= Solution
Let $r_1,r_2$ be the internal <angle of refraction> values measured from the two face normals, and let $i,e$ be the external incidence and emergence <angles>. Geometry of an <optical prism> gives $r_1+r_2=A$ and total deviation $\delta=(i-r_1)+(e-r_2)=i+e-A$.
For an isosceles <optical prism> whose internal beam is parallel to its base, reflection symmetry gives $r_1=r_2=A/2$. Applying <Snell's law> at both faces, with air <refractive index> approximated by one, gives $\sin i=n\sin(A/2)=\sin e$. For transmitted forward rays, $i=e$, so $i=(\delta+A)/2$. Hence
$$
\boxed{n=\frac{\sin[(\delta+A)/2]}{\sin(A/2)}.}
$$
This is the symmetric <minimum deviation> geometry at a single <wavelength>. Transmission requires $n\sin(A/2)\leq1$; otherwise the proposed symmetric path cannot enter and leave through the two faces.
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