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04:29
@JohnRennie Hi :-)
@Mayank hi :-)
yesterday, by John Rennie
n  angle
1  2θ
3  4θ
5  6θ
7  8θ
9  10θ
etc
@JohnRennie If we take Theta=90.
Then there will be 2 reflection
And 1 image
I'm a bit busy this morning. I'll ping you when I'm free.
Ok
:-)
05:26
@JohnRennie Hello sir :)
@Jasmine hi :-)
Are you available for a question now
If it's not too long ...
Okay
I need the energy expression of this system
Initially the balls are both displaced by the angle θ and are motionless. So the total energy is the PE stored in the springs.
"Small displacement" ?
Isn't (b) asking for the total energy given that the two balls start motionless at an angle θ = π/6?
05:36
@JohnRennie yes
And with the balls motionless the total energy is equal to the PE.
Sir when you are done can you kindly see my question?
@JohnRennie Okay
@ManasDogra you can't use conservation of energy as extending the string involves work
Then how does equating the initial and final potential energies give the correct result?
What is the correct method?
05:42
I'm a bit busy this morning. I'll have to get back to you when I have time to go through this in detail.
Ok sir...thanks in advance
 
1 hour later…
07:13
@JohnRennie Are you free now?
@Mayank hi, yes, we can talk now.
3 hours ago, by Mayank
yesterday, by John Rennie
n  angle
1  2θ
3  4θ
5  6θ
7  8θ
9  10θ
etc
When Ø=90°
Then there will be 2 reflection
And no of image will be 1
When I was attempting the derivation of the equation I started with the object on one of the mirrors. I did this because it made the working simple, and in most cases we get the correct result.
The problem with this approach is that if Ø=90° then the first image is also on the mirror, and that means the second reflection doesn't create a new image. That's why in this case we only get the object and one image. If you move the object off the mirror then it does work i.e. you get three images in addition to the object.
I guess there must be other angles where this happens i.e. an image ends up exactly on the mirror so the next reflection leaves it in the same place. It would be interesting to have a play and see if I can find any other examples of this.
So the equation n = 360/Ø - 1 is occasionally wrong in special cases where an image ends up exactly on a mirror. It is generally correct though.
07:20
Hmm
When the 360/ø is odd and the object is not placed at symmetry(angle bisector) then n=360/Ø
@JohnRennie But the formula gives no of images =3
I'm losing track of this now. I think I need to find some mirrors so I can try it for myself.
I've drawn two mirrors at 120° so 360/Φ = 3. But I'm not sure how to get this to show the number of images ...
07:36
Point source+ extended rays
@JohnRennie
Did you select extended rays view?
Red one are images?
Hmm, so you can get either 2 or 3 images by moving the source around.
If I turn on "Seen by observer" I only ever get two images ...
@JohnRennie Why's that?
To be honest I'm getting thoroughly confused now.
I would need to find some mirrors and try it for myself.
07:49
:-/
I don't think formula is consistent.
It is just a short cut
Which gives approximation
 
2 hours later…
09:43
Hi everyone! For the following question:
I found the resistance of the voltmeter to be 50000 ohms and that of ammeter to be 0.02 ohms. So option A is incorrect and B is correct.
How do we show that option 4 is also correct?
09:58
If anyone gets any idea please ping me. Thanks!

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