Refraction through Prisms

When a ray passes through a prism, the angle of deviation $\delta$ depends on the angle of incidence $i$. As $i$ increases from a small value, $\delta$ first decreases, reaches a least value, then increases again. This least value is the angle of minimum deviation $\delta_{min}$.

At minimum deviation the ray passes symmetrically through the prism: the angle of incidence equals the angle of emergence ($i=e$), the refracted ray inside is parallel to the base, and the refraction angles are equal ($r_1=r_2=A/2$). The refractive index is

$\mu=\frac{\sin\left(\dfrac{A+\delta_{min}}{2}\right)}{\sin\dfrac{A}{2}}$

Graph of $\delta$ against $i$: a U-shaped (smooth parabola-like) curve, with $i$ on the horizontal axis and $\delta$ on the vertical axis. The curve falls from a high deviation at small $i$, has its lowest point $\delta_{min}$ at $i=e$, and rises again for larger $i$. The same deviation is obtained for two different values of $i$ (the second being the emergence angle) except at the minimum.

(b) angle of incidence

At minimum deviation the ray passes symmetrically through the prism, so the angle of emergence equals the angle of incidence ($e=i$) and the ray inside is parallel to the base.

A ray falling normally on a shorter face goes straight in and strikes the hypotenuse at an angle of incidence of $45^\circ$. For total internal reflection there, $45^\circ$ must be at least the critical angle $C$ for glass-to-water.

$\sin C=\frac{n_w}{n_g}\le\sin45^\circ=\frac{1}{\sqrt2}$

$n_g\ge\sqrt2\,n_w=1.414\times1.33\approx1.88$

So the minimum refractive index of the prism is about 1.88 (1.89 using $n_w=4/3$).

(d) Inverted and same size

Light enters the hypotenuse face normally (no deviation), meets each of the two equal faces at $45^\circ$, which is greater than the critical angle of glass ($\approx42^\circ$), and is totally internally reflected twice. It leaves through the hypotenuse reversed in direction. The image is the same size as the object, and inverted.

(a) $\delta_{min}=r$

For a thin prism $\delta=(\mu-1)A=0.5A$. At minimum deviation the ray passes symmetrically, so $r=\dfrac{A}{2}$, i.e. $A=2r$. Hence $\delta_{min}=0.5\times2r=r$.

(a) It is reflected totally from the second surface and emerges perpendicular from the third surface.

The ray enters the first face normally, so it goes undeviated inside the prism. At the second face, the angle of incidence is (60^\circ), which is greater than the critical angle (45^\circ), so total internal reflection occurs. The reflected ray then strikes the third face normally and hence emerges perpendicular to it.