When the least deviation equals the refracting angle, cos(θ/2) = n/2, so θ = 2 arccos(n/2). For n = 1.5 this is θ = 82.819° at your index n = 1.5, and 82.8° (about 83°) for n = 1.5.
Problem statement
The least deflection angle of a certain glass prism is equal to its refracting angle. Find the refracting angle.
We assume the glass has refractive index n = 1.5 (ordinary glass) and the prism is in air. This means Snell's law reads sin α1 = n sin β1. The slider lets you change n.
We assume light of one colour. This means n has a single value.
We assume the ray crosses the prism in a plane perpendicular to its edge. This means the problem stays two-dimensional.
Symbols: θ is the refracting (apex) angle. α1 and α2 are the angles of incidence and emergence. β1 and β2 are the angles inside the prism. α is the deviation, and Dmin is its least value.
The big idea in plain words
As the angle of incidence changes, the deviation first falls and then rises. The lowest point is the least deviation. It happens when the ray crosses the prism symmetrically: it enters and leaves at the same angle and runs parallel to the base inside. At that moment the geometry is simple, so one Snell equation links the least deviation, the apex angle and n. Setting the least deviation equal to θ then gives a short equation for θ.
The solution in each frame
Lab frame
The prism is at rest. The full solution is in the next section.
Other frames
No other frame is needed. Nothing moves, so one frame is enough.
The solution step by step
For any ray through a prism, the two normals meet at the apex angle, so θ = β1 + β2. The total turn of the ray is the exact deviation
The least deviation occurs on the symmetric path, where α1 = α2 and so β1 = β2 = θ/2. (Another path with a different α1 gives a larger deviation, which the first graph shows.) On this path α = 2α1 − θ, so α1 = (α + θ)/2. Snell's law at the first face, sin α1 = n sin β1, becomes
The problem says the least deviation equals the refracting angle, α = θ. Put this in:
Use sin θ = 2 sin(θ/2) cos(θ/2) and divide both sides by sin(θ/2), which is not zero:
For n = 1.5: cos(θ/2) = 0.75, so θ/2 = 41.4° and θ = 82.8°, about 83°. For your slider n = 1.5, θ = 82.819°. For the prism now on screen, the least deviation is 82.825°.
A separate check does not use the symmetric-path formula. It searches over all angles of incidence for the smallest deviation, and then adjusts the apex angle until that smallest deviation equals the apex angle. It finds θ = 82.8192°, the same as 2 arccos(n/2) = 82.819°.
Five methods
1. Force method
Not needed. This is a ray-optics problem about directions, not forces.
2. Energy method
Not needed. Only the directions of the ray matter.
3. Graph method
The first graph plots the deviation against the angle of incidence. It dips to a lowest point, the least deviation, at the symmetric path. The second graph plots Dmin against θ and the line Dmin = θ. They cross at the answer.
4. Calculus method
Setting dα/dα1 = 0 for α = α1 + α2 − θ, with sin α1 = n sin β1 and sin α2 = n sin β2, gives α1 = α2. This is why the symmetric path is the least deviation.
5. Check method
Limits: n must be below 2, because cos(θ/2) cannot exceed 1. As n tends to 2, θ tends to 0. As n tends to 1, θ tends to 120°. Units: angles in degrees.
Fastest: θ = 2 arccos(n/2). Most reliable: the numerical search for the smallest deviation.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Refractive index n | Yes | The answer θ = 2 arccos(n/2), the angles inside, the deviation, the drawing and all graphs |
| Apex angle θ | Not the answer, but the picture | The shape of the prism, the angles inside, the deviation and the least deviation |
| Angle of incidence α1 | No | The path of the ray and the deviation now. The least deviation does not depend on it. |
| Given-answer mode | Sets θ and α1 | The prism is set to the answer angle and the ray to the symmetric path |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Refractive index | n | 1.5 |
| Apex angle on screen | θ | 82.82° |
| Angle of incidence | α1 | 82.82° |
| Least deviation | Dmin | 82.825° |
| Answer | 2 arccos(n/2) | 82.819° |
Exam traps
- Using any angle of incidence. The least deviation is only on the symmetric path, where α1 = α2.
- Putting θ in place of θ/2 in Snell's law. The angle inside at each face is θ/2, not θ.
- Dividing by sin(θ/2) without noting that it is not zero for a real prism.
- Giving θ/2 = 41.4° as the answer. The refracting angle is twice that, 82.8°.
- Using n of 2 or more. Then cos(θ/2) would be 1 or more and no such prism exists.