Why Does Light Reflect? (Waves - Physics)

[@Kralasaurusx]

This is so much more satisfying than the introductory physics explanation that just says "light refracts because... uhh, it travels from one material density to another" which really doesn't explain much. The vibrating-electrons-with-destructive-interference explanation makes a lot more sense.

[@PhysicsMadeEasy]

Thank you Kralasaurux, Yes, the change of direction of propagation of rays during refraction, comes from the combination of the destructive interferences between incident light and reemitted light within the material, and the phase shift of the reemitted light due to the delay of that process. Dig a little, and you can then relate this to the speeds of light in the materials (like most high school text books do while only providing geometrical explanations).

[@harleyquinn8202]

Except that electrons are not particles but waves too and they do not vibrate

[@douglasclerk2764]

If this is really what they taught in your introdutory physics course you should demnd your money back.

[@kazukawasaki97]

​@@harleyquinn8202electrons are subject to collapse in superposition when light hits it, don't know how everything works heard it somewhere makes sense though,cathode rays are electrons acting like accelerated beams

[@roshinis804]

I love your explanation. I tried searching many videos but they didn't explain it this way. High school textbooks don't teach this and that's why I'm here. And you are replying to everyone's comment. That's amazing

[@PhysicsMadeEasy]

Hi Roshinis, Thank you for your encouraging words. Yes, that's what I am trying to do with this channel: make students understand the true basics, which are often overlooked by school teachers. Because, I hold the firm belief that with a solid grip on foundational concepts, solid houses can be built. haaa, I appreciate you noticing that I respond to all the comments. It goes with the mission I gave myself, and the channel is still small enough so that I can do so. If I were playing with the big boys, I probably wouldn't be able to do so, that's why I am not pushing my channel too hard: it needs to remain something enjoyable for the viewer and the creator!

[@luukderuijter1332]

I'd love to see the math behind why the light is destructively interfered when reflectes

[@PhysicsMadeEasy]

Hi Luuk, well such mathematics a little beyond my competence. But I can give you a good pointer if you want to dig deeper: You must use a mathematical approach based on Feynman's of path integral formulation of quantum mechanics: the actual maths involves integrating over all possible paths, taking into account the interferences that occur for each them, and all of this within a complex plane (we are talking about oscillating phenomena after all). The name of this technique is called the sum-over-paths method. But my knowledge on this topic stops there. Enjoy the voyage

[@itsiwhatitsi]

This is a very underrated video expalin a lot of interesting things that are hard to find all at once

[@topjeeneethindi1064]

You are great sir ..... how do I thank you . You are a great Personality sir . Love you sir .

[@luiszenonmago8730]

Simple and straight to the point, exactly how I like it. well done sir!

[@PhysicsMadeEasy]

Thank you Luis, I am glad you enjoyed my way of showing things!

[@captainbuggernut9565]

This is an excellent explanation. My under ten is doing this now and this explains more in 5 minutes than three pages of a gcse textbook. I've subscribed. Cheers

[@PhysicsMadeEasy]

Thank you for your kind words. Under ten, studying GCSE physics?

[@sergiocorrales2009]

Got curious about this topic on a quiet Friday evening. I’m glad that I found your video. Thank you for sharing your knowledge with the world. Cheers!

[@PhysicsMadeEasy]

Hello Sergio, thank you very much for your feedback :-)

[@vibecheck663]

Thank you so much for the explanation. Having access to information like this are why I love the internet!

[@PhysicsMadeEasy]

These are very kind words Cyberville! It's my little contribution to the Cyberworld :-)

[@wisdominphysics]

Dear professor,i feel your explanation regarding reflection of light is the most satisfactory explanation.we school teachers normally tell when light incident on mirrors just rebouncing back to the sdame medium.But your explanation seems with almost depth but needs mathematical justification,anyhow to the learners this could be awsome introduction.THANK YOU SIR

[@PhysicsMadeEasy]

Thank you ! I guess the maths become tricky when deriving the angle at which constructive interferences would occur, thus proving the reflection rules. This video was inspired from Feynman's views on the subject (Check his lectures for more info, these are free and downloadable on the web).

[@majashishsinghrawat5507]

Wonderfully explained ❤❤

[@varun9346]

Sir you are very very very underrated

[@asimt2507]

Please create a video for Refraction as well.

[@PhysicsMadeEasy]

Noted

[@adrinfpv]

We are still waiting :(

[@franciscomacedo8895]

This was revolutionary for me, learned this last year but never truly understood. Thank you sir, have a wonderful day

[@sonicstorm4354]

This truly is a video to solve all other videos

[@PhysicsMadeEasy]

:-)

[@hermann7609]

Very well explained, and I love the enthusiasm, thanks a lot !

[@PhysicsMadeEasy]

You are welcome Hermann. How can one not be enthusiast when explaining the majestic magic of nature ;-)

[@atripathi7063]

SIR i have a question How do electrons oscillate while they are revolving on their orbits And if they receive energy from outside in form of light , they must jump to higher energy orbital and in this process do they oscillate ?

[@PhysicsMadeEasy]

Hi, you are considering the simplistic model of atoms in a monoatomic gas form, where each atom is indépendant from the others... In liquids, or solids, or even molecular gases, you cannot consider Bohr's model in such a straight forward manner: Actually here, you need to consider electrons as clouds of probability interacting with each other: the motion of the electrons is not a circular one. And anyway, even if they had a definite trajectory, they could still vibrate around that trajectory. Think about green house gases like CO2 that absorb IR radiation. It is the bonds between the atoms of the gas that absorb the EM wave, by vibrating (they enter resonance because the natural frequency of the oscillator that is the bond is the same as that of the EM wave). The bond being formed by the electrons, you can say that the electrons vibrate too (and electrons are charged: this is why they emit IR light in all directions too, including back towards the surface causing the green house effect) I hope this answers your question.

[@atripathi7063]

@@PhysicsMadeEasy sure SIR thankyou

[@mahesh_gupta1]

Excellent sir.

[@MTalha-s2m]

wow

[@spencerwenzel7381]

I have a question if you have time. If we don't ignore the magnetic field, would the following occur? 1) The electron oscillates due to the oscillating electric field 2) The motion is perpendicular to the EMR's magnetic field 3) A magnetic force would cause the electron the follow a spiral path (if I have my hand rules correct) Thanks!

