@JohnRennie Got a question about this problem.

It keeps saying i forgot the rotation of sphere? I just used this equation :

i.imgur.com/D2vOI3Y.png . Just did Mh / ( Mh + Ms+ Md/2 )

Anything wrong here?

hello😊 @JohnRennie

@JohnRennie Good morning!

@user8718165 @Dante morning :-)

@user8718165 Are you around?

@Dante yes dante:-)

Okie, I'll wait.

@Dante Its okay...I just wished him...currently I've don't have any problems...:-))

@Dante you discuss your doubts:-)

Oh, ok.

@Dante 😊

@JohnRennie brb 3 minsm

@Dante is this prob from hcv?

Idk, I found it in a JEE forum.

its about refraction I think

@JohnRennie What does the depth of pool actually represent? I don't think it's the apparent depth since since it'll have to increase as we move farther

@user8718165 Yeah it is.

is answer (d)?

No, it's a

I assume it means this. The bottom of the pool appears shifted away from the man due to refraction, and the pool appears shallower.

@JohnRennie oh...I was wrong...the refraction effect shouldn't be as prominent as in (d)...there will be comparitively less refraction in (a)...hence a is correct...

can we think like this? @JohnRennie

@JohnRennie Whenever your done with @user8718165 would appreciate if u looked at my problem

@user8718165 the only safe way to approach problems like this is to do the calculation. General arguments can be very risky. Having said that, the calculation looks hard ...

Isn't the length what I've marked we call apparent depth?

In that case, apparent depth should always be > real depth right?

No, wait,

The path length through the water is obviously greater than the depth of the water because the light ray is at an angle. However humans are perfectly capable of realising that they are looking at an angle and adjusting for this.

@JohnRennie Is there some trick to know the correct answer in such probs without going thro' the lengthy calculation?

@user8718165 no

@JohnRennie :-(

Yes, so, this is the apparent depth actually right?

@Dante The path length through the water appears to be reduced by a factor of $n$ and the refraction rotates it upwards.

okay,

In principle calculating how the bottom and far side of the pool are transformed is just geometry, but it looks a somewhat tedious calculation.

The answer is obviously either A or D, but I don't think it's obvious which of those two it is.

I think I got something,

I'll think for a while and come back, not able to work on the geometry, if anyone else is in the queue, you can ask.

@amanuel2 hi, you there?

@JohnRennie I got it :-)

@Dante cool :-)

Found a function for apparent depth in terms of angle of incidence and then differentiated it.

Was bit of a trigonometry though, that was the lengthiest part.

differentiated it twice*

@Dante If you were faced with that question in the exam I guess you'd have to find a short cut. You wouldn't have time for the full analysis.

Actually.

How can we decide the concavity of the graph intuitively?

I would probably consider the behaviour near vertical i.e. near the man. The graphs A and B have different behaviour for the end near the man and it should be straightforward to do a calculation for incident angles near zero.

@JohnRennie a doubt...

@user8718165 yes

Yeah, but our confusion is between A and D right?

I'll brb.

@Dante sorry, typo, I meant A and D not A and B

@JohnRennie after passing thro' a lens all the light rays (from a point on the object) have to meet (after passing through the lens) in order to create a sharp image of the point...right?

@user8718165 I'm busy for five minutes. Back soon.

@JohnRennie Ok sir

Help me with this after you are done with others @JohnRennie

@user8718165 yes. If you consider all the light rays from a single point on the object those rays all have to converge on a single point in the image.

@tatan consider first the collision between C and A. If this is an ideal elastic collision C stops and transfers all its momentum to A. So immediately after the collision A is moving with speed $v$.

The collision is just a way of setting up the initial conditions for the two masses on the spring i.e. B stationary and A moving at speed $v$.

yes

then?

@tatan you'll encounter lots of these two masses on a spring problems.

@JohnRennie so if we keep our eye there imgur.com/a/5tzaHb8 the image of that point won't be sharp since there will be 3 low brightness images of the point in our eye and we won't see that clearly...blurred...is this correct sir?

