Refraction Through a Prism

The Geometry of a Prism

A ray crosses a triangular prism (angle A): refraction at face AB (angles i, $r_1$) and at face AC (angles $r_2$, e). The angle between the emergent and incident rays is the angle of deviation d.

Two lines of pure geometry:

1. In the quadrilateral formed by the faces and normals, $\angle A + \angle QNR = 180^{\circ}$; the triangle of normals gives $r_1 + r_2 + \angle QNR = 180^{\circ}$. Hence

$$\boxed{r_1 + r_2 = A}$$

2. Total deviation = the two bends summed: $d = (i - r_1) + (e - r_2)$, so

$$\boxed{d = i + e - A}$$

Plot d against i: a U-shaped curve with a single minimum. Every deviation except the minimum corresponds to two incidence angles (i and e swap — the path is reversible); at the bottom of the U they coincide.

Minimum Deviation: The Prism's Sweet Spot

At minimum deviation $D_m$: $i = e$ and $r_1 = r_2$ — the ray runs parallel to the base inside the prism. Then $r = A/2$ and $i = (A + D_m)/2$, and Snell's law at either face delivers the jewel:

$$\boxed{n_{21} = \frac{\sin\left[\frac{A + D_m}{2}\right]}{\sin\left[\frac{A}{2}\right]}}$$

Measure A and $D_m$ on a spectrometer table and the refractive index follows — the standard laboratory method.

Thin (small-angle) prisms: sines become angles, and

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

— thin prisms deviate little, and the deviation is independent of i (for small i).

[NEET Important] The two prism facts asked every year: at minimum deviation the internal ray is parallel to the base, and i = e. [JEE Tip] A favourite special case: A = 60 degrees and $D_m = 30$ degrees give $n = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \sqrt{2}$ — memorise the trio (60, 30, $\sqrt 2$).

Solved Examples

Example 1: The classic 60-30 prism [NEET Numerical]

A prism of angle 60 degrees shows minimum deviation 30 degrees. Find its refractive index.

Solution:

1. $n = \frac{\sin[(A+D_m)/2]}{\sin[A/2]} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}}$.
2. $= \frac{0.707}{0.5} = 1.414 = \sqrt{2}$.
3. The (60, 30, $\sqrt2$) trio — the most reused prism numbers in exam history.

Example 2: Deviation from the angles [NEET Numerical]

Light enters a 60-degree prism at i = 45 degrees and emerges at e = 35 degrees. Find the deviation.

Solution:

1. $d = i + e - A = 45 + 35 - 60$.
2. $d = 20^{\circ}$.
3. No Snell's law needed — the relation is pure geometry.

Example 3: Inside the prism [JEE Numerical]

Light strikes a 60-degree prism (n = 1.5) at $i = 45^{\circ}$. Find $r_1$ and hence $r_2$.

Solution:

1. Snell at entry: $\sin r_1 = \frac{\sin 45^{\circ}}{1.5} = \frac{0.707}{1.5} = 0.471 \Rightarrow r_1 = 28.1^{\circ}$.
2. $r_2 = A - r_1 = 60 - 28.1 = 31.9^{\circ}$.
3. The pair always sums to A — use it instead of re-running Snell inside.

Example 4: Thin-prism deviation [NEET Numerical]

A thin prism of angle 6 degrees (n = 1.5) deviates light by how much?

Solution:

1. $D_m = (n-1)A = 0.5 \times 6$.
2. $D_m = 3^{\circ}$ — small prisms, small bends, independent of incidence angle.
3. (Two such prisms opposed make a 'prism pair' deflecting without dispersion tricks — instrument-makers' staple.)

Example 5: At minimum deviation, find i [JEE Numerical]

For the 60-degree, $\sqrt2$-index prism at minimum deviation, what is the angle of incidence?

Solution:

1. $i = \frac{A + D_m}{2} = \frac{60 + 30}{2}$.
2. $i = 45^{\circ}$ (= e, by symmetry).
3. And inside, $r_1 = r_2 = A/2 = 30^{\circ}$ with the ray parallel to the base — the full minimum-deviation portrait.

Example 6: Index from the lab [JEE Numerical]

A spectrometer measures A = 60.0 degrees and $D_m = 38.0$ degrees for a glass prism. Find n.

Solution:

1. $n = \frac{\sin 49^{\circ}}{\sin 30^{\circ}} = \frac{0.755}{0.5}$.
2. $n = 1.51$ — crown glass.
3. This is exactly how Table 9.1's indices are measured.

Example 7: Two angles, one deviation

Why does each deviation (except $D_m$) correspond to two incidence angles?

Solution:

1. $d = i + e - A$ is symmetric in i and e: swapping them leaves d unchanged.
2. Physically, light paths are reversible — run the ray backwards and i, e exchange roles with the same d.
3. Only at the minimum do the two solutions merge ($i = e$): the U-curve's single bottom.

Example 8: Grazing the limit [JEE Numerical]

Under what condition does a ray fail to emerge from the prism's second face AC?

Solution:

1. The ray fails to emerge when it hits AC beyond the critical angle: $r_2 > i_c$.
2. Since $r_2 = A - r_1$, small i (small $r_1$) pushes $r_2$ up — too-shallow entry causes TIR at the second face.
3. Emergence requires $A - r_1 \le i_c$, i.e. $A \le r_1 + i_c \le 2i_c$: a prism with $A > 2i_c$ transmits nothing at any incidence — a neat JEE trap.

Example 9: Parallel to the base — why it matters

State two practical consequences of the minimum-deviation symmetry.

Solution:

1. Measurement: at $D_m$ the geometry is unique and stable (d changes only second-order with i), so spectrometer readings are sharpest there — ideal for measuring n.
2. Alignment: the internal ray parallel to the base means symmetric entry/exit — rotating the prism slightly barely changes d near the minimum (the turning point).
3. Exams reward the phrase: 'at minimum deviation the refracted ray inside the prism is parallel to the base; i = e and $r_1 = r_2 = A/2$.'

Example 10: Building the n-formula

Derive $n = \sin[(A+D_m)/2]/\sin(A/2)$ in three lines.

Solution:

1. At minimum: $r_1 = r_2 = r$, and $r_1 + r_2 = A$ gives $r = A/2$.
2. $D_m = 2i - A$ (from $d = i + e - A$ with $i = e$) gives $i = (A + D_m)/2$.
3. Snell at the first face: $n = \frac{\sin i}{\sin r} = \frac{\sin[(A+D_m)/2]}{\sin(A/2)}$. Done — and examiners award each line.