[@PhysicsMadeEasy]

Hi Spencer, That’s a clever one, I never thought about that. I applied the hand rule, you are corrected, it would imply that the electron would spiral along the path of the beam. My counter argument is that for this to occur, the electrons need to be isolated. It couldn’t happen in a solid material. Take a conducting material where electrons are free to move. When a surface is exposed to light, that would mean that electrons would move away from the surface and accumulate lower, creating inside the material an electric field (and a potential difference) that would oppose that change in position, and pull the electrons back closer to the surface,

[@spencerwenzel7381]

@@PhysicsMadeEasy I was wondering what the EMR-electron interaction would look like in a solid. That answered my question. Thank-you for the response.

[@shrutparikh4167]

Great explanation, thank you

[@PhysicsMadeEasy]

You are welcome Shrut :-)

[@mwerensteijn]

Thank you very much!

[@shreyalmaloo9090]

That's a brilliant video. But I had a few questions 1)I just wanted to clarify that if reflection and refraction are caused by the radiation emitted by electrons so do the incident light rays (can't say photons since obv thus explanation is given treating light as a wave)basically get absorbed by the electrons? 2)Also can you prove it mathematically why the only the light ray which we call the reflected ray undergoes constructive interference

[@PhysicsMadeEasy]

Hi Shreya, 1/ Well, imagine you have a rope and somewhere on the rope there is an object. You hold one side of the rope and send a pulse. The pulse arrives to the object and makes it rise into the air (giving it potential energy). How would the pulse look like afterwards ? It would have the same width but a lower amplitude. Electrons are not absorbing the wave like a photon would be absorbed, they are taking some of its energy when being acted upon by an electric force that makes them oscillate. (you can see the wave as energy moving from one point to another, and electrons on its way are picking some of it). 2/ Yes it can be proven, but it would need quite some work to put this in video… Check this lecture from Feynman for more details : www.feynmanlectures.caltech.edu/II_33.html

[@sergiofalcao3691]

Amazing. Now I know how light reflexion works! Thank you so much!

[@PhysicsMadeEasy]

Hi Sergio, yes Physics is full of these everyday common experiences, that actually show up as being utterly awesome once you get how they occur :-)!

[@anujarora0]

But protons are also charged particles so do they also emit light in the presence of electric field of the incident light? BTW great video keep up the good work 👏👏

[@PhysicsMadeEasy]

Absolutely Anuj. A proton is also a charged particle with a magnitude equal to that of the electron and of opposite sign. Thus, a proton under incident light, will also experience a force of same magnitude and opposite direction to that of an electron. But remember that a proton is 2000 times heavier than an electron therefore it will experience an acceleration which is 2000 times smaller: its oscillation will have a much smaller amplitude… Consequently the contribution of protons to reflection / refraction will be very small compared to that of electrons. This is why I neglected this contribution in the video.

[@anujarora0]

@@PhysicsMadeEasy Thank you for clearing my doubt😊

[@atripathi7063]

@@PhysicsMadeEasy hmmm...

[@josephcoon5809]

@@anujarora0 However, there are all kinds of quarks an virtual particles within a proton and neutron which can interact with gamma rays better than elemental electrons.

[@shashwattripathi5872]

​@@josephcoon5809 As protons are made up of quarks, quarks would also be made up of more fundamental particles, and so on. What you think, how deep does this fascinating thing goes?

[@YouTubist666]

Excellent explanation.

[@PhysicsMadeEasy]

Thank you for your feedback

[@bunny4173]

It was a beautiful explanation ....

[@PhysicsMadeEasy]

Thank you!

[@stefanusayudha1853]

ive been searching for this answer for years.. thx sir

[@PhysicsMadeEasy]

Actually, I asked myself this question a little before making the video... reflected on it (no pun intended haha), checked my ideas using Feynman lessons on electromagnetism. I was correct, so produced and published. I enjoyed that process very much! I am glad you enjoyed the result!

[@stefanusayudha1853]

@@PhysicsMadeEasy yes.. there is alot of question for me.. my whole life believe that light travel slower inside a transparent substance.. but what does "slower" mean.. does the momentum decrease, why does light still show the same spectrum when it out then . but this videos im pretty much agree.. thx you so much..

[@DaleWester]

Fascinating video! I have 2 questions. 1. Can you explain what the phase difference between the incident light and reflected light and 2. Can you explain how and which color(s) are reflected back to the viewer as in the subtractive color theory. Example: Today I have on a red shirt. It is red because it absorbs green and blue and reflects red. How? Why red? What determines the color that gets reflected? I have a color theory presentation to make on Tuesday May 14, 2024 and would love to be able to explain this as if I know what I'm talking about. Thanks for any help.

[@PhysicsMadeEasy]

Hi dale, 1: Light reflects at a boundary between two media. If the refraction index (other name for optical density) of the medium light is reflected upon is larger than that of the medium it came from , then you have a phase change of PI (the wave is said to be inverted), if not there is no phase change. 2 : Your shirt is red because it reflects red and absorbs the other colors. The reason is quantum mechanical and is related to the electronic structure of the molecules or atoms the material is made of. How the electrons are arranged in a molecule or atom will imply that some wavelength of light are absorbed and not others. The Pigment used in you shirt is such that it absorbs colors which are not red. You could use other pigments which are made of molecules absorbing other colors (other wavelengths), and choose another color for your shirt. A good example is chlorophyll in plants that absorb Blue and Red and use the energy from that light to transform CO2 in the atmosphere into organic matter. The result is that they reflect green. That's why plants are green. Dig into this, it could be a good content for your presentation. Good luck!

[@SaritaKumari-ic1fb]

Kindly make a similar video on , Why Light bends on entering another medium?

[@PhysicsMadeEasy]

Thank you for your suggestion. On the list. Refraction is quite simple to understand: if you consider that EM Waves need to change speed when crossing through another material (because of the effect of the charges of this material on the incident electric field), parallel beams of light need to bend to prevent a discontinuity of the wave...

[@danielkursitis9031]

This was amazing! Keep it up!

[@PhysicsMadeEasy]

Thank you Daniel :-)

[@ayaanstudying1534]

i am speechlesss sir thank u sooo much make a video on total internal refraction

[@PhysicsMadeEasy]

Hello Ayaan, thank you for your kind words. I suppose you meant 'total internal reflection'. I'll put it on my list.

[@snoo333]

thank you. i have to learn more about how light works and how we can harness it to create images. I also have to learn about how waves work. man, i should have considered learn how to cook bread. fascinating topic.

[@PhysicsMadeEasy]

Haha, cooking bread is fun too: One of my real life students actually became a baker (and I know he still is interested in physics, kind of like an intellectual hobby now, fun to think about)!

[@augijyotbali2131]

Wow this is SO BEAUTIFUL!!!!!!!!!! Thank you so much sir ! Keep sharing your knowledge! We need it !

