Deviation and dispersion in a prism

In summary, the conversation discusses the reason for a prism's minimum deviation position in dispersing light. The angle of deviation is directly proportional to the angle of incidence and depends on the color of light, leading to more dispersion for larger angles of deviation. The proof and factors affecting deviation are also mentioned and the conversation ends with a numerical approach to understanding the concept.
  • #1
Misr
385
0
Hello,world
[PLAIN]http://img718.imageshack.us/img718/1433/unledkzg.jpg
Could you explain the reason for this?This would help me so much
Thanks in advance
 
Last edited by a moderator:
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  • #2
Misr said:
Could you explain the reason for this?

The reason for what, specifically?
 
  • #3
The reason for what, specifically?
it seems that no body can see the picture
Sorry I will reupload it again
 
  • #4

[PLAIN]http://img718.imageshack.us/img718/1433/unledkzg.jpg
[/PLAIN]
 
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  • #5
Misr said:
it seems that no body can see the picture
Sorry I will reupload it again

No, it was just me at work. I can see the picture fine here at home. I don't think we understand what you are asking. You're going to have to type out a question, otherwise we have no idea what you are confused with.
 
  • #6
Oh ok
Small deviation means large dispersion and vice versa,so why?
Is that the reason why the prism should be in the minimum deviation position to disperse light?
 
  • #7
Misr said:
Hello,world
[PLAIN]http://img718.imageshack.us/img718/1433/unledkzg.jpg
Could you explain the reason for this?This would help me so much
Thanks in advance

Misr said:

[PLAIN]http://img718.imageshack.us/img718/1433/unledkzg.jpg
[/PLAIN]

Angle of deviation + Angle of prism = Angle of incidence + Angle of emergence

Let Angle of prism be constant ie k

Angle of deviation =[itex]\partial[/itex]
Angle of incidence=i
Angle of emergence=e

Now

[itex]\partial[/itex] = k - i+e

Now realize that

[itex]\partial[/itex] [itex]\propto[/itex] i

Forget about e now because each colour has different e

In your first image of large deviation , and small i the e will be very small tally the relation.
In your second image of small deviation , and large i the e will be very large tally the relation.

If e will be large so more dispersion and vice versa.

If you want to know the proof of relation then feel free to ask.

In case of minimum deviation i=e , so ∂ = k - 2i ,
there is maximum dispersion also.

Claculate dispersion like this :
ev+ei+eb+eg+ey+eo+er where v,i,b,g,y,o,r are different colours : 7
 
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  • #8
I don't understand those strange equations you are writing Could you use simpler equations?What kind of equations are they?
I just understand that Angle of deviation + Angle of prism = Angle of incidence + Angle of emergence
Thanks very much!
 
  • #9
I think this script doesn't work on my computer
Could you use images or words please?
 
  • #10
ok I can see the equations now but I have some questions:
∂ = k - i+e
We studied that alpha=i+e-k not k-i+e so how do u explain that?
Now realize that

∂ ∝ i
how do u consider that alpha is directly proportional to i?This is not a direct relation
Forget about e now because each colour has different e
I haven't known that
I can't imagine how I can forget about e , doesn't e depend on i not on the color?
If you want to know the proof of relation then feel free to ask.
which relation?Anyways I certainly want to know the proof
Thanks very much for your help
 
  • #11
Misr said:
ok I can see the equations now but I have some questions:

We studied that alpha=i+e-k not k-i+e so how do u explain that?
how do u consider that alpha is directly proportional to i?This is not a direct relation

I haven't known that
I can't imagine how I can forget about e , doesn't e depend on i not on the color?

which relation?Anyways I certainly want to know the proof
Thanks very much for your help

Sorry , my fault , its ∂ = -k + i+e
I mistyped it !

I think the following sites may crystal clear your concepts better than I can :

The proof of relation ∂ = -k + i+e : http://www.askiitians.com/iit_jee-Ray_Optics/Prism
The factors affecting deviation in prism :http://en.wikipedia.org/wiki/Minimum_deviation
Relation between dispersion and deviation in a prism :http://www.physicsclassroom.com/class/refrn/u14l4a.cfm

Minimum deviation : http://www.mtholyoke.edu/~mpeterso/classes/phys103/geomopti/MinDev.html

Hope this helps :)
 
  • #12
Ok,I'll check them and complete this discussion
Thanks very much
 
  • #13
Still confused Can't anyone in the world solve my problem :(
 
  • #14
Can A prism disperse light when it is not the in the positsion of minimum deviation?
 
  • #15
I don't have a formal proof or a conceptual explanation, but sometimes I understand things while working on a numerical problem. Here's what I would do in this case:

1) Using Snell's law, I compute the angle of deviation when it is minimum (when the ray of light is horizontal inside the prism). I use a typical refractive index: 1.50 .

2) For a different color, I repeat step 1 with a slightly different refractive index (let's say 1.45).

3) Subtracting the values obtained in steps 1 and 2, I find the angle of dispersion.

