Why is angle of incidence equal to angle of reflection?












1














In the Law of Reflection, the angle of incidence is equal to angle of reflection. Why is this true? This is clearly true experimentally, but how does one prove this true mathematically?










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  • It is mostly due to a perfect plane of the reflecting surface. If the surface is not plane, but wavy, you will have a "diffusion" rather than a reflection to some certain angle.
    – Vladimir Kalitvianski
    1 hour ago










  • math.stackexchange.com/questions/1395215/…
    – N. Steinle
    1 hour ago










  • The math behind the proofs linked in the above comments is somewhat straightforward; I'm not sure you're saying what you mean with 'prove this (mathematically)'. It's more likely that you're hoping for a derivation from first principles and fundamental characteristics of light and matter; wording which suggests the same would be helpful.
    – Chair
    1 hour ago
















1














In the Law of Reflection, the angle of incidence is equal to angle of reflection. Why is this true? This is clearly true experimentally, but how does one prove this true mathematically?










share|cite|improve this question









New contributor




Garima is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • It is mostly due to a perfect plane of the reflecting surface. If the surface is not plane, but wavy, you will have a "diffusion" rather than a reflection to some certain angle.
    – Vladimir Kalitvianski
    1 hour ago










  • math.stackexchange.com/questions/1395215/…
    – N. Steinle
    1 hour ago










  • The math behind the proofs linked in the above comments is somewhat straightforward; I'm not sure you're saying what you mean with 'prove this (mathematically)'. It's more likely that you're hoping for a derivation from first principles and fundamental characteristics of light and matter; wording which suggests the same would be helpful.
    – Chair
    1 hour ago














1












1








1







In the Law of Reflection, the angle of incidence is equal to angle of reflection. Why is this true? This is clearly true experimentally, but how does one prove this true mathematically?










share|cite|improve this question









New contributor




Garima is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











In the Law of Reflection, the angle of incidence is equal to angle of reflection. Why is this true? This is clearly true experimentally, but how does one prove this true mathematically?







optics reflection






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edited 49 mins ago









N. Steinle

1,310117




1,310117






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asked 2 hours ago









Garima

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141




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Garima is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.












  • It is mostly due to a perfect plane of the reflecting surface. If the surface is not plane, but wavy, you will have a "diffusion" rather than a reflection to some certain angle.
    – Vladimir Kalitvianski
    1 hour ago










  • math.stackexchange.com/questions/1395215/…
    – N. Steinle
    1 hour ago










  • The math behind the proofs linked in the above comments is somewhat straightforward; I'm not sure you're saying what you mean with 'prove this (mathematically)'. It's more likely that you're hoping for a derivation from first principles and fundamental characteristics of light and matter; wording which suggests the same would be helpful.
    – Chair
    1 hour ago


















  • It is mostly due to a perfect plane of the reflecting surface. If the surface is not plane, but wavy, you will have a "diffusion" rather than a reflection to some certain angle.
    – Vladimir Kalitvianski
    1 hour ago










  • math.stackexchange.com/questions/1395215/…
    – N. Steinle
    1 hour ago










  • The math behind the proofs linked in the above comments is somewhat straightforward; I'm not sure you're saying what you mean with 'prove this (mathematically)'. It's more likely that you're hoping for a derivation from first principles and fundamental characteristics of light and matter; wording which suggests the same would be helpful.
    – Chair
    1 hour ago
















It is mostly due to a perfect plane of the reflecting surface. If the surface is not plane, but wavy, you will have a "diffusion" rather than a reflection to some certain angle.
– Vladimir Kalitvianski
1 hour ago




It is mostly due to a perfect plane of the reflecting surface. If the surface is not plane, but wavy, you will have a "diffusion" rather than a reflection to some certain angle.
– Vladimir Kalitvianski
1 hour ago












math.stackexchange.com/questions/1395215/…
– N. Steinle
1 hour ago




math.stackexchange.com/questions/1395215/…
– N. Steinle
1 hour ago












The math behind the proofs linked in the above comments is somewhat straightforward; I'm not sure you're saying what you mean with 'prove this (mathematically)'. It's more likely that you're hoping for a derivation from first principles and fundamental characteristics of light and matter; wording which suggests the same would be helpful.
– Chair
1 hour ago




The math behind the proofs linked in the above comments is somewhat straightforward; I'm not sure you're saying what you mean with 'prove this (mathematically)'. It's more likely that you're hoping for a derivation from first principles and fundamental characteristics of light and matter; wording which suggests the same would be helpful.
– Chair
1 hour ago










2 Answers
2






active

oldest

votes


















3














Well this can be proven in many ways. If you think of light as waves, then use Huygen's Principle.



