Refraction of light

1. Overview

Refraction is the change in direction of a light ray when it passes from one transparent medium to another, such as from air into glass. This occurs because light changes speed when it enters a material of different optical density, a fundamental principle that allows lenses, cameras, and the human eye to function.

Key Definitions

• Normal: An imaginary line drawn at 90° (perpendicular) to the surface where the light ray hits the boundary.
• Angle of Incidence ($i$): The angle between the incident (incoming) ray and the normal.
• Angle of Refraction ($r$): The angle between the refracted ray and the normal.
• Critical Angle ($c$): The specific angle of incidence that results in an angle of refraction of 90°, where the light travels along the boundary.
• Total Internal Reflection (TIR): When light traveling from a denser medium hits a boundary at an angle greater than the critical angle, and all light is reflected back into the medium.
• Refractive Index ($n$): A measure of how much a medium slows down the speed of light.

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Core Content

The Passage of Light Through Boundaries

When light travels between two different mediums:

• Air to Glass (Less dense to more dense): Light slows down and bends towards the normal. ($i > r$)
• Glass to Air (More dense to less dense): Light speeds up and bends away from the normal. ($r > i$)
• Along the Normal: If light enters at 90° to the surface, its speed changes but its direction does not.

Experiment: Investigating Refraction

1. Place a transparent rectangular block on a piece of paper and trace its outline.
2. Shine a thin beam of light (from a ray box) into the side of the block at an angle.
3. Mark the path of the incident ray and the emergent ray with dots.
4. Remove the block, connect the dots, and draw the path of the ray inside the block.
5. Draw a normal at the point of entry and use a protractor to measure the angle of incidence ($i$) and angle of refraction ($r$).
6. Repeat for different shapes like semi-circular blocks to observe the critical angle.

Internal Reflection and Total Internal Reflection (TIR)

This occurs only when light moves from a more dense medium (glass/water) toward a less dense medium (air).

1. Angle $i <$ Critical Angle: Most light refracts out, some reflects internally.
2. Angle $i =$ Critical Angle: The refracted ray travels at 90° along the boundary.
3. Angle $i >$ Critical Angle: Total Internal Reflection occurs; no light escapes.

Everyday Examples:

• Prisms in Binoculars: Use TIR to turn light 180° to shorten the length of the device.
• Diamonds: The sparkle is caused by multiple internal reflections due to a very small critical angle.

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Extended Content (Extended Only)

The Refractive Index ($n$)

The refractive index is a ratio that describes the optical density of a material. It has no units.

1. Using Speeds: $$n = \frac{\text{speed of light in medium 1}}{\text{speed of light in medium 2}}$$ (Usually, medium 1 is a vacuum or air, where light is fastest).

2. Using Snell’s Law: For light entering a medium from air: $$n = \frac{\sin(i)}{\sin(r)}$$

3. Using the Critical Angle: When the angle of refraction is 90°: $$n = \frac{1}{\sin(c)}$$

Worked Example 1: Finding the angle of refraction

A ray of light hits a glass block with an angle of incidence of 45°. If the refractive index of glass is 1.5, calculate the angle of refraction.

1. $n = \sin(i) / \sin(r)$
2. $1.5 = \sin(45) / \sin(r)$
3. $\sin(r) = \sin(45) / 1.5 = 0.707 / 1.5 = 0.471$
4. $r = \sin^{-1}(0.471) = 28.1^\circ$

Worked Example 2: Finding refractive index from critical angle

The critical angle for light in a glass block is $39^\circ$. Calculate the refractive index.

1. $n = 1 / \sin(c)$
2. $n = 1 / \sin(39) = 1 / 0.629 = 1.59$

This calculation appears in almost every exam session. Sometimes you are given $n$ and must find $c$ instead: rearrange to $\sin(c) = 1/n$, then $c = \sin^{-1}(1/n)$.

Why a ray entering along the normal does not change direction

If a ray hits a boundary at $90^\circ$ to the surface (i.e., along the normal), the angle of incidence is $0^\circ$. Since $\sin(0) = 0$, there is no bending — the ray passes straight through. This is a common 1-mark question.

Optical Fibres

Optical fibres are thin strands of glass or plastic that use Total Internal Reflection to transmit pulses of light over long distances.

• Telecommunications: Light pulses carry data (internet, phone signals) much faster and with less signal loss than copper wires.
• Medicine: Used in endoscopes to see inside the human body.

Why are optical fibres so useful for telecommunications? Optical fibres have become the backbone of modern internet and phone networks for several reasons. Data encoded as light pulses can travel extremely quickly through the fibre, allowing very fast download and upload speeds. Because the light stays trapped inside the glass by TIR, very little energy is lost over long distances — much less than with traditional copper cables. The signal is also very difficult to intercept from outside, making the connection more secure. Glass fibres are also lighter and thinner than copper wires, so more data channels can fit in the same physical cable.

