@mechanist I think it must mean that the glass plate is after S₁ i.e. in the liquid.
The glass plate needs to have a refractive index nₚ that cancels out the extra path length d sinφ at S₂ so we need: p (nₚ - n) = d sinφ where n = n₀ + kt
i.e. the refractive index of the plate has to be greater than the refractive index of the liquid by d sinφ/p
@sanya I drew a diagram to show the sphere after one rotation. We are looking down the z axis so k is out of the page towards us.
I am assuming angles are measured anti-clockwise from the x axis as usual.
The question doesn't say what angle P starts at. It says the latitude is 37°, so on our diagram the distance from the axis is 0.8R, but it doesn't say what longitude P is at.
If P started at an angle of zero the centripetal acceleration is in the negative i direction and the tangential acceleration in the positive j direction.
I was solving this question and the acceleration comes out to be a constant. But from my rudimentary understanding of Lenz's Law the wire should stop accelerating after a while and just come to a halt to resist any change in flux. And thus, acceleration should be a variable which tends to zero after some amount of time. This is what I'm thinking intuitively but it's not matching the result. Where am I going wrong?
@JohnRennie sir please take a look at this doubt whenever you are free
@JohnRennie, I thought about your answer and I think I got it. When both sides of the slits have air, the placement of the slab does not matter, but when one has $\mu_{A}$ and the other has $\mu_B$, then $\delta x=(\mu_{slab}-\mu_{A/B})t$, depending on which side it's placed?
Always thought it was just $(\mu-1)t$ no matter the conditions...
@JohnRennie I think my image did not get uploaded, ill try again but basically in the image provided w the question, they have taken the z and x plane, and the y axis is out of the page..
@mechanist For SHM the acceleration would have to be of the form a=-(omega)^2(x). But here the acceleration is constant. Can you please send that question though if you have it handy? :)
@KavinIshwaran ε - q/c =0 is pretty much energy conservation. I don't get what you mean. Can you please post the solution?
@Swan idk I don't think energy is conserved while charging a capacitor. Even in the simplest case of connecting it to a battery, only half the energy makes it, the rest is lost to resistance. So energy conservation cannot be assumed, and the behavior of the system must be defined by assuming a resistance $r$ then finding out what happens when $r\to 0$.
A capacitor is only reversibly charged when the potential difference between the capacitor and the battery is raised incrementally(i.e. a variable potential battery) and the system is given infinite time to equilibriate. Which I don't think will happen here, there's no way this process is reversible.
@mechanist No, ofc there would be energy loss in the form of electromagnetic pulse when the capacitor will get charged. I just meant that the KVL equation (ε - q/c =0 ) is the energy conservation equation in the sense that KVL is based on energy conservation, that's all.