A much easier proof can also be developed if you consider light as rays propagating in a line. For this we can use Fermat's Principle.



However if you think of light as particles then a much more intuitive proof can be created by considering a ball being hit on the ground. The part of it's velocity parallel to the ground will not change and the part perpendicular to the ground will flip(assuming an Elastic Collision).






share|cite|improve this answer





























    2














    The answer by harshit54 is very concise and clear and gives you answers in multiple layers of understanding. However, to quote Leonardo DiCaprio: we need to go deeper. Not because we must, but because we can! There's a TL;DR below.



    A beam of light can be thought of as a stream of energy packets (photons, which are the quanta of light - lots of interesting words to look up in the dictionary already). Now let's zoom in and look at what happens when the photon hits any material. It runs into a wall of atoms - lots of nuclei surrounded by electrons (also energy packets - there's more to it but let's not write out all of quantum mechanics here). When a photon hits an electron, its energy gets absorbed and the electron goes into a higher energy state. This does not last long; the electron left an "empty" energy state below it, which is an energetically more favorable position for it. Thus, there is a chance that it spontaneously jumps back to a lower energy state. This chance increases over time, so it's pretty certain that it will jump back quite quickly. When it does, it needs to get rid of its extra energy. This energy is released as a photon!



    If this is the only electron in the neighbourhood that releases a photon, it will go in any random direction. HOWEVER! There's a catch. Hint: this is where the wave nature of light comes into play. Let's assume that the beam of light hits the reflecting surface directly from above, so the angle of incidence is 0 degrees. Now you have many electrons that are being bombarded by even more photons, all emitting photons in many directions. The photons that are emitted at an angle however, will be out of phase with eachother (since there is a distance between the electrons, if two photons are emitted at any angle at the same time, there will be a slight delay between them). Photons that are out of phase will tend to cancel eachother out. Photons that are in phase (all the photons that are emitted upwards) will constructively interfere with eachother.



    Now something interesting happens - something that also explains why lasers work. When an electron emits a photon, and there are many photons around it who all have the same phase and direction, the emitted photon will copy the phase and direction of the photons around it! So very quickly, all photons that are emitted in random directions die out and only photons that are emitted perfectly in phase with eachother remain.



    Now tilt your light beam at an angle. No longer the photons that are emitted upwards are in phase with eachother, but only the photons that are emitted at the exact same angle as the incident photons are in phase. So they remain!



    So why does this not happen at any surface? Well, the above only applies to surfaces with lots of electrons, found in materials where electrons are free - for example metals! Surfaces where all electrons are bound will not absorb the photons immediately - they'll penetrate the first few layers of atoms unhindered until by chance they are absorbed. When a new photon is emitted, it will run into other atoms (it's not at the surface anymore!) and keep the reaction going until at the surface, photons are reflected in random directions. Combine this with the fact that without free electrons it is VERY difficult to smooth a surface, it will give you no chance for a decent reflection.



    TL;DR




    1. Photon energy is absorbed by electron


    2. Energy is emitted by electron in the form of a new photon


    3. Photons that are out of phase with eachother die out


    4. Only photons that are emitted at the same angle as the angle of incidence are in phase


    5. Those remain. Reflection!


    6. The above only works in materials with lots of free electrons, like metals







    share|cite|improve this answer





















    • this means non metals will not reflect light? And what about metalloids ?
      – Garima
      3 mins ago











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

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    3














    Well this can be proven in many ways. If you think of light as waves, then use Huygen's Principle.



    A much easier proof can also be developed if you consider light as rays propagating in a line. For this we can use Fermat's Principle.



    However if you think of light as particles then a much more intuitive proof can be created by considering a ball being hit on the ground. The part of it's velocity parallel to the ground will not change and the part perpendicular to the ground will flip(assuming an Elastic Collision).






    share|cite|improve this answer


























      3














      Well this can be proven in many ways. If you think of light as waves, then use Huygen's Principle.



      A much easier proof can also be developed if you consider light as rays propagating in a line. For this we can use Fermat's Principle.



      However if you think of light as particles then a much more intuitive proof can be created by considering a ball being hit on the ground. The part of it's velocity parallel to the ground will not change and the part perpendicular to the ground will flip(assuming an Elastic Collision).






      share|cite|improve this answer
























        3












        3








        3






        Well this can be proven in many ways. If you think of light as waves, then use Huygen's Principle.



