Wednesday, 2 November 2016

optics - why does the graph of deviation angle in a prism doesn't get a symmetry?


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.



Answer



enter image description here


δ=minimum deviation angle.


i1=incident angle for minimum deviation.


i1+=incident angle i1 plus a variation θ.


i1=incident angle i1 minus a variation θ.


δ+= deviation if incident angle equals i1+.


δ= deviation if incident angle equals i1.


Result : δ+δ.





The drawing is sketched with the following data and calculations: A=prism angle=60on1=refraction index of surroundings=1.00n2=refraction index of prism=1.50n=relative refraction index=1/1.50=0.6667i1=incident angle of minimum deviation=arcsin(nsinA2)=48.59ominimum deviation path (red) :48.59o30o30o48.59oδ=minimum deviation angle=2i1A=37.18oθ=variation angle=20oi1+=i1+θ=68.59oi1+ deviation path (blue) :68.59o38.36o21.64o33.58oδ+=deviation angle if incident angle equals i1+=42.17oi1=i1θ=28.59oi1 deviation path (green) :28.59o18.60o41.40o82.70oδ=deviation angle if incident angle equals i1=51.29o




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There exists symmetry, but not in the sense of the question : Let a prism of angle A and a first experiment F with incident angle (i1)F on the left and emergent angle (i2)F from the right. The deviation angle is δF=(i1)F+(i2)FA. If in a second experiment B the incident angle is (i1)B=(i2)F then for the emergent angle of B we have (i2)B=(i1)F and for the deviation angle δB=(i1)B+(i2)BA=(i2)F+(i1)FAδF

This must be expected since (reversing or ignoring the direction of light rays) we have one and the same experiment, the first being its Front view and the second its Back view.



This symmetry is depicted as the symmetry of the graph of the function i2(i1) with respect to the main diagonal of the i1i2 plane, see the last Figure above.




Related : Analytic solution for angle of minimum deviation?.



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