# Why should angle of incidence be equal to angle of emergence for minimum deviation in a triangular prism?

When I was solving questions on Ray optics I encountered a question wherein I had to prove that when a ray of light undergoes minimum deviation through a triangular prism then both the angles of refraction that is the first one at the incident point and the second one at the emergent point should be equal.

Since I was not able to prove this, I looked into the solution for this problem and it was written that for minimum deviation to take place in a triangular prism the angles of incidence that is the angle which the incident ray makes with the normal at that point and the angle of emergence that is the angle which the emergence ray makes with the normal at the point of emergence should be equal. I could not access why this would happen.

• Welcome kay , you should provide more context to be helped. Your question may have many interpretations. You may edit your question to improve it
– user46925
Jan 23, 2016 at 19:34
• @igael can you tell me now?
– kay
Jan 23, 2016 at 19:48

For a prism (or any linear optics) you should be able to reverse the direction of light and get the same result. This means that if a given angle of incidence $\alpha$ results in a certain exit angle $\beta$, then incident angle $\beta$ will result in exit angle $\alpha$.

Now a maximum or minimum in the deviation occurs when a small change doesn't change the outcome. By symmetry this has to happen when $\alpha=\beta$.

• I'm not sure this is convincing. There could be two minima, or more, in which case your argument fails. Dec 21, 2018 at 8:30

When the light enters the prism, it is slowed, proportional to the index of refraction.

The part of its speed normal to the face is the sine of the angle of incidence before it hits, and the sine of the angle it is refracted to inside the glass.

For the resulting speed vector to be the appropriate length, this has to be reduced by a factor of the reduction in speed, so the sine of the angle inside is the original incoming angle divided by the index of refraction.

When it exits, it speeds back up the exact same amount. So the sine of the angle as to be increased by the same factor.

Since we have divided and then multiplied the sine of the angle by the same number, the angle at emergence is the same as the original angle of incidence. You can see that $\delta=(i_1-r_1)+(i_2-r_2)$ so $\delta =(i_1+i_2)-\alpha$ /($\alpha=r_1+r_2$) When the angles of emergence are equal and incidence also the desviation is minimum $$i_1=i_2$$ $$r_1=r_2$$$$\delta = \delta _{min}$$ $$\delta_m=2i-\alpha$$ $$r_1=r_2=\alpha/2$$ $$i_1=i_2=(\delta_m+\alpha)/2$$ After some calculation we conclude that $$n=\frac{sin(\delta_m+\alpha)}{sin(\alpha /2)}$$ the answer to why they must be equal the angle can be deduced by trigonometry If $r_1>i_1$ $sin(r_1)>sin(i_1)$ same thing for $r_2$ and $i_2$ so the $\delta$ is always going to be bigger than $\delta_m$

The picture was from :Tutorvista.com