Refraction through a Prism

When light passes through a flat glass slab, it shifts sideways but emerges in the same direction it entered. A prism is different. Because a prism has tilted faces that are not parallel to each other, the two refractions do not cancel out. Instead, they add up to bend the light through a definite angle. Understanding exactly how this bending works, and how it depends on the shape of the prism and the angle at which light enters, is the focus of this topic.

How Light Travels through a Triangular Prism

Fig 9.21: A ray of light passing through a triangular glass prism

Consider a triangular glass prism ABC, with vertex A at the top. A ray of light PQ hits face AB and enters the glass. At Q, the light bends toward the normal (since glass is optically denser than air). The angle of incidence at this first face is $i$, and the angle of refraction inside the glass is $r_1$.

The refracted ray QN now travels through the glass until it hits the second face AC at point N. Here the light is going from glass back into air, so it bends away from the normal. The angle at which the ray hits the second face (measured from the normal inside the glass) is $r_2$, and the angle at which it leaves the prism is called the angle of emergence $e$.

The overall effect of these two refractions is to deflect the light from its original direction. The angle between the original incident ray (extended forward) and the final emergent ray is called the angle of deviation, $\delta$. This is the quantity we want to express in terms of the prism geometry and incidence conditions.

The Geometry Inside the Prism: Linking $r_1$, $r_2$, and $A$

The prism angle $A$ (the angle at vertex A, between the two refracting faces AB and AC) controls how much the two faces tilt relative to each other. There is a clean geometric relationship that ties $A$ to the two internal refraction angles.

Look at the quadrilateral AQNR formed by the apex A, the first refraction point Q, a point R inside the prism where the normals at Q and N would meet, and the second refraction point N. At Q and N, the normals are perpendicular to the prism faces, so the angles at Q and N in this quadrilateral are each 90 degrees. Since the four angles of any quadrilateral add up to 360 degrees:

$\angle A + \angle QRN + 90° + 90° = 360°$

This simplifies to:

$\angle A + \angle QRN = 180° \qquad \text{(i)}$

Now consider the triangle QNR formed by the refracted ray path inside the prism. The three angles of this triangle must add to 180 degrees:

$r_1 + r_2 + \angle QNR = 180° \qquad \text{(ii)}$

The angle at R in the quadrilateral is the same as the angle $\angle QNR$ in the triangle (they share the same vertex and sides). Comparing equations (i) and (ii):

$\angle A + \angle QRN = r_1 + r_2 + \angle QRN$

Cancel $\angle QRN$ from both sides:

$r_1 + r_2 = A \qquad \text{(9.34)}$

This is an important result. It says that no matter what the angle of incidence is, the sum of the two internal refraction angles always equals the prism angle. The prism geometry locks this relationship in place.

Deriving the Deviation Formula

Each face of the prism contributes to the total deviation. At the first face, the ray bends by an amount $(i - r_1)$, which is the difference between the incident angle and the refracted angle. At the second face, the ray bends by $(e - r_2)$.

The total deviation is the sum of these two contributions:

$\delta = (i - r_1) + (e - r_2)$

Rearranging:

$\delta = i + e - (r_1 + r_2)$

We already know from Eq. (9.34) that $r_1 + r_2 = A$. Substituting:

$\delta = i + e - A \qquad \text{(9.35)}$

This is the deviation formula for a prism. It tells us that the deviation depends on the angle of incidence $i$, the angle of emergence $e$ (which itself depends on $i$ through Snell’s law applied at both faces), and the fixed prism angle $A$.

How Deviation Changes with Angle of Incidence: The $i$-$\delta$ Curve

Fig 9.22: Plot of angle of deviation ($\delta$) versus angle of incidence ($i$) for a triangular prism

If you plot $\delta$ against $i$, you get a U-shaped curve. Here is what makes this curve interesting:

• Two values of $i$ for each $\delta$: For any given deviation (above the minimum), the curve shows two different angles of incidence that produce the same deviation. This happens because the deviation formula $\delta = i + e - A$ is symmetric in $i$ and $e$. If a certain pair $(i, e)$ gives a particular deviation, then swapping them (incident angle becomes $e$, emergence angle becomes $i$) gives the same $\delta$. Physically, this corresponds to tracing the same ray path backwards through the prism: the light follows the exact same route in reverse.

• One special point, the minimum: There is one unique angle of incidence where the deviation is the smallest. At this point the two values of $i$ merge into one. This minimum deviation is denoted $D_m$.

The Minimum Deviation Condition

At the bottom of the U-shaped curve, something special happens. Since $i$ and $e$ are interchangeable in the deviation formula and the curve has a single minimum:

$\text{At } \delta = D_m: \quad i = e$

When the angle of incidence equals the angle of emergence, the light path through the prism is perfectly symmetric. The ray enters and leaves at the same angle relative to their respective faces.

This symmetry has a direct consequence. Applying Snell’s law at both faces: since $i = e$ and the medium is the same on both sides, we must also have $r_1 = r_2$. The refracted ray inside the prism makes equal angles with both faces.

What does this look like geometrically? When $r_1 = r_2$, the refracted ray inside the prism travels parallel to the base of the prism. This is the condition for minimum deviation.

Deriving the Prism Formula for Refractive Index

The minimum deviation condition gives us clean expressions that lead to a powerful result.

Finding $r$ in terms of $A$: Since $r_1 = r_2 = r$ at minimum deviation, and we know $r_1 + r_2 = A$:

$2r = A$

$r = \frac{A}{2} \qquad \text{(9.36)}$

Finding $i$ in terms of $A$ and $D_m$: Since $i = e$ at minimum deviation, the deviation formula $\delta = i + e - A$ becomes:

$D_m = 2i - A$

Solving for $i$:

$i = \frac{A + D_m}{2} \qquad \text{(9.37)}$

Applying Snell’s law: At the first face, Snell’s law gives $n_1 \sin i = n_2 \sin r_1$. The refractive index of the prism relative to air is:

$n_{21} = \frac{n_2}{n_1} = \frac{\sin i}{\sin r}$

Substituting the expressions for $i$ and $r$:

$n_{21} = \frac{\sin\left[\dfrac{A + D_m}{2}\right]}{\sin\left[\dfrac{A}{2}\right]} \qquad \text{(9.38)}$

This is the prism formula. It gives the refractive index of the prism material in terms of two experimentally measurable angles: the prism angle $A$ and the minimum deviation $D_m$. Both can be measured precisely using a spectrometer, making this one of the most accurate methods for determining refractive index in a laboratory.

The Thin Prism Approximation

What happens when the prism angle $A$ is very small? Such a prism is called a thin prism. Since $A$ is small, the minimum deviation $D_m$ will also be small (as we will see from the result).

When angles are small (measured in radians), $\sin \theta \approx \theta$. Applying this approximation to the prism formula:

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

The factors of $1/2$ cancel:

$n_{21} \approx \frac{A + D_m}{A} = 1 + \frac{D_m}{A}$

Rearranging to solve for $D_m$:

$D_m = (n_{21} - 1) \, A$

This result tells us something practical: for a thin prism, the deviation depends only on the prism angle and the refractive index, not on the angle of incidence. Since $n_{21} - 1$ is always a fraction less than 1 for common glass (for example, 0.5 when $n = 1.5$), the deviation is even smaller than the prism angle itself. Thin prisms bend light by only a tiny amount.