[@Toddg1234Mr]

Look at the "Double Slit Experiment" by Todd Grigsby. It uses an interferometer to show that light waves can bounce off another. This applies to the double slit experiment thus the title.

[@sharemarketsimplified3970]

We Need similar video on why light refracts?

[@PhysicsMadeEasy]

It's on the list, but you know, if you understand well why light reflects, you can also get an idea of why it refracts (the light emitted by electrons oscillating due to the incident light is emitted in all direction, including towards the material. That is the refraction part ;-).

[@gowrissshanker9109]

I like your way of explaining sir,As you said In your vedio,if possible please put an vedio how light undergo refraction ? Thank you sir

[@PhysicsMadeEasy]

Hi Gowri, thanks. It's on the list, once I am able to resume producing videos (hopefully by beg 2022).

[@tanmaychaudhary6629]

i love how u say theta

[@Arjunanke]

Physics made easy perfectly named! ❤

[@albertrenshaw4252]

This is an incredible explanation although I'm slightly lost on how color fits into all of this. Why do certain atoms (or, their electrons) 'reflect' certain wavelengths of light but not others, does it have to do with how far apart the atoms are from each other?

[@PhysicsMadeEasy]

Hi Albert, this is not the same phenomena! You see your reflection on a mirror, in a window or a piece of polished metal: that is reflection. You do not see your reflection on a wall painted in red! In the case of the wall, it appears red because it absorbs the complementary color(s). Only red is reflected. The absorption of the other colors occur due to atoms / molecules in the paint (pigments) absorbing the photons which have a wavelength corresponding to these absorbed colors.

[@Trarore.orginal]

This video is criminally underrated

[@PhysicsMadeEasy]

Yeah... KZbin's algo is not always on my side haha... I am glad you found the video and enjoyed it though! Don't hesitate to share it with anyone that could be interested! Let's counter KZbin's Algo!

[@virajgems]

Thank you Sir..

[@PhysicsMadeEasy]

You are welcome Santhiviraj :-)

[@zakirhussain-js9ku]

Nice video. 2 questions. 1. How much energy of the photons is lost in destructive interference and absorbed by the reflecting medium? 2. Since the energy of incident photons is absorbed and re-emitted by the medium is their any time lag between incident and reflected light.

[@PhysicsMadeEasy]

Hi Zakir, 1/ you are applying the concept of photon to classical theory... 2/ In terms of waves, yes, there can be a phase difference between incident and re-emited wave

[@moaazelgarhikhalifa4248]

Wonderful Amazing

[@kirankumar6557]

Thank you sir

[@shashwattripathi5872]

In the refracted light, as the energies of incident light and that due to oscillating electrons add up, why it isn't that refracted light is brighter?

[@PhysicsMadeEasy]

By making the electrons vibrate, the incident light gives them part of its energy. so the sum reflected light energy + refracted light energy should be equal to at maximum to the energy of the incident light. It will be less because some of the light can be absorbed by the material.

[@cmwehrmann1]

What is the selection criteria for refraction in terms of electronic states in something like a dielectric? For remission, wouldn't there be some color preference based on the quantized electronic states and since glass doesn't have a band gap that's in the visible part of the EM spectrum, where does the color come from?

[@PhysicsMadeEasy]

Hi Caleb, Since you ask a question quite advanced, I suppose you realized that the model I propose in the video is classical in nature. It does not involve quantum mechanical notions, like allowed electronic states or band diagrams. Insulators (and semiconducting materials) have a valence band, a forbidden band and a conduction band. The valence band is usually full, and the conduction band empty. Provide quanta of energy sufficient for electrons to jump to the conduction band, and you get a conductor (holes in the valence band and electrons in the conduction band). However, if the material contains impurities, you can have some allowed states in the forbidden bands that show up. These can be quite specific and lead to fluorescence of phosphorescence at specific energies: specific colors of longer wavelengths are emitted immediately (fluorescence) or with a delay (phosphorescence). In the case of colored glass, the principle is different: the perceived color is due to impurities that absorbe the complementary color (usually the impurities are complex compounds: metallic ions surrounded by molecular ions) . I hope this helps!

[@zakirhussain-js9ku]

When a charge oscillates in free space EM wave are produced. Where does the E&M fields of the wave come from, there is nothing around the charge. The charge never loses any of its electric or magnetic field.

[@PhysicsMadeEasy]

Hi Zakir, Suppose you are at a fixed point, and an electron moves in front of you left and right, and oscillates. How will you perceive the field of that electron? As it oscillates in front of you, its distance from you oscillates too (thus also the magnitude of the field strength), as well as the direction of its field strength... Such oscillation of the field strength vector is an EM wave (the Magnetic field being induced by that oscillation too). So you see, even if the absolute electric field generated by the electron does not change, when the electron moves, it changes for you (= for a fixed point).

[@Dabforn]

Before hitting the surface, there is air, where there are also atoms and electrons. Are those also affected by oscillating electric field and create their own EM waves?

[@PhysicsMadeEasy]

I would say yes, but the density of matter is much larger than air, thus much more effective, so we can perceive it as reflection or refraction. Think about why the sky is blue... The sun sends white light (more precisely a Black body spectra), higher wavelength like blue are scattered by air particles (it's a way to say that they are absorbed, and re-emmitted in all direction), that's why the sky is blue. The green house effect is the same idea, the CO2 in the atmosphere absorbs infrared EM waves emitted by the Earth towards space, and reemit it in all directions (including back towards Earth)... To go deeper check my videos on Black Bodies: kzbin.info/www/bejne/e4jUq5pmntaBqtk kzbin.info/www/bejne/q5jWp3R6n5aqqtU

[@samuelxd291]

Thanks professor it always puzzled me how the light bounces back 😄

[@PhysicsMadeEasy]

Hi Samuel. Thank you for letting me know, that now, you know :-)

[@albertrenshaw4252]

I'd love a more intuitive understanding of why certain crystalline structures can reflect light perfectly (like a mirror), why do their oscillating electrons "shoot" light out in a clean directional manner and not scatter it out in a diffused manner?

[@PhysicsMadeEasy]

Think interference. Before taking your shower, check your reflection in the mirror. It's clear. Because all rays reflect in the same directions (teh directions for the which the rays are not cancelled by destructive interference). Now after your shower, tiny water droplets cover your mirror, your reflection is all blurry (or even just looks grey). The reason is that the surface of the mirror is now roughened by the droplets. therefore, when light reflects, the rays surviving the destructive interference are not aligned, and so go in all directions.