It's not too hard to calculate the equation of motion for the whole system, but you don't need to do that for this problem. Conservation of energy and momentum are enough.

For example just after the impact A has momentum $mv$ and B is stationary so the total momentum is $mv$. Since there are no external forces acting the total momentum has to be constant at $mv$.

ok

@tatan Consider (A): for the total KE to be zero both A and B must be stationary. Yes?

yes

But that would make the total momentum zero, which would violate conservation of momentum. So (A) must be false.

yes

Now to check (B). Suppose we write the position of B as $x_b$ and the position of A as $x_a$, then the length of the spring is the difference between the two positions: $L = x_b - x_a$. Yes?

yes

Now differentiate wrt time to get:

$$ \frac{dL}{dt} = \frac{dx_b}{dt} - \frac{dx_a}{dt} = v_b - v_a $$

@tatan OK so far?

yes

At the maximum compression of the spring $dL/dt = 0$, so that means at the maximum compression $v_b = v_a$. yes?

yes

And the total momentum has to be constant at $mv$, so we can work out the velocity at the point of maximum compression. Then from that calculate the KE.

yes.. I can do now

thank you

Cool :-)

Remember these basic ideas because you'll be seeing lots of this kind of problem.

ok

@user8718165 when you introduce an eye things get complicated because the eye has a lens and a screen (the retina) in it. If you put a screen in at the point where you've drawn the eye than the image on the screen will indeed be blurred.

@JohnRennie ok a screen...

@user8718165 then yes the image on the screen will be blurred because your original point is now spread out on the screen.

@JohnRennie but is it the correct way to think like that?

@JohnRennie Okay sir thank you!

Yes, that's a perfectly reasonable way to think about it

@JohnRennie Sir I'm having a lot of trouble visualizing diffraction.Tht's imp for JEE. I also don't know why diffraction happens in the first place...what is its reason?Could you please help?

@JohnRennie Hello

I wouldn't worry about trying to visualise diffraction.

For JEE you just need to be able to do the calculations and they are generally straightforward.

@JohnRennie yes but is there no reason why does diffraction even happen? Do we just have to accept it?

@user8718165 on phone ...

@user8718165 suppose you have a wave. The general expression for a wave is $y(t,x) = A \sin(\omega t - kx)$. OK so far?

@JohnRennie

@pi-π hi

@JohnRennie, If the alpha decay of Uranium-238 is energetically allowed what prevents Uranium-238 from decaying all at one and why is its half life so long?

@pi-π there is a kinetic barrier to the decay.

@JohnRennie, Could you please elaborate?

For the nucleus to emit an alpha particle requires it to pass through an intermediate state that has a higher energy than the initial or final states.

The barrier exists because the nucleus has to change shape during the process of ejecting the alpha particle and that shape change raises the energy of the nucleus.

@JohnRennie, The nucleus being required to pass through the higher energy acts a barrier to the decay. Yes?

@pi-π yes. In order to decay the system has to quantum tunnel through the barrier and that has a very low probability. That's why it takes so long to happen.

@JohnRennie, Got it.

@JohnRennie, Why is electron caputure common in case of heavy nucleus?

@pi-π is electron capture especially common in heavy nuclei?

@JohnRennie, Don't know actually. My book has got a question like that...

Electron capture tends to happen when the nuclei has a relatively low neutron to proton ratio, because electron capture converts a proton to a neutron.

I guess it competes with beta plus decay because that also converts a proton to a neutron. It's just a question of which decay mechanism is more probable.

It isn't obvious to me which would be faster when comparing light and heavy nuclei.

@JohnRennie Hi

Are degrees of freedom of tetraatomic gas with all modes unlocked 18 ?

Hii @JohnRennie... Sorry i left

Are your around??

@JohnRennie My bad had to sleep

@pi-π , It is energy difference between mother and daughter nuclie that matters and decides whether a nuclie would decay via Beta plus decay or electron capture. If the energy difference is high then beta plus decay would take else electron capture.