@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
@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...
@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 ...
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.
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 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 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 $v$.
@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 $mv$.
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?
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.
@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 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, 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?
OK, so $\ce{Fe}$ is the most 'stable element'. As such, why do all elements above it not decay into $\ce{Fe}$? In all cases, would it not lead to an increase in binding energy and therefore energy been released, meaning it is energetically feasible, and should happen spontaneously (given enough t...
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.
@pi-ฯ 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.
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.
@pi-ฯ , 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.