[@alvischu727]

In " 3:06 Reflexion laws, where do they come from?", why does the wave in other directions form destructive interference?

[@PhysicsMadeEasy]

Hi Alvis, I Suppose it takes quite some calculation / modelling to prove that. I remember taking this result (the constructive interference occurs only when incident and reflected angle are identical and the lines supporting them form a plane with the normal), from Feynman lectures.

[@surendrakverma555]

Very good 😊🙏🙏🙏🙏

[@Hanuman..72368]

Very helpful 👌

[@PhysicsMadeEasy]

Thank you!

[@leon1508]

It really helped ♥️

[@PatrickStar-mt5nb]

Hi! I would like to ask you about the electric and magnetic waves, how can it be that their maximum is aligned. Energy never diappears, right? So, if that's the case then shouldn't there be a pi/2 difference?

[@PhysicsMadeEasy]

Energy does not disappear, it spreads out. The derivative in space of E is related to the derivative in time of B, and the derivative in time of E with the derivative in space of B... These relations are Maxwell equations. When you solve them for empty space, you do find two sine functions perpendicular to each other and in phase... If you want to dig, check the link to Feynman's lecture after this comment. This lecture will not straightforwardly solve these equations for you, but it will give you the tools to do so (and actually, some good videos doing that job are out there, I've seen some ). www.feynmanlectures.caltech.edu/II_20.html

[@sussybaka6597]

Sir a little question This phenomenon of reflection should only occur in metals because non metals doesn't have free electrons so how they osscialte Pls clear my doubt sir by the way thanks for your crystal clear explanation 😄

[@PhysicsMadeEasy]

Hi Priyanshu, in non metals, the electrons are blocked in bonds, so they cannot travel around easily. But, they still can oscillate around their equilibrium position...

[@shareefhuzaf1554]

Sir , please give a class on luminescence

[@PhysicsMadeEasy]

Thank you for the suggestion. I'll put it on the list!

[@or5026]

Thanks for explaining, that in fact none light is reflected, but object emmits light, because it is affected by equivalent charge of visible light value.

[@seekingrafael5630]

I love this!!

[@PhysicsMadeEasy]

Thanks for letting me know. I am glad you enjoyed the voyage!

[@mizzmusicthief]

Neat!

[@swarupkumarsatpathy]

That's great Sir...

[@renebeland4837]

Thank you so much for your explanations! I have a request: Can you explain how radio waves bend back down to earth at certain frequencies and not others? Also is a radio wave traveling through space different than the one that bent back to earth?

[@PhysicsMadeEasy]

Good question Rene. This is due to refraction. You recall from high school, when a beam of light changes media (medium = environment in which the beam travels) from air to water, then the beam bends closer to the normal ( = line perpendicular to) of the interface. From water (high refractive index) to air (low refractive index), it bends away from the normal. For radio waves, it is the same thing. The beam goes from dense air to thin air, so there is a change of media, thus of refraction index. In that case, it is progressive, so the bend is more like a gentle curve. The radio wave antenna emits in all direction, so the most horizontal beams will gently bend to travel a large fraction of the earth perimeter! The refractive index is dependent on the wavelength (this is called chromatic dispersion). This implies that some frequencies are more bent than others. Hence the path of only some radio frequencies will match the curvature of the Earth. If you want to reflect on chromatic dispersion, look at how rainbows are formed ;-). I hope it helps!

[@tundralilac274]

But if most of the reflected light rays were canceled out by destructive interference, wouldn't only a tiny fraction of the incident ray's energy end up in the reflected (plus the refracted beam)?

[@PhysicsMadeEasy]

Hi, good question (It made me scratch my head :-) ) Laws of conservation of energy imply that the intensity of the incident wave be shared between various phenomena: Absorption of the EM wave, transmission of the EM wave and reflection of the EM Wave. Your question deals with the share that is reflected. When there is destructive interference, the energy does not disappear, it is just distributed elsewhere. There is actually a law called Lambert Cosine Law that implies that the energy is distributed according to the angle of reflection. Basically, all of the fraction of the incident energy involved in the reflection, is attributed to the rays that are interfering constructively.

[@ShellYoung]

Just click it: 2:45

[@zeatoen2896]

So whether an object is coloured or reflective and opaque or transperent depends on "structure" of the particles Is this right?

[@PhysicsMadeEasy]

Yes, the optical properties of an object depends _ on the material used, thus the structure that take its fundamental constituents (atoms), for example the molecular structure of pigments _ the type of bonding (how the electrons are organised in material) _ its shape: think how a polished surface reflects light compared to a rough surface

[@Mylittledistraktions]

Love this

[@PhysicsMadeEasy]

me too :-), that is why I made a video!

[@physicslab5787]

Very nice video

[@PhysicsMadeEasy]

Thank you

[@sammalik6355]

Where is the video for refraction?

[@PhysicsMadeEasy]

It should have come (I do videos by series, and obligations in life usually interrupts the video production period). When I did this one, I was in a "wavy" mood... It may come one day, yet there are many other topics I would like to talk about too. So we'll see :-)!

[@denissanterre3092]

I am looking for why some molecules reflect some energies (colors) and not others. I did not find the answer here. As far as this video goes I have a different take on this. I imagine the electrons as spring boards to the photon. The electron would absorbs one photon and then "spring" another of the same energy in the direction of the incident angle. Photon traveling through the material would not be absorb by the electron. However the forces of the atom would slow the transmission causing it to bend. Much like a magnet dropped down a copper pipe. Photons which are absorbed by the electrons with enough energy to move the electron to a higher valance would be held at the level until the electron returns to its normal valance. That photon would have to have a lower energy than the original photon. It would be unique to that valance. Any energy which is not transmitter would be transferred by means of the electron to the proton and conserved as motion we call heat. The details of this is what I am looking for.

[@PhysicsMadeEasy]

Hi Denis, The color of a body is related to the energy level of molecules and atoms at the surface of that body. If a photon carrying the energy corresponding to the difference between two levels, it will be absorbed. The other photons will just follow the process of reflection described in the video. A good example is chlorophyll. This molecule absorbs red and Blue, green is reflected, this is why plants are green… As for refraction (photons travelling in the material), you can view this like the photons trying to travel in a form of melasse (interaction of the EM wave with the EM fields formed by the charged particles in the material). They are "forced" to slow down, and a geometrical consequence of this, is that the ray of light bends.

[@denissanterre3092]

@@PhysicsMadeEasy This is what I thought how refraction worked with the EM between the photon and the atoms. This is the first time I have seen it written down. Some videos omit this or others are completely wrong. I think I revisit the chlorophyll videos to understand better about reflection. Thank you for sharing your knowledge. It is the kindness of people like you who are making the world a better place.