        A much easier proof can also be developed if you consider light as rays propagating in a line. For this we can use Fermat's Principle.



        However if you think of light as particles then a much more intuitive proof can be created by considering a ball being hit on the ground. The part of it's velocity parallel to the ground will not change and the part perpendicular to the ground will flip(assuming an Elastic Collision).






        share|cite|improve this answer












        Well this can be proven in many ways. If you think of light as waves, then use Huygen's Principle.



        A much easier proof can also be developed if you consider light as rays propagating in a line. For this we can use Fermat's Principle.



        However if you think of light as particles then a much more intuitive proof can be created by considering a ball being hit on the ground. The part of it's velocity parallel to the ground will not change and the part perpendicular to the ground will flip(assuming an Elastic Collision).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        harshit54

        43510




        43510























            2














            The answer by harshit54 is very concise and clear and gives you answers in multiple layers of understanding. However, to quote Leonardo DiCaprio: we need to go deeper. Not because we must, but because we can! There's a TL;DR below.



            A beam of light can be thought of as a stream of energy packets (photons, which are the quanta of light - lots of interesting words to look up in the dictionary already). Now let's zoom in and look at what happens when the photon hits any material. It runs into a wall of atoms - lots of nuclei surrounded by electrons (also energy packets - there's more to it but let's not write out all of quantum mechanics here). When a photon hits an electron, its energy gets absorbed and the electron goes into a higher energy state. This does not last long; the electron left an "empty" energy state below it, which is an energetically more favorable position for it. Thus, there is a chance that it spontaneously jumps back to a lower energy state. This chance increases over time, so it's pretty certain that it will jump back quite quickly. When it does, it needs to get rid of its extra energy. This energy is released as a photon!



            If this is the only electron in the neighbourhood that releases a photon, it will go in any random direction. HOWEVER! There's a catch. Hint: this is where the wave nature of light comes into play. Let's assume that the beam of light hits the reflecting surface directly from above, so the angle of incidence is 0 degrees. Now you have many electrons that are being bombarded by even more photons, all emitting photons in many directions. The photons that are emitted at an angle however, will be out of phase with eachother (since there is a distance between the electrons, if two photons are emitted at any angle at the same time, there will be a slight delay between them). Photons that are out of phase will tend to cancel eachother out. Photons that are in phase (all the photons that are emitted upwards) will constructively interfere with eachother.



            Now something interesting happens - something that also explains why lasers work. When an electron emits a photon, and there are many photons around it who all have the same phase and direction, the emitted photon will copy the phase and direction of the photons around it! So very quickly, all photons that are emitted in random directions die out and only photons that are emitted perfectly in phase with eachother remain.



            Now tilt your light beam at an angle. No longer the photons that are emitted upwards are in phase with eachother, but only the photons that are emitted at the exact same angle as the incident photons are in phase. So they remain!



            So why does this not happen at any surface? Well, the above only applies to surfaces with lots of electrons, found in materials where electrons are free - for example metals! Surfaces where all electrons are bound will not absorb the photons immediately - they'll penetrate the first few layers of atoms unhindered until by chance they are absorbed. When a new photon is emitted, it will run into other atoms (it's not at the surface anymore!) and keep the reaction going until at the surface, photons are reflected in random directions. Combine this with the fact that without free electrons it is VERY difficult to smooth a surface, it will give you no chance for a decent reflection.



            TL;DR




            1. Photon energy is absorbed by electron


            2. Energy is emitted by electron in the form of a new photon


            3. Photons that are out of phase with eachother die out


            4. Only photons that are emitted at the same angle as the angle of incidence are in phase


            5. Those remain. Reflection!


            6. The above only works in materials with lots of free electrons, like metals







            share|cite|improve this answer





















            • this means non metals will not reflect light? And what about metalloids ?
              – Garima
              3 mins ago
















            2














            The answer by harshit54 is very concise and clear and gives you answers in multiple layers of understanding. However, to quote Leonardo DiCaprio: we need to go deeper. Not because we must, but because we can! There's a TL;DR below.



