Physics · Simulation and virtual lab Free
Light: reflection, refraction and the critical angle
Shine a ray box at a mirror or a glass block, measure the angles with a protractor and discover the laws of reflection and refraction.
- You will learn to
- Verify the laws of reflection with a plane mirror.
- Show that sin i ÷ sin r is constant and find the refractive index from a graph.
- Find the critical angle and observe total internal reflection.
- You will measure
- Angles of incidence and refraction (°); then n and the critical angle.
- Time needed
- About 35–45 min
- Equipment
- Nothing extra. Real version: ray box, plane mirror, rectangular and semicircular glass blocks, protractor, white paper, optical pins.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To verify the laws of reflection and refraction, to find the refractive index of glass from a graph of sin i against sin r, and to find the critical angle.
What to do
- Choose the Plane mirror. Set i = 20°, 40° and 60° and record each time. Compare r with i.
- Choose the Rectangular glass block (glass). Set i = 10° and tap Record i and r.
- Repeat for i = 20°, 30°, 40°, 50°, 60° and 70°. Look at the emergent ray: what do you notice?
- Look at the graph of sin (angle in air) against sin (angle in the block): find the refractive index n from the gradient.
- Choose the Semicircular block. Increase i slowly from 20° until the refracted ray disappears. Record the angles just before and after.
- Hide the angle values and practise reading the angles on the protractor yourself. Then try the water box.
Results
Questions
State the two laws of reflection.
The angle of reflection equals the angle of incidence, and the incident ray, the reflected ray and the normal all lie in the same plane.
Why does a ray bend towards the normal when it enters glass from air?
Light travels more slowly in glass than in air, so when it enters the glass at an angle it changes direction towards the normal.
State Snell's law.
For two given media, the ratio sin i ÷ sin r is constant; this constant is the refractive index n.
A ray enters glass (n = 1.50) at i = 40°. Find the angle of refraction.
sin r = sin 40° ÷ 1.50 = 0.643 ÷ 1.50 = 0.429, so r = 25.4°.
Why is the emergent ray from a rectangular block parallel to the incident ray?
The two faces of the block are parallel, so the bending at the second face exactly reverses the bending at the first face. The ray is only shifted sideways (lateral displacement).
Find the critical angle of glass with n = 1.50.
sin c = 1 ÷ n = 1 ÷ 1.50 = 0.667, so c = 41.8°.
For teachers: expected results and how the model works
Glass n = 1.50: i = 10° → r = 6.6°; 30° → 19.5°; 50° → 30.7°; 70° → 38.8°. The gradient of sin i against sin r is 1.50 (perspex 1.49, water 1.33). Critical angles: glass 41.8°, perspex 42.2°, water 48.8°. With “Hide” selected learners must read the protractor (5° ticks, labels every 30°) and will typically be within ±1°. The emergent ray is drawn parallel to the incident ray (dashed line shows the undeviated path).
Exam practice Exam set
National-exam style questions on this simulation (16 marks). Try each one on paper first, then open the marking scheme.
A ray of light passes from air into water (n = 1.33) at an angle of incidence of 50°. (a) Calculate the angle of refraction. (b) Draw a labelled diagram showing the incident ray, the normal and the refracted ray. (4 marks)
Marking scheme: (a) sin r = sin 50° ÷ 1.33 = 0.766 ÷ 1.33 = 0.576 (1), r = 35.2° (1). (b) Normal drawn at 90° to the surface (1); refracted ray bent towards the normal, with arrows and labels (1).
State two conditions for total internal reflection to occur. (2 marks)
Marking scheme: Light must travel from an optically denser medium to a less dense medium (1), and the angle of incidence must be greater than the critical angle (1).
A learner obtained these results with a glass block: i = 20°, r = 13°; i = 40°, r = 25.5°; i = 60°, r = 35°. (a) Calculate sin i ÷ sin r for each pair. (b) State what the results show. (4 marks)
Marking scheme: (a) 0.342 ÷ 0.225 = 1.52 (1); 0.643 ÷ 0.431 = 1.49 (1); 0.866 ÷ 0.574 = 1.51 (1). (b) The ratio is constant (about 1.5): Snell's law is obeyed and n of glass ≈ 1.5 (1).
Explain how an optical fibre carries light around a bend. (3 marks)
Marking scheme: Light enters the glass core and strikes the boundary with the cladding at an angle greater than the critical angle (1); it is totally internally reflected (1); this repeats many times along the fibre, so the light follows the fibre with very little loss (1).
The critical angle of a diamond is 24.4°. Calculate its refractive index and explain why diamonds sparkle. (3 marks)
Marking scheme: n = 1 ÷ sin 24.4° = 1 ÷ 0.413 = 2.42 (1, 1). The very small critical angle means most light that enters is totally internally reflected many times before leaving, so the diamond sparkles (1).
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Light: reflection, refraction and the critical angle
Aim: To verify the laws of reflection and refraction, to find the refractive index of glass from a graph of sin i against sin r, and to find the critical angle.
Apparatus (real laboratory): ray box, plane mirror, rectangular and semicircular glass blocks, protractor, white paper, optical pins.
What to do (with the simulation on www.donatstem.co.rw)
- Choose the Plane mirror. Set i = 20°, 40° and 60° and record each time. Compare r with i.
