@Scáthach the sound is reflecting off the cardboard and travelling back towards the source, and this creates interference. So the changes in intensity are due to constructive and destructive interference as the cardboard is moved.
I'm not sure I'd describe it as a standing wave because it only has one boundary (at the cardboard). It's just down to the path difference between the incident (red) and reflected (blue) ray.
@Scáthach Yes. If the phases of the incident and reflected waves are the same at the point D then they reinforce and you get constructive interference. If the phases are in antiphase (a phase difference of $\pi$) at D they cancel each other out and you get destructive interference.
With the detector at the initial position the path lengths from S1 and S2 are the same so the sound waves are in phase. The question asks how far the detector has to move along the line for the path difference to become half a wavelength.
Initially $p=mv {cos(45) i+sin(45) j}$ and finally $p=mv{- cos(45) i+sin(45) j}$. So, $P_initial-p_final=2mv cos(45) i $ where i,j are unit vectors in positive x and y directions
If you throw it at an angle of 45 then only it can rebound at an angle. I mean in that case the normal to the wall at the point of impact bisects the angle between initial and final velocity vector
Its not mentioned what L/2 and d mean but assuming L is length of the board and d is the distance moved its pretty clear that if you move a distance of L/2 (relative to the board) to the center the board also moves backward by a distance of d (relative to the ground)... so, relative to the ground your net displacement is L/2-d . I mean $x_{y,g}=x_{y,b}+x_{b,g}$ where x is displacement and y means you, b means board and g means ground.
(Note that this is a vector equation.. so be careful about signs)
Yes. If the polariser is horizontal it allows only light from the source polarised in the horizontal plane (I forget what fraction of the total that is). Then all the light allowed through the polariser passes through the analyser.
The source consists of all polarisations. The polariser allows through only the subset of the light from the source that is polarised in the horizontal plane.
The dark red arrow just shows the direction of travel of the light ray. The direction of polarisation is always normal to the direction of travel.
If you think of the light wave as a lateral wave then the displacement is always normal to the direction of travel (because that's what a lateral wave is). The direction of polarisation is just the direction of the displacement.
The unpolarised light is a superposition of many individual waves with their directions of polarisations in random directions. The polariser allows through just the waves with their polarisations in the direction that matches the polariser.
(it's a little more complicated than that, but that's the basic idea)
@Fawad It tells you the modulation frequency, 250kHz, is 10% of the carrier so the carrier is 2500kHz. That means the station occupies the range from 2500-250kHz to 2500+250kHz. OK so far?
In real life you would allocate the same bandwidth (250kHz) to stations regardless of their base frequency. Anyway, for the purposes of this question is doesn't matter.
Whether you allocate 250kHz or 290kHz if you place the new station at 2900kHz its range is going to overlap the existing station.
There's a really simple way to understand this using a diagram. The diagram itself is very simple, but explaining how and why the diagram works will take a little while.
I can try and explain if you would like, or I can just show you the diagram.
We can represent a wave as a vector of fixed length rotating anticlockwise. So S1, S2 and S3 are shown as three vectors. We are told the phase difference between them is 120° and that means on our diagram they are spaced 120° apart.
The total wave is given by adding up the three vectors, and obviously when you add the three vectors they cancel each other out and you get zero.
Suppose you have a vector of length $A$ rotating anticlockwise at an angular frequency $\omega$. Then at a time $t$ that vector makes an angle $\omega t$ with the horizontal.
So the vertical component (the $y$ component) of the vector is $A\sin\omega t$, and that is the equation for a wave. OK so far?
If we add our two waves we get $A_1\sin(\omega t) + A_2\sin(\omega t + \phi)$, and this is the same as the $y$ component of the vector we get from adding our two vectors.
So we can add our two waves by first adding the vectors, then taking the $y$ component of the result.
Likewise if we have three waves we can calculate the sum by adding the three vectors representing those waves and then taking the $y$ component of the result.
@JohnRennie If we add our two waves we get $A_1\sin(\omega t) + A_2\sin(\omega t + \phi)$, and this is the same as the $y$ component of the vector we get from adding our two vectors.
Sometimes this is useful and sometimes not - it depends on the question. In this case it is very useful because the sum of our three vectors is zero, and the $y$ component of a zero length vector is zero.
@Scáthach In this question adding the y components gives $A\sin(\omega t) + A\sin(\omega t + 120°) + A\sin(\omega t + 240°)$. Is it obvious to you what the result of this sum is going to be? (because it isn't obvious to me :-)
The phases at $P$ will depend on the relative phases of the three sources and the distances between them, but in this case we don't care as we are told what the phases at $P$ are.
Which of the following combination of statements is true regarding the interpretation of the atomic orbitals ? (a) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum. (b) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number. (c) According to wave mechanics, the ground state angular momentum is equal to $\frac{h}{2\pi}$ (d) The plot of $\psi$ Vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value.
(a) is kind of true. Only the s orbitals have non-zero values at the nucleus so the higher angular momentum p, d, etc do stay away from the nucleus more than the s orbital. But is it correct to say d orbitals have lower density near the nucleus than p orbitals. I'm not sure.
I'm not sure what (b) means, though I'd guess (b) and (d) are basically the same question.
For (c), it depends on the atom. For a hydrogen atom the ground state has zero angular momentum not $\hbar$.
@Abcd in that case it can't be answered as the atomic orbitals differ from atom to atom. There is no analytic expression for the orbitals except for hydrogenic atoms.
So the distance $x_v$ is just $v$ (7 in this case) times the spacing between divisions on the vernier scale. And that spacing is $0.9 MSD$ (0.9 mm in this case).
On the main scale we started at $MSR$ and moved on division $v$ times, so $x_m = MSR + v MSD$.
And $x_r = v VSD = v (0.9 MSD)$
So $D = x_m - x_r = MSR + v MSD - v (0.9 MSD) = MSR + v (0.1 MSD) $
In the algorithm I described above we used the variable $v$ that is an integer. Since $v$ is an integer the smallest difference we can measure is between $v$ and $v+1$.
If you increase $v$ by one you increase $D$ by $MSD - VSD$
That's why $MSD-VSD$ is the smallest distance we can distinguish.
@JohnRennie 0 error for above caliper means "This much it will always show extra, so subtract it from the value you get while using the caliper" . Am I right?
That first sentence should be: Imagine the main scale continues to the left of zero, so the first division to the left of zero is -0.1, then -0.2 then etc etc
So moving from 1 to 0 on the vernier scale cannot move you any farther left of 0 on the main scale than -0.9MSD i.e. the position of the vernier zero on the main scale has to be greater than or equal to -0.9 MSD on the main scale.
@JohnRennie imgur.com/a/zZ6p8BJ What's the reason intensity of $K_{\beta}$ rays is less? I had thought higher shells will have more electrons, so they'll have higher probability to jump to lower shell. But, they'll have to travel long way down. Does the second factor dominate which results in this? ^
There isn't a simple explanation. The transition probability depends on how much the electric field of a photon mixes the two orbitals involved. You end up having to calculate an integral that is vaguely like:
$$ \int \psi_{final}^* x \psi_{initial} $$
It turns out that this is greater for the 2p to 1s orbitals than for the 3p to 1s.
I suppose if you want a arm waving argument you could say that the 3p and 1s overlap less than the 2p and 1s because the 3p is on average farther away from the nucleus than the 2s, and the 1s orbital is concentrated round the nucleus.
@Abcd and the match on the vernier scale is at 7. So the distance from zero to the match on the vernier scale is 7 times the vernier spacing i.e. 7.7 mm.
