Photo Reflection Theory
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AGARWAL, G. -. (2005). Photo Reflection Theory. National Undergraduate Research Clearinghouse, 8. Available online at http://www.webclearinghouse.net/volume/. Retrieved June 27, 2017 .

Photo Reflection Theory
GAURAV -. AGARWAL
VIDYA VIKASH ACADEMY SCIENCE

Sponsored by: GAURAV AGARWAL (gauravagarwal@sancharnet.in)
ABSTRACT
This research was conducted in order to answer the following questions:1. How do coloured materials reflect light and how does radiations cause heating effect and what is the affect of the light in our retina?2. What actually is a coloured body and what are its properties?3. Explain Blackbody Radiation?

4. Is it possible to have a blackbody?

An experiment was conducted in order to answer the following questions and the effect of light on coloured materials were observed and according to that a theory was thought of and with the results a mathematical formula was devised and it was found to in perfect agreement with the experimental value. This theory strongly supports Plank’s theory and gives us a broader outlook. I am really sure that this theory can rival Einstein’s photoelectric effect.

For more information of the theory kindly log on to my websitehttp://kolkata.sancharnet.in/gauravagarwal/Or, email me at gauravagarwal@sancharnet.in

Gaurav Agarwal.


 


PHOTO REFLECTION THEORY

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Photoreflection Theory

Explanation for reflection of lightfrom a coloured material and blackbody radiation.

Photoreflection

Photoreflection

Definition:When photons of total frequency ‘ni’strike a target the total energy is the algebraic sum of the reflectivepotential ‘Wrp’ of the target and the heat liberated due tothe collision ‘D’.

Or,it can also be defined as:

Thesummation of the products of frequency and intensities of all the photons isequal to the algebraic sum of the refractive potential which is the energyreleased in the form of energy of a discrete frequency and heat energy which isabsorbed by the material.

Ek=Wrp+ D……………….(1).

SIihni=Wrp+ D…………...(2).   

Reflectivepotential: It is the property of a substance by which it liberates photons of specificfrequencies and a specific total energy. It is equal to the total energy of thephotons of the specific frequencies present in the incident radiation which isreflected by the body during collision. It is to be noted that reflectivepotential varies with the incident radiations. It is equal to the product of thefrequencies, the intensity, and the Plank’s constant ‘h’ that gives us thetotal energy of radiations reflected.

Wrp=SIihni- D……………...(3).

Also,Wrp=SIrhnr…………...(4).

Whereh=Plank’s constant

nr=Frequency of reflected radiation(s).

ni=Frequencyof incident radiation(s).

Ii=Intensityof incident radiation(s).

Ir=Intensityof reflected radiation(s).

So,the expression can be written as:

hSIrnr=hSIini-D………….(5).


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What actuallyis a colored body and what are its properties?

A coloured body is a materialhaving a specific reflective potential `Wrpand iscapable of emitting radiations of the frequency of the specific colour theyhave.

Howdo coloured bodies emit the specific radiations of the colour they possess?

When photons of a specific total frequencystrike a coloured material having a specific reflective potential ittries to emit photoelectrons but it is not possible when the work function ismore than that of the energy contained in the photon and hence an inelasticcollision takes place between the radiation and the material.

Since we know that visible lightconsists of a mixture of photons of various frequencies so the desired frequencyis obtained due to the property of material i.e. reflective potential, which isperceived by the pigment of the retina as the specific colour. The rest of thephotons are retarded to 0 velocity and since photons have zero rest mass they donot affect the mass of the material and the energy is converted into heatenergy. It is because of this property of materials that light of only aspecific frequency is liberated from coloured material and this is why blackcoloured bodies get hot easily than white coloured bodies.

Conclusion

So from the theory we canconclude that:

The nature of light radiationsemitted by a material depends upon the reflecting potential of the material.

The energy of the photonsabsorbed by the material is converted to heat energy.

This theory explains blackbodyradiation that is a property of a material to completely absorb or emitradiations.

Howdoes this theory explain Blackbody Radiation?

When photons are incident on ablackbody all the radiations are absorbed and are stored by the material in theform of heat.

We also know that a blackbody isa complete emitter of radiations stored in it. So, when the material is heatedthe photons gain energy and become mobile thereby producing the photons thatwere stored in it and hence the heat energy is converted into radiations.

For a Blackbody

Wrp=0

Therefore all the energysupplied is converted to heat energy.

Isit possible to have a blackbody?

A blackbody is a completeabsorber and emitter of radiations but according to the equation h=nWrp+ .Ditis seen that the energy of radiations emitted depends upon the incidentradiations so for a blackbody the emitted radiations must be zero and in orderto achieve it we need to have a substance having a reflective potential equal tothe energy of radiation.

So it is impossible to have abody that can absorb all the radiations but it is possible to have a body thatcan absorb or emit a discrete frequency of radiations.


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Example to illustrate photo reflective effect

 

1.     An incident radiation consists of the mixture of following radiations ofthe intensities and frequency  whenobserved under a spectrometer:

i.                   radiation 1: I1=200, n1=7.5 x 1014 Hz.

ii.                 radiation 2: I2=350, n2=6 x 1014 Hz.

iii.               radiation 3: I3=150, n3=5 x 1014 Hz.

iv.               radiation 4: I4=250, n4=5.5 x 1014 Hz.

 

When the radiation is incidenton a material radiation 2 is reflected totally what is the heat produced?

 

Ans:

 

We know that;

SIihni=Wrp+ D.

Or,D=hSIini-hSIrnr.

Now since all the radiations ofthe radiation 2 is reflected

nr=6x 1014 Hz.

 

Ir=350,

 

 

D=h(200x7.5x 1014+350x6x1014+150x5x1014+250x5.5x1014-350x6x1014)

D=6.625x10-34(3.625x1017).

D=2.4015625x10-16J.

 

So, the heat absorbed by the bodyis the above.


Gaurav Agarwal.

 

Submitted 9/30/2005 7:29:50 AM
Last Edited 10/17/2005 12:10:42 PM
Converted to New Site 03/09/2009

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