- Choose the Rectangular glass block (glass). Set i = 10° and tap Record i and r.
- Repeat for i = 20°, 30°, 40°, 50°, 60° and 70°. Look at the emergent ray: what do you notice?
- Look at the graph of sin (angle in air) against sin (angle in the block): find the refractive index n from the gradient.
- Choose the Semicircular block. Increase i slowly from 20° until the refracted ray disappears. Record the angles just before and after.
- Hide the angle values and practise reading the angles on the protractor yourself. Then try the water box.
Results
| # | Set-up | i (°) | r (°) | sin i | sin r |
|---|
| 1 | | | | | |
| 2 | | | | | |
| 3 | | | | | |
| 4 | | | | | |
| 5 | | | | | |
| 6 | | | | | |
| 7 | | | | | |
| 8 | | | | | |
| 9 | | | | | |
Graph
Plot sin (angle in air) on the vertical axis against sin (angle in the block) on the horizontal axis. Draw the best straight line through the origin and find its gradient n.
Questions
- State the two laws of reflection. [2]
- Why does a ray bend towards the normal when it enters glass from air? [2]
- State Snell's law. [2]
- A ray enters glass (n = 1.50) at i = 40°. Find the angle of refraction. [2]
- Why is the emergent ray from a rectangular block parallel to the incident ray? [2]
- Find the critical angle of glass with n = 1.50. [2]
- A ray of light passes from air into water (n = 1.33) at an angle of incidence of 50°. (a) Calculate the angle of refraction. (b) Draw a labelled diagram showing the incident ray, the normal and the refracted ray. [4]
- State two conditions for total internal reflection to occur. [2]
- A learner obtained these results with a glass block: i = 20°, r = 13°; i = 40°, r = 25.5°; i = 60°, r = 35°. (a) Calculate sin i ÷ sin r for each pair. (b) State what the results show. [4]
- Explain how an optical fibre carries light around a bend. [3]
- The critical angle of a diamond is 24.4°. Calculate its refractive index and explain why diamonds sparkle. [3]
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Light: reflection, refraction and the critical angle
Expected results
Glass n = 1.50: i = 10° → r = 6.6°; 30° → 19.5°; 50° → 30.7°; 70° → 38.8°. The gradient of sin i against sin r is 1.50 (perspex 1.49, water 1.33). Critical angles: glass 41.8°, perspex 42.2°, water 48.8°. With “Hide” selected learners must read the protractor (5° ticks, labels every 30°) and will typically be within ±1°. The emergent ray is drawn parallel to the incident ray (dashed line shows the undeviated path).
Answers
- State the two laws of reflection. [2]
The angle of reflection equals the angle of incidence, and the incident ray, the reflected ray and the normal all lie in the same plane.
- Why does a ray bend towards the normal when it enters glass from air? [2]
Light travels more slowly in glass than in air, so when it enters the glass at an angle it changes direction towards the normal.
- State Snell's law. [2]
For two given media, the ratio sin i ÷ sin r is constant; this constant is the refractive index n.
- A ray enters glass (n = 1.50) at i = 40°. Find the angle of refraction. [2]
sin r = sin 40° ÷ 1.50 = 0.643 ÷ 1.50 = 0.429, so r = 25.4°.
- Why is the emergent ray from a rectangular block parallel to the incident ray? [2]
The two faces of the block are parallel, so the bending at the second face exactly reverses the bending at the first face. The ray is only shifted sideways (lateral displacement).
- Find the critical angle of glass with n = 1.50. [2]
sin c = 1 ÷ n = 1 ÷ 1.50 = 0.667, so c = 41.8°.
- A ray of light passes from air into water (n = 1.33) at an angle of incidence of 50°. (a) Calculate the angle of refraction. (b) Draw a labelled diagram showing the incident ray, the normal and the refracted ray. [4]
(a) sin r = sin 50° ÷ 1.33 = 0.766 ÷ 1.33 = 0.576 (1), r = 35.2° (1). (b) Normal drawn at 90° to the surface (1); refracted ray bent towards the normal, with arrows and labels (1).
- State two conditions for total internal reflection to occur. [2]
Light must travel from an optically denser medium to a less dense medium (1), and the angle of incidence must be greater than the critical angle (1).
- A learner obtained these results with a glass block: i = 20°, r = 13°; i = 40°, r = 25.5°; i = 60°, r = 35°. (a) Calculate sin i ÷ sin r for each pair. (b) State what the results show. [4]
(a) 0.342 ÷ 0.225 = 1.52 (1); 0.643 ÷ 0.431 = 1.49 (1); 0.866 ÷ 0.574 = 1.51 (1). (b) The ratio is constant (about 1.5): Snell's law is obeyed and n of glass ≈ 1.5 (1).
- Explain how an optical fibre carries light around a bend. [3]
Light enters the glass core and strikes the boundary with the cladding at an angle greater than the critical angle (1); it is totally internally reflected (1); this repeats many times along the fibre, so the light follows the fibre with very little loss (1).
- The critical angle of a diamond is 24.4°. Calculate its refractive index and explain why diamonds sparkle. [3]
n = 1 ÷ sin 24.4° = 1 ÷ 0.413 = 2.42 (1, 1). The very small critical angle means most light that enters is totally internally reflected many times before leaving, so the diamond sparkles (1).