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According to (i1+i2)-A , the graph of i1- and the angle of deviation shows below.

enter image description here

This graph is not symmetrical. It usually oriented to right. According to above equation I can generally understand why doesn't it get a symmetry. But I can't see a way of describe this ,theoretically why is it not symmetrical.

I searched this in so many resources, but I couldn't find a satisfied answer. So, I hope a theoretical answer, for this problem, not a mathematical description.

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    $\begingroup$ I take it that this is about a prism of triangular cross-section. To get a feel for what's happening, I suggest that you sketch the paths of rays with different values of $i_1$ through a prism and calculate $D$ for some of them. You might first consider $i_1=0$ and $i_1=89°$. $\endgroup$ – Philip Wood Jun 2 '18 at 18:00
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    $\begingroup$ It is the mathematical way, that substituting values to above formular. Without that, isn't there a way to describe it theoretically. $\endgroup$ – Osal Thuduwage Jun 3 '18 at 1:49
  • $\begingroup$ Alright, just do the sketching and don't calculate anything! The sketching by itself should convince you that you won't get the same value of $D$ for values of $i_1$ that are equally above and below $i_1=\frac{A+D_{min}}{2}$. This is, of course, the requirement for the curve to be symmetrical. Incidentally, perhaps you should ask yourself first of all: why do you expect the curve to be symmetrical? I see no reason at all for expecting it to be so. $\endgroup$ – Philip Wood Jun 3 '18 at 7:24
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enter image description here

$\color{red}{\delta^{*}}$=minimum deviation angle.

$\color{red}{\mathrm{i}1^{*}}$=incident angle for minimum deviation.

$\color{blue}{\mathrm{i}1^{\boldsymbol{+}}}$=incident angle $\mathrm{i}1^{*}$ plus a variation $\:\theta$.

$\color{green}{\mathrm{i}1^{\boldsymbol{-}}}$=incident angle $\mathrm{i}1^{*}$ minus a variation $\:\theta$.

$\color{blue}{\delta^{\boldsymbol{+}}}$= deviation if incident angle equals $\mathrm{i}1^{\boldsymbol{+}}$.

$\color{green}{\delta^{\boldsymbol{-}}}$= deviation if incident angle equals $\mathrm{i}1^{\boldsymbol{-}}$.

Result : $\delta^{\boldsymbol{+}}\ne\delta^{\boldsymbol{-}}$.


The drawing is sketched with the following data and calculations: \begin{align} \mathrm A & = \text{prism angle}= 60^{\rm o} \tag{01}\\ n_{1} & = \text{refraction index of surroundings}= 1.00 \tag{02}\\ n_{2} & = \text{refraction index of prism}= 1.50 \tag{03}\\ n & = \text{relative refraction index}=1/1.50=0.6667 \tag{04}\\ \mathrm i1^{\boldsymbol{*}} & = \text{incident angle of minimum deviation}= \arcsin\left(n\sin\tfrac{\mathrm A}{2}\right)=48.59^{\rm o} \tag{05}\\ & \text{minimum deviation path (red) } : 48.59^{\rm o} \Longrightarrow 30^{\rm o} \Longrightarrow 30^{\rm o} \Longrightarrow 48.59^{\rm o} \tag{06}\\ \delta^{*} & = \text{minimum deviation angle}=2\cdot\mathrm i1^{\boldsymbol{*}}-\mathrm A=37.18^{\rm o} \tag{07}\\ \theta & = \text{variation angle}=20^{\rm o} \tag{08}\\ \mathrm{i}1^{\boldsymbol{+}} & = \mathrm i1^{\boldsymbol{*}}+\theta=68.59^{\rm o} \tag{09}\\ \mathrm{i}1^{\boldsymbol{+}} & \text{ deviation path (blue) } : 68.59^{\rm o} \Longrightarrow 38.36^{\rm o} \Longrightarrow 21.64^{\rm o} \Longrightarrow 33.58^{\rm o} \tag{10}\\ \delta^{\boldsymbol{+}} & = \text{deviation angle if incident angle equals }\mathrm{i}1^{\boldsymbol{+}} =42.17^{\rm o} \tag{11}\\ \mathrm{i}1^{\boldsymbol{-}} & = \mathrm i1^{\boldsymbol{*}}-\theta=28.59^{\rm o} \tag{12}\\ \mathrm{i}1^{\boldsymbol{-}} & \text{ deviation path (green) } : 28.59^{\rm o} \Longrightarrow 18.60^{\rm o} \Longrightarrow 41.40^{\rm o} \Longrightarrow 82.70^{\rm o} \tag{13}\\ \delta^{\boldsymbol{-}} & = \text{deviation angle if incident angle equals }\mathrm{i}1^{\boldsymbol{-}} =51.29^{\rm o} \tag{14} \end{align}


enter image description here enter image description here enter image description here enter image description here enter image description here