[@satyambhartiya2308]

Thankyou for the explanation sir. I have some doubts regarding this. Light is the radiation emitted by the combination of electric and magnetic field, so should't it be called electromagnetic radiation rather than electromagnetic wave? When light hits charged particles, they oscillate; it happens by the energy transmission(h*mu) of light into charged particles or by the force of the electric field component component of light. Also does Electric field and magnetic field have energies of their own, if yes, form where did it come from and how does they transmit energy when they hit an object?

[@PhysicsMadeEasy]

Hello Satyam, “so shouldn't it be called electromagnetic radiation rather than electromagnetic wave?” An electromagnetic wave is an electromagnetic radiation… these terms are synonyms. “When light hits charged particles, they oscillate; it happens by the energy transmission(h*mu) of light into charged particles. “ Be careful, E=hf, is quantum physics formula. In this video, we stay purely classical. Light is an electric field that is oscillating (The magnetic field is induced by this oscillation and is not of importance for us here). So, when a charge is hit by light, it is affected by an electric force that is oscillating, thus the charge starts oscillating too. A charge creates an electric field around it. Because that charge is now oscillating, the electric field it generates oscillates too, which is equivalent to say that this oscillating charge emits light (here in this video, reflection). “Also does Electric field and magnetic field have energies of their own”: I see this written a lot, and conceptually I disagree with the idea that a field “has” energy when we remain within classical boundaries: It creates potential (the potential energy that a particle would have when placed within it per unit of charge), but that’s all. A field is a mathematical concept, it is abstract. Energy is concrete because it is the capacity to do work (thus generate action), and in my humble opinion it can be held only by charge carriers, in other words, particles. However, as soon as we enter the realm of quantum physics, this perception changes radically, because a particle cannot be dissociated from its field, so saying that a particle has energy is equivalent to say that its field has energy. But this is another story. I hope this helps,

[@or5026]

Hi tech question: how human body can become "of non reflective to ligh properties"? There are tales that advanced Yogies for second or two could become invisible. From what is told in a video, I understand, Yogies can create a barier from all electromagnetic fields and his body. 2nd question: of what nature that barrier might be?

[@PhysicsMadeEasy]

"From what is told in a video, I understand, Yogies can create a barier from all electromagnetic fields and his body": If that was the case, the yogie would be visible as a dark patch (like Chinese shadows). What he would need to do is, on the contrary, be a perfect transmitter of electromagnetic fields: If all light passes through him without interaction, then he would be invisible... Unfortunately for his objective, the yogie is also made of matter, and matter interacts with light. ;-)

[@josephcoon5809]

Thank you for the vid. Ive been looking for an explanation for why reflected light is polarized parallel to the surface while refracted light is polarized perpendicular to the surface. I had thought that it might have something to do with a precession like effect that an electron’s magnetic field would have on the polarization mode of on incident light, but most content that I have seen treats valence electrons in a surface like a “sea” of electrons for metallic crystals. I assume transparent materials act more like a cloud... Anyhow, +1 Like +1 Sub Thank you for your contribution.

[@PhysicsMadeEasy]

That is a tricky one to answer within the scope of wave classical theory. Maybe a starting point for you: physics.stackexchange.com/questions/233746/why-is-reflected-light-polarised

[@josephcoon5809]

@@PhysicsMadeEasy I appreciate the link. I had actually thought about it a bit after my initial comment, and I realized that the electron cloud is heavily constrained to two dimensions of low energy movement; parallel to the surface. For electrons to vibrate in the third axis, they would be moving toward and away from the atoms’ nuclei which would be energy level changes, or quantum changes. In short, x/y-axis vibrations would be low energy classical motion and z-axis vibrations would be higher energy quantum motion. I still have to consider why refractions are perpendicular, though. Perhaps it is the energy that is not absorbed in the reflection passing though the interface? The whole point of this research is regarding Polarization Mode modulation within fiber optics. My thought is that passing the pulse between two flat waveguides would completely polarize the pulse before insertion into the conventional circular waveguide. This method would allow 8-16 polarization modes within the same wave guide which can offer 8-16 distinct carrier waves within the same frequency. This would also reduce Polarization Mode dispersion issues in current optical circuits which have to be accounted for through placing new wave guides and conditioning the path to eliminate birefringence and core deformations; or integrating PMD correcting electronics within the system. I figure the second option can be modified to create far higher bandwidth while reducing the need for upgrading older/deformed wave guides.

[@mdnurainansari3414]

Can you explain why light refract in another video please

[@PhysicsMadeEasy]

Hi, Thanks for the suggestion. One day maybe, now I am just over my head with other things (Especially students because we are entering an exam period). In a nutshell, when the beam of light enters the material, it will interact with the charge carriers of the material, lose energy, the wavelength increases, and thus its propagation slows down (it's a far fetched analogy but you can see like a kind of friction). The, a little geometry, and you realize that the direction of the beam must change to respect continuity of the wave: This phenomena is called refraction.

[@theamazinggoldfish8713]

QUESTION please, does light hitting light (from different angles),... cause deflection?

[@PhysicsMadeEasy]

Light hitting light? It wouldn't create a deflection, like if the light was hitting a different medium and refract. On the other hand, there would be interference. (the two EM waves would add up to each other at the point their meet), but then go on their merry way.

[@Md.AbulHasan-s1w]

Since electron itself create light when incident light heated on the surface, with the same frequency of incident light, then the reflecting light is for the electrons oscillation rather than the incoming light. So, how the incoming light reflected?

[@PhysicsMadeEasy]

I didn't understand your question, could you rephrase it? thank you

[@tanushtaneja]

Why do the electrons move at the same frequency though, if they have mass and can have bonds?

[@PhysicsMadeEasy]

Hi Tanush, Imagine you have a little sister that enjoys the swings. Together you got to the park, your sister sits on the swings. and you start pushing her so that she oscillates. You are applying a force regularly, at a certain frequency. What is the frequency of the oscillation of your sister: of course the same, and that even though she is bonded to something (the rope of the swing) and has inertial mass. When an EM wave meets an electron, the electron (your sister) will be subjected to an electric force that will make it oscillate the same way as the EM wave (you)...

[@tanushtaneja]

@@PhysicsMadeEasy Thank you so much, you provide better explanations than my teachers! You're amazing!!

[@Bugfvbnb]

why does the scattering light is constructive at the opposite angle direction but destructive at others? (what makes them half a period apart?)