4) I repeat step 1, 2 and 3 for a random case where the deviation is not minimum (angle of incidence = 45 degrees, for example). As a result, I should find a smaller angle of dispersion.
 
  • #16
Misr said:
Still confused Can't anyone in the world solve my problem :(

Why ? Have you gone through those sites which I gave in post #14? Your question is not refined and hence it is quite hard for me to explain it to you.

Misr said:
Can A prism disperse light when it is not the in the positsion of minimum deviation?

Yes , why not? I recommend that you should go through those sites again and read them carefully one by one !

[Post by YPelletier]
I don't have a formal proof or a conceptual explanation, but sometimes I understand things while working on a numerical problem. Here's what I would do in this case:

1) Using Snell's law, I compute the angle of deviation when it is minimum (when the ray of light is horizontal inside the prism). I use a typical refractive index: 1.50 .

2) For a different color, I repeat step 1 with a slightly different refractive index (let's say 1.45).

3) Subtracting the values obtained in steps 1 and 2, I find the angle of dispersion.

4) I repeat step 1, 2 and 3 for a random case where the deviation is not minimum (angle of incidence = 45 degrees, for example). As a result, I should find a smaller angle of dispersion.

This is what I typed in my earlier post. I am not sure what OP is searching for ? I gave answer and even gave reference sites?

What is the question Misr ? Please type your question again so that I can be clear what answer do you want exactly ?
 
  • #17
I don't have a formal proof or a conceptual explanation, but sometimes I understand things while working on a numerical problem. Here's what I would do in this case:

1) Using Snell's law, I compute the angle of deviation when it is minimum (when the ray of light is horizontal inside the prism). I use a typical refractive index: 1.50 .

2) For a different color, I repeat step 1 with a slightly different refractive index (let's say 1.45).

3) Subtracting the values obtained in steps 1 and 2, I find the angle of dispersion.

4) I repeat step 1, 2 and 3 for a random case where the deviation is not minimum (angle of incidence = 45 degrees, for example). As a result, I should find a smaller angle of dispersion.
You mean I should use numbers?I don't really understand what you are trying to say
could you use an example?

What is the question Misr ? Please type your question again so that I can be clear what answer do you want exactly ?
I want to prove that small deviation means large dispersion and vice versa
I recommend that you should go through those sites again and read them carefully one by one !
I already read the physicsclassroom tutorial in your link and I even printed it few weeks ago but no use..I can't find the answer to my question
 
  • #18
I don't have a formal proof or a conceptual explanation, but sometimes I understand things while working on a numerical problem. Here's what I would do in this case:

1) Using Snell's law, I compute the angle of deviation when it is minimum (when the ray of light is horizontal inside the prism). I use a typical refractive index: 1.50 .

2) For a different color, I repeat step 1 with a slightly different refractive index (let's say 1.45).

3) Subtracting the values obtained in steps 1 and 2, I find the angle of dispersion.

4) I repeat step 1, 2 and 3 for a random case where the deviation is not minimum (angle of incidence = 45 degrees, for example). As a result, I should find a smaller angle of dispersion.
Ok ,how toi do that?
I tried this on another problem
I wanted to show that the thin prism is always in the position of minimum deviation whatever the angle of incidence
but I failed , so could you help with this idea ?
 
  • #19
When white light falls on a prism do all the wavelength of white light have the same angle of incidence?
 
  • #20
Misr said:
When white light falls on a prism do all the wavelength of white light have the same angle of incidence?

When white light falls on prism , it is dispersed after the refraction not at that flick of second. At the line of separation between two media it is after all a monochromatic beams of light. It will have same angle of incidence though.

Prove it experimentally , your question :small deviation means large dispersion and vice versa

Take angle of incidence = 40o

Compute by Snell's law :

sin 40o/sin r = 1.52

Make use of trigonometric tables.
Find for r.
Now find angle of deviation using this formula :

d=i-r or d=r-i

Whatever comes greater.

Now use this formula :
e= d+A-i

Where A is angle of prism. Make it constant ie 60o.
Here consider e as angle of dispersion.
Find e or angle of dispersion.

Repeat it again and again using different angle of incidence. 50o , 60o , 70o etc.

Note your observations.

:wink:

(My exams are going on and that is why busy.)
 
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  • #21
That's great but I have some questions
d=i-r or d=r-i

Whatever comes greater.
Isn't d=i+angle of emergence-A?
e= d+A-i
How did you derive this formula?
Thanks very much and good luck with your exams
 
  • #22
Misr said:
That's great but I have some questions

Isn't d=i+angle of emergence-A?

How did you derive this formula?
Thanks very much and good luck with your exams

Its sheer logic. Suppose you have a prism on which a monochromatic ray of light makes an angle of incidence of i with respect to normal at point of incidence.It then refracts towards the normal and makes an angle of refraction say r. Since that ray refracts towards the normal , then obviously i>r , right ? So how much does the ray deviate an angle ? Its i-r ! So d= i-r

If r>i then obviously d=r-i !

