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

[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. $$\eqalign{ \delta &= \vartheta - \alpha -\beta \cr &= \arcsin(n\sin\psi) -\alpha -\beta \cr &= \arcsin\big(n\sin(\alpha+\varphi)\big) -\alpha -\beta \cr &= \arcsin\left(n\sin\left(\alpha +\arcsin\frac{\sin \beta }{n}\right)\right) -\alpha -\beta \cr &=\arcsin\left(n\left(\frac{\sin\beta}{n}\cos \alpha+\sqrt{1-\frac{\sin^2\beta}{n^2}}\sin\alpha \right)\right) -\alpha -\beta \cr &=\arcsin\left(\sin\beta\,\cos\alpha+\sqrt{n^2-\sin^2\beta}\,\sin\alpha\right)-\alpha -\beta \cr }$$
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.