[@Bugfvbnb]

is that because the light hit the closest electron first?

[@Bugfvbnb]

but what if the light is perpendicular to the plane?

[@PhysicsMadeEasy]

Hi Zuxian, The interferences between all radiators (= the electrons put in motion by the incident EM Wave that in turn become a source of EM Wave) lead to the laws of reflections. The origin of these laws (how all radiators and their emission interact/interfere with each other) is quite complicated, and too long to explain and develop in a youtube comment. I invite you to refer to Feynman lectures, Volume 1, Chapter 31, “the origin of the refractive index” as a solid starting point if you want to dig into this (these are free on internet).

[@kinggore5229]

Nice.

[@melonmusk4242]

I might be dumb to ask this, but everything don't reflect the light but they have charge (electron, protons),don't these charges oscilltes in the influence of EM wave?? Or they do reflect but we can observe on regular surfaces cause reflection is regular and not on irregular surfaces cause it is diffused?? Please help I'm confused. ❤

[@PhysicsMadeEasy]

Hello Melon Musk (??? 😉) My first words are going to be very “teacherly”, but nonetheless true: no question is dumb! There are also other phenomena that occur: refraction, and absorption. The incoming light will be affected more or less by each (reflection, refraction, abortion) depending on the electronic structure of the material and its orientation. For a more detailed understanding, quantum physics can be more satisfying. Regarding the roughness of the surface, yes, this has a huge effect: Think about the mirror in your bathroom. It reflects your face clearly, but what about after a hot shower? Vapor condenses into little drops that condense on the mirror surface. So now, the mirror surface is full of tiny bumps, so that each point has a normal direction which is different: Light gets reflected, but in all directions. Rays are all mumbled up, and you see a uniform grey surface.

[@melonmusk4242]

@@PhysicsMadeEasy thank you ❤ You are great. I want to be smart like you.

[@faizanhameed6363]

You are love❤

[@Coolboi-c5n]

Hi sir I have a question -- U told us that some ray of light gets reflected and some gets refracted then Does the intesity of that light reduces ???

[@PhysicsMadeEasy]

Yes, it reduces: Intensity is an energy (per second and unit area), and Energy is conserved. Some of it will be used to put in motion the charged particles, so the sum of reflected and refracted intensities has to be less than the incident intensity.

[@Coolboi-c5n]

@@PhysicsMadeEasy thank for ur reply and by the way video was awesome . 👍👍👍👍

[@no-de3lg]

Why are the incidents light and reflected must be in the same plane ? Also reflected light are scattered in all directions 180 degrees? Or 360 degrees and is one electron scattering light in an infinite directions just one particle I mean

[@PhysicsMadeEasy]

It is quite complicated to explain, but roughly, the reflected beam has the geometrical characteristics it has (same plane as normal and incident light, and with a reflected angle = incident angle) because all rays in other directions are subjected to destructive interference. If you considered only one electron, you would be correct: the electron is a point source. But you have to consider all the electrons for which the emissions interfere with each other.

[@atripathi7063]

I have a doubt Sir , as showed in the vedio electrons emit light but their is a theory that light is the movement of photons . I am curious about the dual nature of matter too. How could something be a wave and particle at the same time

[@PhysicsMadeEasy]

Hello, Because nobody really knows haha! Light behave as a wave in some cases (diffraction, refraction) or as a flow of particles in others (atomic absorption, photoelectric effect). Thus the confusion: is it a particle or a wave? other particles like electrons exhibit the same weird behavior. In one of my first videos (What is light ?) published 4-5 years ago, I suggest that light is both a wave and a particle, but today my opinion has changed: it is neither. When we perceive light as a particle or a wave, it is just how we perceive light from a biased perspective. Sometimes it appears as a wave, sometimes as a particle. Imagine you are looking at a cube from a face, you could think it is a square, but if you a facing a vertex, you could think it is a hexagone…It is neither… it is a cube. The best description of particles/waves is derived from quantum field theory, for which every type of elementary particle is associated with a quantum field that fills the whole universe (photon field for photons, electron field for electrons, up quark field for the up quarks etc.). You can see a quantum field like the surface of a calm lake. A localized fluctuation of that surface is seen as a particle: If you throw a stone in that lake, and then take a photo a few instants later, indeed you will see something localized (particle) with wave properties… The analogy stops there though because the particle will have momentum in a given direction, while on a real lake surface, the fluctuation will propagate in all directions. It’s a fascinating subject, if it interests you, you should dig a little, it’s a lot of fun!

[@atripathi7063]

@@PhysicsMadeEasy THANKS SIR, I am really interested in how it works . My goal is to become a physicist, i am in 9th grade now , so i know it will take some time for me to reach that level but i will try my best You're of a great help to me

[@PhysicsMadeEasy]

Yes, Physics is really a fascinating subject. More you dig, more you understand and more you are amazed. In a way, I envy you, because you have all these awesome things to discover: More you understand and more you realise our world is like science fiction.... Enjoy the process!

[@atripathi7063]

#physics made Easy

[@haddow777]

So, you're claiming that none of the original photons in the first beam are reflected. Instead new photons with the same frequency, or energy, are emitted, and the only ones not cancelled out are the ones in line with the angle of the original beam? I keep looking around for this, and others keep saying the original photons continue on in the new direction. I evwn just googled it and one or the first sources made that claim too. I'm not claiming you're wrong, just pronouncing myself confused.

[@PhysicsMadeEasy]

Hi Haddow. First, please note that I am having classical approach to reflection (in classical physics, light is an EM wave). EM waves (and it’s also true for photons), do not bounce back like balls on a wall. They get either absorbed or scattered. I didn’t say that an EM Wave couldn’t continue through the material. But when passing through it, it will affect the charges around it that will start to oscillate and you know that oscillating charges generate EM waves… 3 main scenarios, (this is very simplified of course): 1/ The EM wave gets fully absorbed at the surface (which corresponds to full damping of an oscillation by its charged environment, which in turn emits light). In that case the material can be seen as white body: It reflects everything. Such Materials are metals for example, with electrons free to move around, thus to free to oscillate. 2/The EM wave passes with no loss of intensity. That’s a perfectly transparent material like a pure mineral crystal like diamond. Usually these are insulators or eventually semi conductors with large forbidden band (again diamond). The EM wave (or photon) is deflected due to a different speed of light in that material : that’s Refraction. If the EM wave is of short wavelength, close to interatomic distances (X-Rays for example), it can get scattered, that’s diffraction. 3/ And then you have the intermediate cases: where some gets absorbed (and thus reflected if the absorption/dampening occurs at the surface, and some passes through via refraction… Does this clarify things ?

