The relative refractive index is n1/n2 = 1 / √(η² − 1). For η = 1.28 this gives n1/n2 = 1.2516. For η = 1.28 it is 1.25.
Problem statement
Two optical media have a plane boundary between them. Let θcr be the critical angle of incidence of a beam. Let θ1 be the angle of incidence at which the refracted beam is perpendicular to the reflected one. The beam comes from the optically denser medium. Find the relative refractive index of the media if sin θcr / sin θ1 = η = 1.28.
We assume the beam travels in the denser medium 1 (index n1) towards the rarer medium 2 (index n2). This means n1 is greater than n2 and a critical angle exists.
We assume the boundary is flat. This means the normal is the same everywhere and Snell's law applies at O.
We assume the reflected angle equals θ1. This means the angle between the reflected and refracted rays is β = 180° − θ1 − θ2.
Symbols: N = n1/n2 is the relative index we want. θ2 is the angle of refraction. β is the angle between the reflected ray OB and the refracted ray OC.
The big idea in plain words
Two special angles are linked to one number N. The critical angle is where the refracted ray just grazes the boundary. The second angle is where the reflected and refracted rays are at 90°. Both come from Snell's law. Their sines have a ratio η that depends only on N. So if η is given, N can be found.
The solution in each frame
Lab frame
The boundary is at rest. The full solution is in the next section.
Other frames
No other frame is needed. The media do not move, so the angles are the same for every observer at rest.
The solution step by step
At the critical angle the refracted ray grazes the boundary, so θ2 = 90°. Snell's law n1 sin θcr = n2 sin 90° gives
When the reflected and refracted rays are perpendicular, θ1 + θ2 = 90°, so θ2 = 90° − θ1. Snell's law at that angle reads
From tan θ1 = 1/N, a right triangle with sides 1 and N has hypotenuse √(1 + N²). So sin θ1 = 1 / √(1 + N²). Divide the two sines:
Square both sides: η² N² = 1 + N². Then N² (η² − 1) = 1, so
With η = 1.28: η² = 1.6384, η² − 1 = 0.6384, √0.6384 = 0.7990, so N = 1.2516. The critical angle is arcsin(0.7990) = 53.035° for your slider and the perpendicular angle is 38.625°. Your current η = 1.28 gives N = 1.2516.
A separate check finds θ1 where β = 90° by bisection, using Snell's law at every step. It gives θ1 = 38.6248°. Then sin θcr / sin θ1 = 1.28, which matches the given η = 1.28.
Five methods
1. Force method
Not needed. This is a ray-optics problem about directions, not forces.
2. Energy method
Not needed. The split of energy between the rays is not used.
3. Graph method
The middle graph plots η against N. It is a falling curve, so each η matches one N. The right graph shows θ2 against θ1 and the line θ2 = 90° − θ1, which meet at the perpendicular angle.
4. Calculus method
Not needed. A small check: dη/dN = −1 / (N² √(1 + N²)), which is negative, so η falls steadily as N rises and the answer is unique.
5. Check method
Limits: as N grows large, η tends to 1, so η must be above 1. When N = 1 the media are equal and η = √2, so η must also be below 1.414. The given 1.28 is inside this range. Units: all quantities are ratios, so there are no units to track.
Fastest: the formula N = 1 / √(η² − 1). Most reliable: the numerical solution of β = 90°.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| η | Yes | N, the critical angle, the 90° angle, the arcs, the rays and all graphs |
| Angle of incidence θ1 | Not the answer, but the picture | θ2, β, the arcs, the three rays, and total internal reflection past θcr |
| Wanted n1/n2 (given-answer mode) | Sets η | The model solves η for the wanted ratio |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Given ratio of sines | η | 1.28 |
| Relative index | n1/n2 | 1.2516 |
| Critical angle | θcr | 53.035° |
| Angle where the rays are at 90° | θ1 | 38.625° |
| Current angle of incidence | θ1 | 20° |
Exam traps
- Taking the beam from the rarer medium. A critical angle exists only when the beam starts in the denser medium.
- Writing tan θ1 = N. Here tan θ1 = n2/n1 = 1/N, because the beam starts in medium 1.
- Forgetting that η must be below √2, otherwise N would be below 1 and the media would be the wrong way round.
- Measuring angles from the boundary instead of from the normal.
- Reporting n2/n1 instead of n1/n2. The answer 1.25 is n1/n2.