            A beam of light can be thought of as a stream of energy packets (photons, which are the quanta of light - lots of interesting words to look up in the dictionary already). Now let's zoom in and look at what happens when the photon hits any material. It runs into a wall of atoms - lots of nuclei surrounded by electrons (also energy packets - there's more to it but let's not write out all of quantum mechanics here). When a photon hits an electron, its energy gets absorbed and the electron goes into a higher energy state. This does not last long; the electron left an "empty" energy state below it, which is an energetically more favorable position for it. Thus, there is a chance that it spontaneously jumps back to a lower energy state. This chance increases over time, so it's pretty certain that it will jump back quite quickly. When it does, it needs to get rid of its extra energy. This energy is released as a photon!



            If this is the only electron in the neighbourhood that releases a photon, it will go in any random direction. HOWEVER! There's a catch. Hint: this is where the wave nature of light comes into play. Let's assume that the beam of light hits the reflecting surface directly from above, so the angle of incidence is 0 degrees. Now you have many electrons that are being bombarded by even more photons, all emitting photons in many directions. The photons that are emitted at an angle however, will be out of phase with eachother (since there is a distance between the electrons, if two photons are emitted at any angle at the same time, there will be a slight delay between them). Photons that are out of phase will tend to cancel eachother out. Photons that are in phase (all the photons that are emitted upwards) will constructively interfere with eachother.



            Now something interesting happens - something that also explains why lasers work. When an electron emits a photon, and there are many photons around it who all have the same phase and direction, the emitted photon will copy the phase and direction of the photons around it! So very quickly, all photons that are emitted in random directions die out and only photons that are emitted perfectly in phase with eachother remain.



            Now tilt your light beam at an angle. No longer the photons that are emitted upwards are in phase with eachother, but only the photons that are emitted at the exact same angle as the incident photons are in phase. So they remain!



            So why does this not happen at any surface? Well, the above only applies to surfaces with lots of electrons, found in materials where electrons are free - for example metals! Surfaces where all electrons are bound will not absorb the photons immediately - they'll penetrate the first few layers of atoms unhindered until by chance they are absorbed. When a new photon is emitted, it will run into other atoms (it's not at the surface anymore!) and keep the reaction going until at the surface, photons are reflected in random directions. Combine this with the fact that without free electrons it is VERY difficult to smooth a surface, it will give you no chance for a decent reflection.



            TL;DR




            1. Photon energy is absorbed by electron


            2. Energy is emitted by electron in the form of a new photon


            3. Photons that are out of phase with eachother die out


            4. Only photons that are emitted at the same angle as the angle of incidence are in phase


            5. Those remain. Reflection!


            6. The above only works in materials with lots of free electrons, like metals







            share|cite|improve this answer





















            • this means non metals will not reflect light? And what about metalloids ?
              – Garima
              3 mins ago














            2












            2








            2






            The answer by harshit54 is very concise and clear and gives you answers in multiple layers of understanding. However, to quote Leonardo DiCaprio: we need to go deeper. Not because we must, but because we can! There's a TL;DR below.



            A beam of light can be thought of as a stream of energy packets (photons, which are the quanta of light - lots of interesting words to look up in the dictionary already). Now let's zoom in and look at what happens when the photon hits any material. It runs into a wall of atoms - lots of nuclei surrounded by electrons (also energy packets - there's more to it but let's not write out all of quantum mechanics here). When a photon hits an electron, its energy gets absorbed and the electron goes into a higher energy state. This does not last long; the electron left an "empty" energy state below it, which is an energetically more favorable position for it. Thus, there is a chance that it spontaneously jumps back to a lower energy state. This chance increases over time, so it's pretty certain that it will jump back quite quickly. When it does, it needs to get rid of its extra energy. This energy is released as a photon!



            If this is the only electron in the neighbourhood that releases a photon, it will go in any random direction. HOWEVER! There's a catch. Hint: this is where the wave nature of light comes into play. Let's assume that the beam of light hits the reflecting surface directly from above, so the angle of incidence is 0 degrees. Now you have many electrons that are being bombarded by even more photons, all emitting photons in many directions. The photons that are emitted at an angle however, will be out of phase with eachother (since there is a distance between the electrons, if two photons are emitted at any angle at the same time, there will be a slight delay between them). Photons that are out of phase will tend to cancel eachother out. Photons that are in phase (all the photons that are emitted upwards) will constructively interfere with eachother.



            Now something interesting happens - something that also explains why lasers work. When an electron emits a photon, and there are many photons around it who all have the same phase and direction, the emitted photon will copy the phase and direction of the photons around it! So very quickly, all photons that are emitted in random directions die out and only photons that are emitted perfectly in phase with eachother remain.



