9.6 Refraction through a Prism

Figure 9.23 shows the passage of light through a triangular prism ABC. The angles of incidence and refraction at the first face AB are i and $r_1$ , while the angle of incidence (from glass to air) at the second face AC is$r_2$ and the angle of refraction or emergence e. The angle between the emergent ray RS and the direction of the incident ray PQ is called the angle of deviation, δ

In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°. $$\mathrm{\angle }A+\mathrm{\angle }QNR={180}^{\circ }$$ From the triangle QNR, $$r_1+r_2+\mathrm{\angle }\mathrm{QNR}={180}^{\circ }$$ Comparing these two equations, we get

$$r_1+r_2=A$$

The total deviation $\delta$ is the sum of deviations at the two faces, $$\delta =(i-r_1)+(e-r_2)$$ that is, $$\delta =i+e-A$$

Thus, the angle of deviation depends on the angle of incidence. A plot between the angle of deviation and angle of incidence is shown in Fig. 9.24. You can see that, in general, any given value of $\delta$ , except for i = e, corresponds to two values i and hence of e. This, in fact, is expected from the symmetry of i and e in Eq. (9.35), i.e., $\delta$ remains the same if i and e are interchanged. Physically, this is related to the fact that the path of ray in Fig. 9.23 can be traced back, resulting in the same angle of deviation. At the minimum deviation $D_m$ , the refracted ray inside the prism becomes parallel to its base. We have $$\delta =D_{m^{\prime }},i=e\text{ which implies }{r}_1=r_2$$ Equation (9.34) gives

$$2r=A\text{ or }r=\frac{A}{2}$$

In the same way, Eq. (9.35) gives

$$D_{\mathrm{m}}=2i-A,\text{ or }i=(A+D_{\mathrm{m}})/2$$

The refractive index of the prism is

$$n_{21}=\frac{n_2}{n_1}=\frac{\sin \left[(A+D_m)/2\right]}{\sin [A/2]}$$

The angles A and $D_m$ can be measured experimentally. Equation (9.38) thus provides a method of determining refractive index of the material of the prism.

For a small angle prism, i.e., a thin prism, D m is also very small, and we get $$\begin{array}{l}{n}_{21}=\frac{\sin \left[(A+D_m)/2\right]}{\sin [A/2]}=\frac{(A+D_m)/2}{A/2}\\ D_m=(n_{21}-1)A\end{array}$$

It implies that, thin prisms do not deviate light much.