Recent content by Null_Void

I think there must be one point which is at rest where no friction acts. It is at this point that the directions of the forces change. Using similarity of triangles and equating the velocities at those points, the vertical distance of the point from the tip of Cone-A comes out to be: ##Y =...

I made a mistake. The velocities at M and N can never be zero since it would imply that ##ω_{A} = ω_{B} = 0## which is not true. Thus there must be constant slipping. Do you mean that ##τ_{net} ≠ 0##? implying Angular Momentum is not conserved. But since there are no external torques shouldn't...

I know that friction will act on both the cones until there is no slipping throughout the length in contact. Velocity at points M and N should be Zero in the end since ##V = ωR## and ##R = 0## Total Angular Momentum of the system must be conserved since ##τ_{net} = 0## But I'm not sure how...

I understand what you mean, I think it's similar to the basketball and tennis ball example. But then now that I think about it, since we already keep the wall at rest why can't we simply assume that the magnitude before and after will be the same. Or is it wrong to analyse impulses in different...

I know this is where I'm going wrong, but why wouldn't the horizontal speed be same?

Let the velocity of the wall be ##u## Since there is no impulse in the vertical direction, the vertical motion of the ball is unaffected by the collision. I proceeded by working in the frame of the moving wall, therefore the ball now Initially has : ##V_x = 8 + u ## ##V_y = 6## For the ball...

I'm very confused about the role of the plane mirror, so both, the part cut off by the plane mirror and the virtual image formed of the remaining part of the original real image act as a virtual object for the second lens?

But going by your logic, the object height will ##4mm## and the corresponding magnification at lens-2 will be ##-1/3##. Therefore the final image height would be ##4/3mm##, but the answer is given as ##8/3mm##. Here's the solution:

It should be ##60cm## plus the distance to the imaginary intersection point of the reflected rays, but I'm not sure about how to find the required distance. Any hints?

I think I get it now. I believe my confusion was due to the assumption that virtual objects act like real objects. Nevertheless, your posts cleared it up. Thank you @haruspex and @Steve4Physics for your time and patience!

I think they mean the height of the image formed by lens-2. We are expected to find the magnification from the horizontal distance and then apply it to find the height of the final image

4mm of the object will be formed as a real image at 60cm, while the 2mm part will act as a virtual object, something like this: So yes, the rays will definitely diverge after reflection from the plane mirror. I initially proceeded by considering the real image ##A_1C_1## and virtual object...

@Steve4Physics First of all, thank you for taking the time to go through the video. I'm sorry that was extremely misleading, the video said that the rays appear to converge. I understand the need for a virtual object, but I still don't understand the image formation by a virtual object. Do we...

So I was able to find out quite easily the image distance after the first refraction and the corresponding magnification. Employing Cartesian convention ##1/v - 1/u = 1/f## Substituting the values gives us: ##v_1 = +60cm ## ## m_1 = -3 ## Thus the height of the image below the principal axis...

@Steve4Physics thanks for the resources, they were very helpful. But if you don't mind, I have certain questions from the video. In the end we have a double lens system and we are to find where the final image will be formed and naturally we use the concept of a virtual object to do so. My...