There exists symmetry, but not in the sense of the question : Let a prism of angle A and a first experiment $\:\mathcal{F}\:$ with incident angle $\:(\mathrm{i}_{1})_{\mathcal{F}}\:$ on the left and emergent angle $\:(\mathrm{i}_{2})_{\mathcal{F}}\:$ from the right. The deviation angle is $\:\delta_{\mathcal{F}}=(\mathrm{i}_{1})_{\mathcal{F}}+(\mathrm{i}_{2})_{\mathcal{F}}-\mathrm{A}$. If in a second experiment $\:\mathcal{B}\:$ the incident angle is $\:(\mathrm{i}_{1})_{\mathcal{B}}=(\mathrm{i}_{2})_{\mathcal{F}}\:$ then for the emergent angle of $\:\mathcal{B}\:$ we have $\:(\mathrm{i}_{2})_{\mathcal{B}}=(\mathrm{i}_{1})_{\mathcal{F}}\:$ and for the deviation angle \begin{equation} \delta_{\mathcal{B}}=(\mathrm{i}_{1})_{\mathcal{B}}+(\mathrm{i}_{2})_{\mathcal{B}}-\mathrm{A}=(\mathrm{i}_{2})_{\mathcal{F}}+(\mathrm{i}_{1})_{\mathcal{F}}-\mathrm{A}\equiv \delta_{\mathcal{F}} \tag{015} \end{equation} This must be expected since (reversing or ignoring the direction of light rays) we have one and the same experiment, the first being its $\:\mathcal{F}$ront view and the second its $\:\mathcal{B}$ack view.

This symmetry is depicted as the symmetry of the graph of the function $\:\mathrm{i}_{2}(\mathrm{i}_{1})\:$ with respect to the main diagonal of the $\:\mathrm{i}_{1}-\mathrm{i}_{2}\:$ plane, see the last Figure above.


Related : Analytic solution for angle of minimum deviation?.

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  • $\begingroup$ Cheers!!!! for giving such successful and completed answer. Thanks for spending your time for creating relevant images and graphs..... $\endgroup$ – Osal Thuduwage Jun 6 '18 at 2:52
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Some mathematics is unavoidable. First, in stating clearly what we mean by the curve being symmetrical.

Let us denote by $i^*$ the value of $i_1$ or $i_2$ for which $i_1=i_2$.

[It's easy to show that $i^*=\arcsin (n\ \sin \frac{A}{2})$] in which $A$ is the angle at the 'top' of the prism.]

If the curve were symmetrical we'd have the same angle, $D$, of deviation if$$i_1-i^*=i^*-i_2.$$[We know that $i_2$ is the other angle of incidence that will give the same D.] The equation means that, if the curve is symmetrical, the mean of $i_1$ and $i_2$ is $i^*$.

This is not going to be the case because in order to relate $i_1$ to $i_2$ we have to apply the sine function, and the arcsine function when we apply Snell's law at each surface. Sine and arcsine are non-linear functions: you don't get the same size of change in $\sin \theta$ when $\theta$ changes from (say) 50° to 70° that you get when $\theta$ changes from (say) 50° to 30°. Similarly for arcsin.

I'm sorry if this is too mathematical, but I don't think we can explain the non-symmetry of the curve otherwise (except informally, by ray-sketching). The basic reason for the non-symmetry is the non-linearity of the sine function.

[It might be worth mentioning that the bottom of the curve is pretty symmetrical, that is for small variations of $i_1$ from $i^*$, we do have, approximately,$$i_1-i^*=i^*-i_2.$$I'm afraid the best way by far to explain why this should be the case is to appeal to mathematics, specifically to terms higher than the second in the relevant 'Taylor expansion' being negligible for small excursions from the minimum. Sorry!]

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    $\begingroup$ I am not expert in mathematics, but this is the way of the answer, I expected. The reasons that you show with combining of sine curve and the arc sine curve saying that " The basic reason for the non-symmetry is the non-linearity of the sine function". If you can attach a graph or curve, how does the non linearity of the sine curve affect with the deviation angle symmetry, it will more understandable and more helpful. Thank you very much for spending your time with my question. $\endgroup$ – Osal Thuduwage Jun 3 '18 at 16:35
  • $\begingroup$ You can find a graph of the sine function very easily. Its non-linearity is quite clearly apparent. However I'm convinced that the best way to convince yourself of the non-symmetry is by doing a couple of simple calculations. Choose a refractive index and a prism angle, $A$ (say $n$ = 1.50 and 60°). Calculate $i^*$. [I think it's 48.6° with my values of$n$ and $A$.] Then sketch the ray for $i_1=58.6°$ and calculate $i_2$ using Snell's law (twice) and simple geometry. I bet you'll find that $i_2$ isn't 38.6° ! Good luck! $\endgroup$ – Philip Wood Jun 3 '18 at 17:23
  • $\begingroup$ It looks as if Frobenius has got there first! $\endgroup$ – Philip Wood Jun 3 '18 at 18:08

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