What is the formula for the angle of deviation in a thin prism?

[Misr]

I checked my calculations several times and there's nothing wrong with them
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

 

[xts]

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

Wrong! You made your plot with 0.1° accuracy - that is ok.
But with such accuracy you should get 5.1°, not 5°.

 

[Misr]

Wrong! You made your plot with 0.1° accuracy - that is ok.
But with such accuracy you should get 5.1°, not 5°.

So what I have to do?
Would it make such a great difference if I get 5.1 or 5?
how can I draw an accurate graph?

 

[xts]

So what I have to do?

Calculate everything again with the accuracy you assumed at start. If you assumed accuracy 0.1° - make all calculations and plots with such accuracy.
And make your plots with points placed much densier, starting not from 0, but in range, let's say -40° to 40°.

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

You don't know that.
If you believe the accuracy 0.1° is too high, and lower accuracy (e.g. 1° or 0.5°) would be sufficient - then make a lower accuracy plot and mark points according to this, at the risk that after making the plot you may spot it was not accurate enough and you must do it again with better accuracy. The best practice is to make such plots with best accuracy reacheable with your computing and presentation tools.

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.

how can I draw an accurate graph?

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 learned 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-Excel

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

 

[Misr]

Okay,one more trial
[PLAIN]http://img849.imageshack.us/img849/3051/graphio.jpg
I hope you spend some time checking these calculations
I tried to be more accurate this time

 

[xts]

I hope you spend some time checking these calculations

Don't try to be vicious! It took me about 3 mins to write and run the following C program:

#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

Yes, your plot is OK now within required accuracy.

I tried to be more accurate this time

Great! So now make remaining plots (for other apical angles) and compare all of them.

 

[Misr]

Yes, your plot is OK now within required accuracy.

That's good

Great! So now make remaining plots (for other apical angles) and compare all of them.

I don't need to do that
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?

 

[Misr]

Yes, your plot is OK now within required accuracy.

That's good

Great! So now make remaining plots (for other apical angles) and compare all of them.

I don't need to do that
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?

 

[xts]

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°)

 

[Misr]

Yes.The less the apical angle ,the less the deviation angle depends on the incidence angle
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 parallel 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?

 

[xts]

I think there is no need for such 'explanation', and such reasoning often leads to errors.

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?

 

[Misr]

There is nothing more in that than mathematics.

Yeah.may be you are right
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

 

[xts]

You may combine all those steps together, substituting all angles used in intermediate steps and reduce the final formula a bit. <br /> \eqalign{<br /> \delta &amp;= \vartheta - \alpha -\beta \cr<br /> &amp;= \arcsin(n\sin\psi) -\alpha -\beta \cr<br /> &amp;= \arcsin\big(n\sin(\alpha+\varphi)\big) -\alpha -\beta \cr<br /> &amp;= \arcsin\left(n\sin\left(\alpha +\arcsin\frac{\sin \beta }{n}\right)\right) -\alpha -\beta \cr<br /> &amp;=\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<br /> &amp;=\arcsin\left(\sin\beta\,\cos\alpha+\sqrt{n^2-\sin^2\beta}\,\sin\alpha\right)-\alpha -\beta \cr<br /> }
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.