[@aljawisa]

So why doesn't a static or moving electric field produce colors? How do we not see colors around a charged electrical wire? Article: "Switching colors with electricity."

[@PhysicsMadeEasy]

Hi Aljawisa, the oscillation of the electric field needs to be at a very high frequency for it to reach the visible range (petaherz). See my video What is an EM wave for more details. The article you are referring to involves electrochemical processes. The coloring phenomenon is related to molecular energy levels that can change depend on the oxidation state of the compound… This is Unrelated to the physics presented in this video…

[@aljawisa]

@@PhysicsMadeEasy Thanks.

[@vermillion4971]

Can you tell about why reflecting surfaces can`t reflect EM waves with very high freq. like Extreme X-Ray, Gamma, Ultra high frequency Gamma.

[@PhysicsMadeEasy]

Very interesting question. When reading your question, the explanation that came to my mind is based on quantum physics, not waves (wave theory was used for the video). In order for a particle to interact with another, both particles have to be in a similar range of energy. Here, we are talking about electrons, for which the energy is much less than that of a photon of energy corresponding to hard XRay or gamma, so they will not interact, like they do with visible photons. The gamma photons could actually interact with some heavy nuclei if they have the exact energy (i.e. the reverse of gamma decay), but a nucleus position being well defined, its effective surface will probably be very small, so probably gamma reflection would be marginal. If someone has a wave-based explanation, I would be glad to read it!

[@josephcoon5809]

@@PhysicsMadeEasy The same reason the mesh on your microwave blocks microwaves but allows light through. The electron “mesh” in most materials is too large for the tiny oscillations of x-rays/gamma rays. I read an article about some scientists getting results that were slightly off regarding some gamma ray emissions, and they are hypothesizing that it may be from the denser electron/positron virtual particle pairs in the nucleons that may be providing that finer electron field mesh. However, the index of refraction for gamma rays is, apparently, 1.000001. This causes a tiny fraction of a fraction of deflection for gamma photon.

[@vermillion4971]

@@PhysicsMadeEasy Thanks for your answer, seems like it makes sense, but is it possible that Electrons is just "bouncing off" from "E orbitals" instead of emitting the EM wave with the same frequency(obviously i`m talking about ionising, that`s why UV and above is called Ionising EM Radiation), anyway, your explanation fits with this, because, as you`ve said, the Electron energy on given orbital is less than the energy of EM wave, let`s say Gamma range, and he`s just "bouncing off" ionising the atom and don`t emit any light. Anyway, i think my thoughts about this is not 100% right and i`ll deep dive into it soon. P.S. Sorry for my english, it`s not my native language even close. And thanks for your work on explanation such complicated things of our life. Our World is strange, but is`nt it beautiful? ;)

[@RR-mx4yh]

@@PhysicsMadeEasy This is very interesting, I didn't know about it. My guess is that these radiation frequencies exceed the relaxation frequency of electronic polarization (around 10^17 Hz if I am not mistaken, which is similar to the frequency of x-rays). The oscillation of the field is far too fast for the electrons to follow, therefore there is no generation of the electron's oscillating field. Any thoughts?

[@vermillion4971]

@@josephcoon5809 Then how x-ray and gamma can ionize atoms and react with matter, letting us to detect them, if it didn't even react with electrons on the atom's orbitals? All EM waves more/less can react with electrons in atoms, isn't it? Please correct me if i'm wrong.

[@zeatoen2896]

How can we explain double reflection in eyeglasses

[@PhysicsMadeEasy]

Reflexion occurs at an interface of two materials. You can have reflexions from air to glass but also from glass to air. Additionally, most eyeglasses are covered with a thin film to provide them with useful properties (like anti-reflexion at certain wavelengths: the thin film is engineered so that reflexion at the the air-film and film-glass interfaces are destructively interfering).

[@ZafarKhan-te4dr]

true to its title " Physics-made-easy".

[@PhysicsMadeEasy]

Thank you Zafar!

[@carbon7676]

So light basically reflects by being absorbed and re-emitted?

[@PhysicsMadeEasy]

Exactly, you got it!

[@carbon7676]

@@PhysicsMadeEasy I can't find any other explanation of why light reflects. Many people claim it doesn't get absorbed and re-emitted with deconstructive interference and yet they don't have an explanation to what is really going on, other than it bounces similar to a ball, which is ridiculous. Did you come up with this amazing theory? If not, what's your source?

[@AsadAli-le2ms]

because of their dominance known of speed that's why they reflect not a deflection.

[@nilk8416]

Hello. I wanted to ask you when a stationary source emits light vertically, How is it possible in a reference system in which the source moves, how is it that light is deflected from its vertical way.

[@PhysicsMadeEasy]

I am not sure I understand fully your question. From what I understood, I would answer, because light has a speed. Consider this analogy, it might help: take a stick and plunge it in water: you create circular ripples (the line perpendicular to these ripples are the rays of the waves, like the beam of light). Now create continuous source of ripples, by moving up and down the stick. Same, you get the ripples. Now do this from a moving boat, the shape of the ripples will be change because you are moving: the waves on the surface of the water have a speed that does not depend on the speed of the boat.

[@nilk8416]

@@PhysicsMadeEasy Hello! I'm sorry my comment was not understood. I ask what happens in the case of special relativity. How the electromagnetic wave which consists of two waves perpendicular to each other, one electric and one magnetic and the vector of the motion of the wave is perpendicular to the two waves. I ask how it is possible in a stationary reference system for the electromagnetic wave to be perpendicular to the source, While in the reference system in which the source moves the electromagnetic wave is not vertical but has some inclination forward; Stationary frame of reference 𝆃 ▭ source stationary Moving frame of reference 𝆱 ▭→source is moving How is possible for the moving source to emit light not vertical?

[@zeroonetime]

From 0 to 0. Infinity Squared = 010d. Creation Evolution Entropy.

[@sunitarawat9012]

I don't know why but most of his viewers are 🇮🇳🇮🇳🇮🇳Indians.....

[@PhysicsMadeEasy]

It's true, I noticed this also in my stats. I guess when a seed lands somewhere, it makes babies... lol I also noticed that the Indian schooling system is very elistist. I might be wrong but to me, it doesn't really teach science for the sake of educating and inspiring. It appears to be focused on the selection process... This is maybe why they do not make things easy for the students and do not take the time to focus on the basics. Hence, why my channel is a hit over there. And I am glad that Physics Made Easy is succesful at filling that gap. Because, my experience as a teacher in real life has shown me that with some good basics, getting a solid level in physics is available to all students.