I think you are referring to this formula : e= d+A-i
This formula is same as :
Angle of incidence + Angle of emergence = Angle of deviation + Angle of prism
Angle of emergence = Angle of deviation + Angle of prism - Angle of incidence
Derivation can be found here : http://www.askiitians.com/iit_jee-Ray_Optics/Prism
 
  • #23
So d= i-r
This formula is the angle of deviation before leaving the prism,while the formula I posted is the angle of deviation after leaving the prism
correct??
I think you are referring to this formula : e= d+A-i
That's true
could you have a look at my work
let two rays ,ray1 and ray2
refractive index of ray1=1.5

refractive index of ray2=1.45
the Angle of the prism=60
1-At i=40,I calculated the angle of refraction using Snell's law
I calculated the second angle of incidence from the relation,A=angle of refraction+the second angle of incidence,
I calculated the angle of emergence by using Snell's law again
I calculated the deviation for both rays after exiting the prism from the relation:
e= d+A-i
d of ray1=38.4
d of ray2=33.6
dispersion=38.4-33.6=4.8

2- At i=50
d for ray1=37.2
d for ray2=33.1
dispersion=37.2-33.1=4.1

3-At i=60,
d for ray1=38.8
d for ray2=35
dispersion=38.8-35=3.8

Note that the deviation for both rays in no.(2) is less than the deviation in no. (1)
and also the dispersion of rays in no.(2) is less than the dispersion is less than the dispesion in no.(1)
Hope you can understand what I'm trying to say,
 
  • #24
Misr said:
This formula is the angle of deviation before leaving the prism,while the formula I posted is the angle of deviation after leaving the prism
correct??

That's true
could you have a look at my work
let two rays ,ray1 and ray2
refractive index of ray1=1.5

refractive index of ray2=1.45
the Angle of the prism=60
1-At i=40,I calculated the angle of refraction using Snell's law
I calculated the second angle of incidence from the relation,A=angle of refraction+the second angle of incidence,
I calculated the angle of emergence by using Snell's law again
I calculated the deviation for both rays after exiting the prism from the relation:
e= d+A-i
d of ray1=38.4
d of ray2=33.6
dispersion=38.4-33.6=4.8

2- At i=50
d for ray1=37.2
d for ray2=33.1
dispersion=37.2-33.1=4.1

3-At i=60,
d for ray1=38.8
d for ray2=35
dispersion=38.8-35=3.8

Note that the deviation for both rays in no.(2) is less than the deviation in no. (1)
and also the dispersion of rays in no.(2) is less than the dispersion is less than the dispesion in no.(1)
Hope you can understand what I'm trying to say,

Dah ! I think this clarifies your problem.

Angle of dispersion or Angular dispersion = δvioletred

For violet light (nviolet), refractive index = 1.534
And for red light , refractive index = 1.514
Angle of prism i.e A = 60

At i = 40
Violet light :
sin i/sin r1 = nviolet
sin 40/sin r1 = 1.534

Find for r1 ...

Then A = r1+r2
Find for r2 ...

Then again use snell's law to find for angle of emergence :

sin r2/sin e = 1/1.534
(from denser medium to rarer of course refractive index will be multiplicative inverse...)

Find for e..

Now
A+ δviolet = i+e
Find for δviolet ..

For red light :

sin i/sin r1 = nred
sin 40/sin r1 = 1.514

Find for r1 ...

Then A = r1+r2
Find for r2 ...

Then again use snell's law to find for angle of emergence :

sin r2/sin e = 1/1.514
(from denser medium to rarer of course refractive index will be multiplicative inverse...)

Find for e..

Now
A+ δred = i+e
Find for δred ..

Now Find angle of dispersion by : δviolet - δred
And find angle of deviation δ by : (δviolet + δred)/2


___________________________________________________________________

Repeat this whole procedure at i=50 , 60 , 70 .. and note your observations on δ and angular dispersion.

:)
 
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1. What is deviation in a prism?

Deviation in a prism refers to the bending of light as it passes through the prism. This occurs because different wavelengths of light have different speeds when passing through a medium, causing them to bend at different angles.

2. How does a prism cause dispersion?

A prism causes dispersion by separating white light into its component colors. This is due to the different wavelengths of light being bent at different angles, causing them to emerge from the prism at different points.

3. What is the difference between deviation and dispersion?

Deviation refers to the bending of light within a prism, while dispersion refers to the separation of white light into its component colors. In other words, deviation is the cause of dispersion.

4. How does the angle of incidence affect deviation and dispersion?

The angle of incidence refers to the angle at which light enters the prism. As the angle of incidence increases, so does both the amount of deviation and dispersion. This is because a larger angle of incidence results in a larger difference in the speeds of light passing through the prism, causing a greater separation of colors.

5. What is the relationship between the refractive index and deviation in a prism?

The refractive index of a material is a measure of how much light is bent when passing through it. In a prism, the higher the refractive index, the greater the deviation of light will be. This means that materials with higher refractive indices will cause a greater separation of colors and a more pronounced dispersion effect.

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