- #26

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I don't understand what you mean,I checked the numbers well

Could you write this final formula ?

- Thread starter Misr
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- #26

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I don't understand what you mean,I checked the numbers well

Could you write this final formula ?

- #27

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No. I won't do it for you. If you want to learn something - you must do some calculations yourself. I've already made a sketch for you - which was easy enough that you could do it yourself.I'm not very good at Math...Could you write this final formula ?

Put a little your own effort to find an answer to your question!

Ready answers from others won't teach you nothing. Do it yourself.

- #28

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Could you give me a small hint or tell me what kind of formula you mean

- #29

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which formulae would I need to derive this formula

- #30

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As youdo i need this final formula?I already drew this wrong graph without the formula

- simpler formula (derive one yourself, I've already told you its form: δ as a function δ(β,α,n);

- more careful calculations, if you prefer to make it in multiple steps, rather than using single formula.

Read previous posts. I answered this question already.

- #31

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for incidece=0,refraction=0

then second angle of incidence is eqivalent to the apical angle=10 degrees

by applying Snell's law,we find the emergence angle = 1.5sin10

using the formula in post 18 ,you will find that deviation=5 degrees

I repeat these steps for different angles of incidence and i get different angles of deviation(misfortunately)

You can check the calculations yourself and you'll find that there's nothing wrong with them

as for the simple formula,I can't derive one formula unless I have other variables

n=Sin(angle of incidence)/sin(angle of refraction)=Sin(angle of emergence)/sin(angle of second incidence)=sin(deviation+incidence+Apical angles)/sin(Apical+refraction angles)

So I still have "refraction" angle in the formula

- #32

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Wrong! You made your plot with 0.1° accuracy - that is ok.for incidece=0,refraction=0

then second angle of incidence is eqivalent to the apical angle=10 degrees

by applying Snell's law,we find the emergence angle = 1.5sin10

using the formula in post 18 ,you will find that deviation=5 degrees

But with such accuracy you should get 5.1°, not 5°.

- #33

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So what I have to do?Wrong! You made your plot with 0.1° accuracy - that is ok.

But with such accuracy you should get 5.1°, not 5°.

Would it make such a great difference if I get 5.1 or 5?

how can I draw an accurate graph?

- #34

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Calculate everything again with the accuracy you assumed at start. If you assumed accuracy 0.1° - make all calculations and plots with such accuracy.So what I have to do?

And make your plots with points placed much densier, starting not from 0, but in range, let's say -40° to 40°.

You don't know that.Would it make such a great difference if I get 5.1 or 5?

If you

But if you assume that the plot is made with 0.1°, and you mark 5° where appropriate value is 5.1°, you are not able to say if the plot you just made results from the formula or from your calculation errors, thus you cannot trust it.

Up to you: from computing values for multiple points and plotting them manually with pencil and graph paper, through manual computation of points and plotting them in some graphic program (like you did recently) to using any data/function plotting program you like (the one you were taught at school). If you haven't learnt any at school - you have opportunity to learn one now - there are lots of such programs available free, very popular one is gnuplot (http://www.gnuplot.info/). As the last resort you may use some spreadsheet program to make plots: openoffice.org:Calc or MS-Excelhow can I draw an accurate graph?

Added: Don't ask me to teach you how to use gnuplot! It is pretty well documented, there are examples and good manual.

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- #35

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Okay,one more trial

[PLAIN]http://img849.imageshack.us/img849/3051/graphio.jpg [Broken]

I hope you spend some time checking these calculations

I tried to be more accurate this time

[PLAIN]http://img849.imageshack.us/img849/3051/graphio.jpg [Broken]

I hope you spend some time checking these calculations

I tried to be more accurate this time

Last edited by a moderator:

- #36

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Don't try to be vicious! It took me about 3 mins to write and run the following C program:I hope you spend some time checking these calculations

Code:

```
#define DEGTORAD (2*3.141592654/360)
#define alpha (10*DEGTORAD)
#define n (1.5)
void misr(void) {
for (int i=-40; i<=40; i+=10) {
double beta = DEGTORAD*i;
double phi = asin(sin(beta)/n);
double psi = phi + alpha;
double theta = asin(n*sin(psi));
double delta = theta - alpha - beta;
printf ("%3.0f %2.1f\n", beta/DEGTORAD, delta/DEGTORAD);
}
}
-40 6.6
-30 5.7
-20 5.2
-10 5.0
0 5.1
10 5.5
20 6.2
30 7.6
40 10.3
```

Great! So now make remaining plots (for other apical angles) and compare all of them.I tried to be more accurate this time

- #37

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That's goodYes, your plot is OK now within required accuracy.

I don't need to do thatGreat! So now make remaining plots (for other apical angles) and compare all of them.

I just want to prove that the thin prism is always in the position of minimum deviation whatever the angle of incidence is(this is according to my textbbook)

In other words,I wanted to show that the angle of deviation -in a thin prism-doesn't change by changing the angle of incidence

Because the angle of minimum deviation in a thin prism is only dependant upon the apical angle and the refractive index according to the relation in the main post

deviation=Apical(n-1)

but that is not satisfied in my graph because as you see,deviation changes by changing the angle of incidence.