[@sunitarawat9012]

@@PhysicsMadeEasy I don't know why KZbin not recommend your videos on large scale, your videos are really awesome, and your channel diserve millions of subscribers.. Love you ❤ ..... Love to maths and physics... ❤❤

[@arslanrasit]

Hold on... So no photon is actually getting "reflected", huh? That "reflected" photons are actually brand new!

[@PhysicsMadeEasy]

Hello Rasit, yup, you got it! Nature is full of surprises :-)

[@arslanrasit]

@@PhysicsMadeEasy If I'd had a teacher like you, I wouldn't study linguistics... I mean, I would still have a hard time about math but still... Btw, do you know anyone who teaches math just like you teach physics? I mean, by explaining the logic behind everything so I don't have to "memorize" things but I will actually "understand"?

[@zeatoen2896]

So if there are bunch of neutrons put together there is no way to see them .u just killed it man

[@PhysicsMadeEasy]

Interesting thought Zeatoen... As a thought experiment, I believe they could be detectable very very faintly. Neutrons are globally neutral but are composed of charged quarks (2 downs and 1 up). The quarks could oscillate under the effect of the EM wave. However, the amplitude of the oscillation would probably be very small... In addition the up quark would oscillate in opposition of phase with the 2 downs: the EM wave they would generate would be very close to the opposition of phase nearly cancelling each other (the difference from a total opposition would be linked to their different positions within the neutron). This is just a speculation based on classical physics... At this scale, we are in the realm of quantum mechanics. but it was fun to think about ;-)

[@swarupkumarsatpathy]

Sir when light goes into denser medium then it would only slow down (not changing the direction) ... but why does the light bend?

[@PhysicsMadeEasy]

Hello Swarup, When light goes into a denser medium, it does slow down because of its interaction with the charges in the material. Remember light is an oscillating electric field, charges present in the material have their electric fields that will interact with the incident one. What you perceive once the light enters the material is the sum of all these electric fields. If it arrives perpendicular to the surface, then it will just slow down without bending, but if it arrives with an angle, its direction changes. There are multiple ways to explain this, the one taught in school is geometrical (that allows to derive Snell’s law, the second with wave theory, and I suspect that quantum theory should also be able to help.. In this video, I attempt to explain things using wave theory: I discuss that the direction of a reflection is defined by the constructive interference of all electric fields (incident and induced). You can have a similar reasoning with refraction: the bending of light is due to the direction favored by the constructive interference of all the EM Waves (incident and induced by the charges of the material). I hope this helps!

[@swarupkumarsatpathy]

@@PhysicsMadeEasy Thank You Very Much Sir. It's helpful. I am lecturer. I teach students.. but there a lots of thing which we have to learn.. Thank you for providing these informations with interactive videos...

[@swarupkumarsatpathy]

@@PhysicsMadeEasy If in future you come to India please visit the beautiful state of odisha and e-mail me. [e-mail]

[@dgs327]

But sir electric field is not a medium then how light is a wave

[@PhysicsMadeEasy]

Hi Rohit. the concept of wave is not defined by the medium. It is defined by an oscillation that propagates. If you need a medium to visualise it, think of empty space like a medium. What oscillates is the property of that space. For electric fields it would be the effect of the field (= the electric force) on an electric charge that is oscillating and propagating. Think about gravitational waves, which travel also in empty space. In that case the property that oscillates is the curvature of that space.

[@no-de3lg]

Why no one talking about why the wave interferes or reflects

[@PhysicsMadeEasy]

This is standard wave theory... It is called the principle of superposition. For example, If you add two identical waves but you shift one so that its peaks coincides with the dips of the other, the waves cancel. This is called destructive interference.

[@flyingbirds6794]

I think this explanation is wrong because electrons don't behave like this. They absorb the light and directly emit it, it can ONLY be in certain energy levels. So if your explanation is correct, the electrons absorb and emit only some frequencies, but it can emit all frequencies like in the case of photoelectric effect.

[@PhysicsMadeEasy]

Hello Ugur, I think you are confusing with atomic or molecular energy levels… indeed a photon can be absorbed if it has the exact energy corresponding to the difference between two energy levels, and then is re-emitted in all directions. This explains color: a specific paint for example contains molecules which have the right energy level differences corresponding to this color. But do you reflect in a painted wall ? Here, this is not the same process. First of all, remember that I am within the scope of classical physics, where energy states are continuous (what you are describing is rooted in quantum mechanics). But more intuitively, think about this: Take a piece of metal (a conductor), the electrons are free to move there. The electrons can react to any wavelike electromagnetic stimuli that they are subjected to: that is why you reflect in a (polished) metal (Mirrors are made of a thin layer of aluminum or silver protected by a some glass).

[@flyingbirds6794]

@@PhysicsMadeEasy Thank you for your reply.But there are some wrong points.First,when electrons absorb a photon, they will move to a higher energy level like what we see in photoelectric effect.When we talk about electrons, we must think a little bit Quantum mechanics.And as far as I know, electrons in metals are not %100 free to move,but close to it.And they behave as I said before like in the photoelectric effect. And also electrons can't always vibrate in the frequency of EM wave. And your theory can't solve the color issue. And mirrors are different from metals in this property.

[@PhysicsMadeEasy]

@@flyingbirds6794 Hi Again Ugur, Yes, in metals, there are also energy levels, but they are very close to each other and form a band. The levels are so close to each other, that unless you are in extreme cold conditions, electrons are able to move around in energy freely (within a certain range). So, the approximation of a continuous range of energy for the electrons is acceptable in my opinion. That eliminates what you called the problem with the color… Oh, and btw, The reflective part of a mirror is made of metal… Now for the photoelectric effect, I do not know to what you are referring too in regards to reflection: If the energy of the photons is larger than the work function, yes some electrons will be ejected and in that case, more intense is the light, more will get ejected. But another phenomena occurs: a frequency threshold called metal plasma frequency, where the oscillation of the electrons cannot keep up (The metal starts to act like a dielectric). This plasma frequency changes for each metal but all are located in UV, this is why UV is not reflected. Maybe this is what you meant?

[@flyingbirds6794]

@@PhysicsMadeEasy Yes, electrons are very close to be %100 freely. But as we know from photoelectric effect, they're still slightly bounded to the atoms.And this is the case when an electron absorbs a photon. But your explanation is different. And this reality creates the color problem. If thr reality is what you told, there are different problems. I hope I can told you what I think.

[@uzairahmed1113]

what is the name of theory?

[@PhysicsMadeEasy]

The wave theory of classical physics?