Time period of compound pendulum is 2π √[(k^2 + l^2/ l)/g] where k is radius of gyration and l is distance b/w COM and suspension point. My book says the minimum value is when k=l. Why is this so?
You change your picture on your main site (in your case physics.stackexchange.com). Optionally you can then go to your network profile and Go to Update Profile Info > Physics.SE. Then it will take less than half an hour for your icon to be updated in chat as well.
Don't expect it to change right away, chat is old and slow :P
As far as I am aware, you cannot change it in chat if you don't change it on the main site
@Abcd There are many more light rays other than the two shown. Light spreads out in all directions after passing through the hole (diffraction). The diagram is looking at what happens at any particular point P, where rays from the two slits interfere. These are the only 2 rays which are relevant, all others are ignored. They do not deliberately converge on P they are the only rays of all emerging from the slits which reach P.
@harambe Bullet Hits Mass on a Spring : Yes if the bullet is embedded then there is a new equilibrium position because the mass has changed. The nature of the collision tells you how much energy is added to the system.
@Abcd That is a difficult question to answer. Could John Rennie handle this one?
Huygens' principle is one of those ideas that are really fundamental but also very hard to get an intuitive grasp of. To some extent the simplest course is to just accept it and move on.
When a plane wave passes through a slit that is narrow compared to the wavelength the light on the other side emerges from the slit in all directions i.e. on this diagram the wavefronts would be semicircles.
I wonder if I can find a diagram of this ...
That's just some random diagram I found using Google.
@JohnRennie Leave that. I think I need better understanding of Young's interfernece experiment first to be able to solve questions. Could you explain that?
Should I send the diagrams from my textbook for the Young's experiment? They are nice diagrams
The two slits both cause the light passing through them to spread out in all directions like the diagram I posted above. So at every point $y$ on the screen there will be rays from $S_1$ and $S_2$ that hit the screen at that point.
If the slit is narrow compared to the wavelength then yes the slit behaves like a point source so the light from it spreads out evenly in all directions i.e. as semicircles.
Since the light rays from the slits spread out in all directions evenly there are always rays from the slits to the point $y$ wherever you choose the point on the screen.
The waves are sinusoids traveling left to right (towards the slits). The red lines mark the lines of constant phase, and those are the red lines on your diagram.
Now suppose we have two slits, and we consider some point:
The light rays from the different slits have travelled different distances to reach the point where they meet (the point P on the screen in your diagram).
So or the two light rays we have:
$$ E_1 = E_0 \sin(-kx_1) $$
$$ E_2 = E_0 \sin(-kx_2) $$
And the total electric field at $P$ is $E_1 + E_2$.
Suppose P is the centre point where $x_1 = x_2$, then when we add the fields we just get $2E_0$ i.e. $E_0\sin(-kx) + E_0\sin(-kx)$ where $x_1 = x_2 = x$.
And I said that the effect of the glass plate was to change the optical length of the light ray, where I described the optical length as the number of wavelengths?
In the right hand diagram we want the slab to increase the optical path of $r_2$ by $\lambda$. That way we'll get $r_2 - r_1 = \lambda$ and we get the $1\lambda$ fringe.
@harambe **SHM is scary ... ** : All such problems are solvable with your level of understanding, if you approach them methodically.
(1) Calculate KE of block after collision. (2) Use conservation of energy to find highest point reached Q. (3) Calculate new equilibrium position O which depends on new mass M. (4) Amplitude of oscillation is distance OQ=A.
(5) New angular frequency is $\omega'=\sqrt{k/M}$. (6) New equation of motion is $y=A\sin(\omega' t+\phi)$ measured from the new equilibrium point O. $\phi$ can be found from initial point P where $t=0$ : known distance $OP=-A\sin\phi$.
@Abcd the question is a bit artificial. It assumes you know which fringe is which.
I suppose you could start with a very thin slab so it only moved the fringes very slightly. Then gradually increase the thickness of the slab. That way you could watch the $1\lambda$ fringe move and keep track of it.
@Abcd the optical path length is basically the same as the phase
When you add two waves, how they add depends on the difference in the phase. If they are in phase they add while if they are $\pi$ out of phase they subtract leaving zero.
@harambe Just noticed that the options for Q15 (which follows on from Q14) make use of only one of the options in Q14. Therefore you can get the correct answer to Q14 without any calculations at all!
The optical path length is measured in wavelengths, and if you have different media with different refractive indexes then the wavelength of the light wll be different in the different media.
$nL/\lambda$ is just the number of wavelengths in a length $L$ of the medium with refractive index $n$.
@JohnRennie I clearly don't remember the problem... But it was like a bird collides with a verticaly pivoted stick with a velocity and remains attached. Then the stick rotates... There used to be couple of expressions summing up to give a total moment of inertia. Could you elaborate it?
If the bird can be treated as a point mass the the moment of inertia of the bird is just $mr^2$, where $m$ is the mass of the bird and $r$ is the distance from the centre of rotation.
@JohnRennie Oww... I thought we would advance like finding the moment of inertia for bird and rod system for the too end and add it with the inertia of rest of rod...
@Abcd If the two light rays have the same frequency then the phase difference between them is only a function of their relative optical path lengths i.e. the phase difference doesn't change with time.
And at optical frequencies that happens so fast the eye can't see it. So you don't see an interference patern at all. Just a smooth mixture of red and blue light.
So at the point $P$ the two phases are $\omega t - kx_1$ and $\omega t - kx_2$
But if we stay at point $P$ the distances $x_1$ and $x_2$ are constants, because they are the distances from the slits to the point P. So the two phases are both just:
