horizontally.
 
@JohnRennie okay sir...not even a bit? here it has an upward component so can it go up by dx
 
Right, but we've agreed that for any $dx > 0$ as soon as it has moved by $dx$ the force becomes horizontal.
So it can move upwards by $dx$ but only when $dx \to 0$.
i.e. it can only move upwards by zero.
 
@JohnRennie yeah sir...but if there weren't g...then would it rise sir?
@JohnRennie okay sir
 
@user8718165 No
Suppose you now increase the wall radius to make it slightly bigger than the ball radius, but only slightly bigger.
 
@JohnRennie okay sir
 
8:29 AM
Now the ball can move upwards a non-zero distance $\Delta x$ before it loses contact with the curved part of the wall. So now it can move upwards, but only a bit.
 
@JohnRennie got it sir...
 
In zero-g it would get a small vertical component of velocity, so it would bounce back mostly horizontally but at a small angle to the horizontal.
 
@JohnRennie sir then the ball will move away from the floor....right?
 
Yes
 
@JohnRennie hello sir...I have a last qn
 
8:42 AM
@user8718165 yes?
 
8:53 AM
@JohnRennie sir could you please repeat just once more if the curvature matches...why can't the ball move after collision? I got the dx argument but still confused. Please help sir
@JohnRennie hello sir
@JohnRennie are you working now? :)
 
Discussing Python ...
 
@JohnRennie ah..okay sir...could you please ping me when done sir?
 
9:13 AM
Actually, now I think about it my argument isn't valid. An ideal collision takes no time and the force is infinite, so my argument that the limit of $F \cdot dx = 0$ is not valid.
 
@JohnRennie sir the ball was traveling towards left and the force applied is F...please neglect gravity. Please have a look here Incline is 45 deg.
 
I'll have to go away and think about. But that will have to wait.
 
9:58 AM
@user8718165 I'm not annoyed, it's just that the question is harder than I thought and I will need some time to think about it. But right now I have other stuff to do.
 
 
6 hours later…
3:44 PM
@JohnRennie hello sir :-)
 
@user8718165 hi :-)
 
4:04 PM
@JohnRennie, Hi sir :-) ;May I ask a doubt from a question from Doppler Effect (Sound Waves)?
 
@M.GuruVishnu hi, yes, sorry I missed your ping earlier.
 
@JohnRennie No problem sir. Actually I figured that out after notifying you. That's why I removed. Sorry for disturbing then.
Here afterwards please don't ask "sorry" to me :-)
Question: A source emitting a sound of frequency $f$ is placed at a large distance from an observer. The source starts moving towards the observer with a uniform acceleration $a$. Find the frequency heard by the observer corresponding to the wave emitted just after the source starts. The speed of sound in the medium is $v$. (contd.)
Doubt: In a video lecture, I saw that the same formula for Doppler Shift in frequency could be used when observer, source or both are accelerating. But velocity of source is taken at the instant it emits a particular wave front, and velocity of the observer is taken when that particular wave front reaches him. (contd.)
So, in the above question, the velocity of source at the time of emitting is $0$. And the observer is not moving. So I concluded, the apparent frequency is $f$ which is the same as that of the source. But the answer is incorrect. For your reference in my book it is given as $\frac{2vf^2}{2vf-a}$. Could you please explain or give a hint for this anomaly sir?
 
I'm not sure I understand. The frequency is a function only of velocity, not of acceleration.
 
@JohnRennie Yes sir. So can we say the book is incorrect?
 
Maybe there is some aspect to the question that we have missed.
 
4:16 PM
@JohnRennie I think the observer must also be accelerating. Only then the frequency depends on acceleration.
 
@M.GuruVishnu yes, I agree.
 
@JohnRennie Thank you sir. Let me try the problem with this new assumption and ask if I've any doubts.
 
But then I would expect the frequency heard by the observer to depend on the distance between the source and the observer, because the travel time of the sound will determine how fast the observer is moving.
 
@JohnRennie Oh yes sir. And we are not supplied with distance. May be "a large distance" hints us to do a little bit of approximation. But I am not sure of this because, the doppler shift doesn't depend on the separation between source and the observer.
 
@M.GuruVishnu to be honest I'm disinclined to spend time on a question that isn't clearly stated as it could well just be time wasted.
 
4:23 PM
@JohnRennie Ok sir. Thank you :-)
@JohnRennie, Sir, could you please tell whether this statement "In a video lecture, I saw that the same formula for Doppler Shift in frequency could be used when observer, source or both are accelerating. But velocity of source is taken at the instant it emits a particular wave front, and velocity of the observer is taken when that particular wave front reaches him." from the "Doubt" message is applicable for all cases? Or are there any limits to this?
 
@M.GuruVishnu that seems correct to me. Once the wave has been emitted its frequency (in the rest frame of the air) is fixed and cannot change. So it depends on the velocity of the source at the moment the wave was emitted.
 
Ok sir. Thank you for the clarification.
Bye sir :-)
 
4:52 PM
@M.GuruVishnu Bye :-)
 

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