The beam leaves parallel to its first path, shifted sideways by x = d sin θ (1 − cos θ / √(n² − sin² θ)). For n = 1.5, θ = 60° and d = 6 cm, x = 3.075 cm. For glass with n = 1.5, θ = 60° and d = 6.0 cm it is 3.07 cm.
Problem statement
A light beam falls on a plane-parallel glass plate of thickness d = 6.0 cm. The angle of incidence is θ = 60°. Find the deflection (sideways shift) of the beam that passes through the plate.
We assume the glass has refractive index n = 1.5 and the plate is in air. This means Snell's law reads sin θ = n sin β. The slider lets you change n.
We assume both faces are flat and parallel. This means the angle at which the beam leaves in air is θ again, so it leaves parallel to the first beam.
We assume only the refracted beam is followed. This means the small reflected parts are ignored.
Symbols: θ is the angle of incidence. β is the angle of the beam inside the glass. d is the plate thickness. x is the shift, the shortest distance between the leaving beam and the first beam carried straight on.
The big idea in plain words
The beam bends toward the normal when it enters the glass and bends back by the same amount when it leaves. So it leaves in its first direction, but along a line that has moved sideways. That sideways distance is x. It grows with the thickness, with the angle, and with the bending power of the glass.
The solution in each frame
Lab frame
The plate is at rest. The full solution is in the next section.
Other frames
No other frame is needed. The plate does not move, so one frame is enough.
The solution step by step
Call the entry point O and the exit point M. Snell's law at O gives the angle β in the glass:
The beam crosses the plate along OM. Its length comes from the right triangle with side d:
Inside the glass the beam has turned by the angle (θ − β) from its first direction. The shift x is the distance from M to the first line carried on. In the right triangle O, M, N, with the right angle at N:
Expand sin(θ − β) = sin θ cos β − cos θ sin β and divide by cos β:
From Snell's law sin β = sin θ / n, so cos β = √(n² − sin² θ) / n and tan β = sin θ / √(n² − sin² θ). Put this in:
Numbers for n = 1.5, θ = 60°, d = 6.0 cm: sin θ = 0.866, cos θ = 0.5, n² − sin² θ = 2.25 − 0.75 = 1.5, and √1.5 = 1.225. Then x = 6.0 × 0.866 × (1 − 0.5/1.225) = 6.0 × 0.866 × 0.592 = 3.07 cm. Your current values give x = 3.075 cm, which is 51.2 % of the thickness.
A separate check places O and M as points, with M at 4.243 cm along the plate, and measures the distance of M from the first line with a cross product. It gives 3.0748 cm, the same as the formula, x = 3.075 cm.
Five methods
1. Force method
Not needed. This is a ray-optics problem about directions, not forces.
2. Energy method
Not needed. Only the direction of the beam matters, not its energy.
3. Graph method
The first graph plots x against θ. It starts at zero for a straight beam and rises toward d for a grazing beam. The second graph shows x against n, and the third shows x rising in a straight line with d.
4. Calculus method
Not needed for the answer. For small angles the shift is x ≈ d θ (1 − 1/n), because sin θ ≈ θ and cos θ ≈ 1, which is a quick estimate.
5. Check method
Limits: at θ = 0 the shift is zero. At n = 1 the shift is zero, since there is no glass effect. As θ tends to 90° the shift tends to d. Units: x is in the same unit as d.
Fastest: x = d sin θ (1 − cos θ / √(n² − sin² θ)). Most reliable: the cross-product measurement of the point M.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Refractive index n | Yes | β, the path OM, the shift x, the drawing and all graphs |
| Angle of incidence θ | Yes | β, x, the rays, the arcs and the dots on the graphs |
| Thickness d | Yes | The plate height, OM, x and the graphs |
| Wanted shift (given-answer mode) | Sets θ | The model solves θ for the wanted shift |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Refractive index | n | 1.5 |
| Angle of incidence | θ | 60° |
| Thickness | d | 6 cm |
| Angle in the glass | β | 35.264° |
| Shift | x | 3.075 cm |
Exam traps
- Giving the path length OM as the answer. The shift x is measured across the first beam, not along the glass path.
- Using the angle of incidence θ in place of the turn (θ − β) in x = OM sin(θ − β).
- Using the difference of angles in degrees without the sine: x is OM sin(θ − β), not OM (θ − β).
- Forgetting that the beam leaves parallel to the first beam. The direction does not change, only the position.
- Putting n in the wrong place in Snell's law: for air to glass it is sin θ = n sin β.