Did you understand what I'm trying to say?

- #38

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That's goodYes, your plot is OK now within required accuracy.

I don't need to do thatGreat! So now make remaining plots (for other apical angles) and compare all of them.

I just want to prove that the thin prism is always in the position of minimum deviation whatever the angle of incidence is(this is according to my textbbook)

In other words,I wanted to show that the angle of deviation -in a thin prism-doesn't change by changing the angle of incidence

Because the angle of minimum deviation in a thin prism is only dependant upon the apical angle and the refractive index according to the relation in the main post

deviation=Apical(n-1)

but that is not satisfied in my graph because as you see,deviation changes by changing the angle of incidence.

Did you understand what I'm trying to say?

- #39

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I think I understand what you say, and this is why you should make remaining plots...

From the last plot you may see that apical angle 10° is not quite "thin prism" in the meaning your book use.

Within the chosen accuracy the first rule* min deviation angle = (n-1) apical angle * is fulfilled, but the second rule is not true in this case: deviation angle varies noticeably with incidence angle.

So now make such plots for smaller apical angles: 1°, 2°, 5° and look at them - to see how both your rules fit to reality for smaller and smaller angles.

To make the comparison easier, you may want to scale them, and mark the y-axis not as absolute in degrees, but as a proportion to apical angle (make plots of deviation_angle/apical_angle as a function of incidence_angle for fixed values of n and apical_angle.

I advice to make them with higher accuracy (0.01° rather than 0.1°)

From the last plot you may see that apical angle 10° is not quite "thin prism" in the meaning your book use.

Within the chosen accuracy the first rule

So now make such plots for smaller apical angles: 1°, 2°, 5° and look at them - to see how both your rules fit to reality for smaller and smaller angles.

To make the comparison easier, you may want to scale them, and mark the y-axis not as absolute in degrees, but as a proportion to apical angle (make plots of deviation_angle/apical_angle as a function of incidence_angle for fixed values of n and apical_angle.

I advice to make them with higher accuracy (0.01° rather than 0.1°)

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- #40

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but how can we explain something like that?

I have an explanation that always chases me and i want to know if it is true or not:since it

We know that there's no deviation in a parallelogram since the sides are paralell to each other

The apical angle in a thin prism is very small and the sides are "about to be parallel"

the two sides of the prism coincide if the apical angle angle is zero

for this reason,the deviation angle in a thin prism is very small because it looks like (a little bit)like a parallelogram

what do you think?

- #41

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You've explained this already: starting from Snell's law you've found a formula (well, you've not got a formula in strict sense, but rather algorithm to compute the value for any given angles), and this formula exhibits dependency on incidence angle. There is nothing more in that than mathematics.

How would you explain this deeper?

- #42

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Yeah.may be you are rightThere is nothing more in that than mathematics.

anyways , is there really one formula that explains the relation between (refractive index,angle of incidence,angle of deviation?

can you give just some hints

- #43

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You may combine all those steps together, substituting all angles used in intermediate steps and reduce the final formula a bit. [tex]

\eqalign{

\delta &= \vartheta - \alpha -\beta \cr

&= \arcsin(n\sin\psi) -\alpha -\beta \cr

&= \arcsin\big(n\sin(\alpha+\varphi)\big) -\alpha -\beta \cr

&= \arcsin\left(n\sin\left(\alpha +\arcsin\frac{\sin \beta }{n}\right)\right) -\alpha -\beta \cr

&=\arcsin\left(n\left(\frac{\sin\beta}{n}\cos \alpha+\sqrt{1-\frac{\sin^2\beta}{n^2}}\sin\alpha \right)\right) -\alpha -\beta \cr

&=\arcsin\left(\sin\beta\,\cos\alpha+\sqrt{n^2-\sin^2\beta}\,\sin\alpha\right)-\alpha -\beta \cr

}[/tex]

Then - if you like - you may want to expand the formula to Taylor's series to better see its approximate behaviour for incidence angles close to minimum deflection and for small apical angles.

\eqalign{

\delta &= \vartheta - \alpha -\beta \cr

&= \arcsin(n\sin\psi) -\alpha -\beta \cr

&= \arcsin\big(n\sin(\alpha+\varphi)\big) -\alpha -\beta \cr

&= \arcsin\left(n\sin\left(\alpha +\arcsin\frac{\sin \beta }{n}\right)\right) -\alpha -\beta \cr

&=\arcsin\left(n\left(\frac{\sin\beta}{n}\cos \alpha+\sqrt{1-\frac{\sin^2\beta}{n^2}}\sin\alpha \right)\right) -\alpha -\beta \cr

&=\arcsin\left(\sin\beta\,\cos\alpha+\sqrt{n^2-\sin^2\beta}\,\sin\alpha\right)-\alpha -\beta \cr

}[/tex]

Then - if you like - you may want to expand the formula to Taylor's series to better see its approximate behaviour for incidence angles close to minimum deflection and for small apical angles.

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