light consist of photon, how this photon produce? is that due to oscillation of electric or magnetic produces photon? if light shows destructive interference with the other light, is the photon destroyed?
For example when you see fringes in the YDSE it's because energy has moved from the dark areas to the light areas. If you average out the light intensity you get the same average energy as in the interference hadn't taken place.
How this relates to photons is a bit complicated. Photons are not little balls of light. A photon is quite a complicated object that is usually spread out over a region of space i.e. it is delocalised.
Quantum objects like photons and electrons don't have a position is the same way we think of macroscopic objects like balls having a position.
Instead they are more like fuzzy clouds i.e. spread out over a region of space.
The interference affects the way the photon is spread out. If you think of the photon as a cloud then the interference makes the cloud thicker in some places than in others, but it's still the same delocalised photon.
I asked my teacher about that he said light is an EM wave is not totally adopted, till we get result from. This we can assume, and he also said that, s why we have have dual character of light @JohnRennie
Light is a quantum field, and this can sometimes behave like a wave and sometimes behave like a particle depending on how it is interacting.
@yuvrajsingh I wouldn't say Maxwell was wrong. His equations describe light well in the classical limit. Maxwell didn't know light was quantised because he died before quantum mechanics was discovered.
Reading about photons I hear different explanations like "elementary particle", "probability cloud", "energy quanta" and so forth. Since probably no one has ever seen a photon (if "seen" it supposedly - and rather conveniently - ceases to exist) but many experiments seem to verify its properties ...
The binding energy in the nucleus is typically far higher than visible light energies. Transitions in nuclei typically emit gamma rays not visible light.
Visible light is about the same energy as transitions of the electrons in atoms.
If $x$ and $y$ are equal they have to have the same units. For example you can't say $x$ metres = $y$ seconds. They would have to be $x$ metres = $y$ metres or $x$ seconds = $y$ seconds.
Rather than choose a specific unit of distance like a metre or a mile, we use $[L]$ (for Length) to indicate a unit of distance. So on the left side the dimensions are $[L]$.
So let's look at the right hand side and check that it also has units of distance.
Note this also applies whenever we add values. If we have $x + y$ then $x$ and $y$ must have the same dimensions because you can't add metres to seconds, or seconds to kilograms.
Often we have to do some complicated derivation or integral where it's easy to make a mistake. You can use dimensional analysis on your final result to check for errors.
Any dimensional inconsistencies mean we must have made a mistake somewhere.
If $\ell$ is the length of the pipe then for the fundamental $\lambda = \ell/2 = v/f$
So $f = 2v/\ell$
And we are told that the velocity $v$ is accurately known so the only error is in $\ell$. Note that the radius $r$ doesn't appear in the equation. Giving you the error in the radius is a red herring.
You can do this by differentiating the equation, or you can just shortcut by noting that the percentage error in $f$ is the same as the percentage error in $\ell$.
@JohnRennie So I designed this circuit to switch on the Right circuit if I switch on the Left circuit , I want to know what will be the voltage at the 4th pin of optocoupler ?
The way an optocoupler works is that the light from the diode excites electrons in the transistor and causes a current to flow even though the base current is zero so the transistor should be off.
When you have a current $I_{in}$ flowing in the diode part of the coupler it produces a light intensity proportional to $I_{in}$, and this causes a leakage current in the transistor proportional to the light intensity. So the current $I_{out}$ though the transistor is proportional to $I_{in}$.
@JohnRennie sir suppose a block is moving with some velocity $v$ and a ball hits it with velocity $v'$, $(v'>v)$ The wall is somehow controlled such that its velocity can't be changed at all..what will happen sir..?
@JohnRennie I think the ball will hit the wall with $v'-v$ velocity...
If the wall velocity is the same before and after the collision just work in the rest frame of the wall. In that frame the velocity of the ball just changes sign in the collision.