            Now tilt your light beam at an angle. No longer the photons that are emitted upwards are in phase with eachother, but only the photons that are emitted at the exact same angle as the incident photons are in phase. So they remain!



            So why does this not happen at any surface? Well, the above only applies to surfaces with lots of electrons, found in materials where electrons are free - for example metals! Surfaces where all electrons are bound will not absorb the photons immediately - they'll penetrate the first few layers of atoms unhindered until by chance they are absorbed. When a new photon is emitted, it will run into other atoms (it's not at the surface anymore!) and keep the reaction going until at the surface, photons are reflected in random directions. Combine this with the fact that without free electrons it is VERY difficult to smooth a surface, it will give you no chance for a decent reflection.



            TL;DR




            1. Photon energy is absorbed by electron


            2. Energy is emitted by electron in the form of a new photon


            3. Photons that are out of phase with eachother die out


            4. Only photons that are emitted at the same angle as the angle of incidence are in phase


            5. Those remain. Reflection!


            6. The above only works in materials with lots of free electrons, like metals







            share|cite|improve this answer












            The answer by harshit54 is very concise and clear and gives you answers in multiple layers of understanding. However, to quote Leonardo DiCaprio: we need to go deeper. Not because we must, but because we can! There's a TL;DR below.



            A beam of light can be thought of as a stream of energy packets (photons, which are the quanta of light - lots of interesting words to look up in the dictionary already). Now let's zoom in and look at what happens when the photon hits any material. It runs into a wall of atoms - lots of nuclei surrounded by electrons (also energy packets - there's more to it but let's not write out all of quantum mechanics here). When a photon hits an electron, its energy gets absorbed and the electron goes into a higher energy state. This does not last long; the electron left an "empty" energy state below it, which is an energetically more favorable position for it. Thus, there is a chance that it spontaneously jumps back to a lower energy state. This chance increases over time, so it's pretty certain that it will jump back quite quickly. When it does, it needs to get rid of its extra energy. This energy is released as a photon!



            If this is the only electron in the neighbourhood that releases a photon, it will go in any random direction. HOWEVER! There's a catch. Hint: this is where the wave nature of light comes into play. Let's assume that the beam of light hits the reflecting surface directly from above, so the angle of incidence is 0 degrees. Now you have many electrons that are being bombarded by even more photons, all emitting photons in many directions. The photons that are emitted at an angle however, will be out of phase with eachother (since there is a distance between the electrons, if two photons are emitted at any angle at the same time, there will be a slight delay between them). Photons that are out of phase will tend to cancel eachother out. Photons that are in phase (all the photons that are emitted upwards) will constructively interfere with eachother.



            Now something interesting happens - something that also explains why lasers work. When an electron emits a photon, and there are many photons around it who all have the same phase and direction, the emitted photon will copy the phase and direction of the photons around it! So very quickly, all photons that are emitted in random directions die out and only photons that are emitted perfectly in phase with eachother remain.



            Now tilt your light beam at an angle. No longer the photons that are emitted upwards are in phase with eachother, but only the photons that are emitted at the exact same angle as the incident photons are in phase. So they remain!



            So why does this not happen at any surface? Well, the above only applies to surfaces with lots of electrons, found in materials where electrons are free - for example metals! Surfaces where all electrons are bound will not absorb the photons immediately - they'll penetrate the first few layers of atoms unhindered until by chance they are absorbed. When a new photon is emitted, it will run into other atoms (it's not at the surface anymore!) and keep the reaction going until at the surface, photons are reflected in random directions. Combine this with the fact that without free electrons it is VERY difficult to smooth a surface, it will give you no chance for a decent reflection.



            TL;DR




            1. Photon energy is absorbed by electron


            2. Energy is emitted by electron in the form of a new photon


            3. Photons that are out of phase with eachother die out


            4. Only photons that are emitted at the same angle as the angle of incidence are in phase


            5. Those remain. Reflection!


            6. The above only works in materials with lots of free electrons, like metals








            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 58 mins ago









            enzolima

            597149




            597149












            • this means non metals will not reflect light? And what about metalloids ?
              – Garima
              3 mins ago


















            • this means non metals will not reflect light? And what about metalloids ?
              – Garima
              3 mins ago
















            this means non metals will not reflect light? And what about metalloids ?
            – Garima
            3 mins ago




            this means non metals will not reflect light? And what about metalloids ?
            – Garima
            3 mins ago










            Garima is a new contributor. Be nice, and check out our Code of Conduct.










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