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Key Equations

Equation	Symbols	Units
$n = \frac{\sin i}{\sin r}$	$n$ = Refractive index, $i$ = incidence, $r$ = refraction	$n$ (None), $i/r$ (degrees)
$n = \frac{v_1}{v_2}$	$v_1$ = Speed in air, $v_2$ = Speed in medium	$v$ (m/s)
$n = \frac{1}{\sin c}$	$c$ = Critical angle	$c$ (degrees)

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Common Mistakes to Avoid

• ❌ Wrong: Measuring the angle between the ray and the surface of the block.
  • ✓ Right: Always measure the angle between the ray and the normal.
• ❌ Wrong: Showing light bending away from the normal when entering a denser medium (like glass).
  • ✓ Right: Remember "FAST" (Faster Away, Slower Towards). Light slows down in glass, so it moves towards the normal.
• ❌ Wrong: Drawing the light ray going straight through at an angle without bending.
  • ✓ Right: Light must change direction unless it enters exactly along the normal (0°).
• ❌ Wrong: Assuming TIR can happen when light goes from air into glass.
  • ✓ Right: TIR only happens when light tries to leave a denser medium to enter a less dense one.

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Exam Tips

1. The "Normal" is Priority: In any refraction diagram, draw the normal dashed line first. It is the reference point for all angles.
2. Check your Calculator: Ensure your calculator is in DEG (Degrees) mode, not RAD (Radians), before calculating sines.
3. Emergent Ray Parallelism: If a ray enters and leaves a rectangular block, the final emergent ray should be drawn parallel to the original incident ray.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A ray of light travels from air into a glass block, as shown below. The angle of incidence, $i$, is $50^\circ$. The refractive index of the glass is 1.50.

(a) Define the term refractive index. [1]

(b) Calculate the angle of refraction, $r$, inside the glass block. [3]

(c) State what happens to the speed of light as it enters the glass block. [1]

Worked Solution:

(a)

1. The refractive index is the ratio of the speed of light in a vacuum (or air) to the speed of light in the medium. $\boxed{\text{Refractive index} = \frac{\text{speed of light in vacuum}}{\text{speed of light in the medium}}}$ [Correct definition stated]

How to earn full marks:

• State the definition of refractive index, including the ratio of speeds.

(b)

1. State Snell's Law equation. $n = \frac{\sin i}{\sin r}$ [Correct formula stated]

2. Rearrange the equation to solve for the angle of refraction. $\sin r = \frac{\sin i}{n} = \frac{\sin 50^\circ}{1.50} = 0.5107$ [Correct substitution and rearrangement]

3. Calculate the angle of refraction. $r = \sin^{-1}(0.5107) = 30.7^\circ$ $\boxed{r = 30.7^\circ}$ [Correct value for the angle of refraction, with units]

How to earn full marks:

• State Snell's Law.
• Correct substitution of $i$ and $n$ into the formula.
• Calculate the correct angle of refraction with units.

(c)

1. The speed of light decreases as it enters the glass block. $\boxed{\text{The speed of light decreases}}$ [Correct statement about the speed of light]

How to earn full marks:

• State that the speed decreases.

Common Pitfall: Make sure you use the correct formula for refractive index. Also, remember that the angle of refraction is smaller than the angle of incidence when light travels from air into a denser medium like glass. Double-check your calculator is in degree mode!

Exam-Style Question 2 — Extended Response [8 marks]

Question:

Optical fibres are used in endoscopes for medical imaging.

(a) Describe what is meant by total internal reflection (TIR). [3]

(b) State two conditions necessary for total internal reflection to occur. [2]

(c) The refractive index of the glass used in an optical fibre is 1.60. Calculate the critical angle, $c$, for light travelling within the fibre. [3]

Worked Solution:

(a)

1. Total internal reflection is when light is reflected entirely back into the denser medium. [Correct description of reflection]
2. This happens when light attempts to exit a denser medium and enter a less dense medium. [Correct description of the media involved]
3. The angle of incidence must be greater than the critical angle. [Correct description of the angle of incidence]

How to earn full marks:

• Light is reflected back into the denser medium.
• Light is travelling from a denser to a less dense medium.
• The angle of incidence is greater than the critical angle.

(b)

1. Light must be travelling from a denser medium to a less dense medium. $\boxed{\text{Light must be travelling from a denser medium to a less dense medium}}$ [Correct statement of medium requirement]
2. The angle of incidence must be greater than the critical angle. $\boxed{\text{The angle of incidence must be greater than the critical angle}}$ [Correct statement of angle requirement]

How to earn full marks:

• State the condition of light travelling from a denser to a less dense medium.
• State the condition of the angle of incidence being greater than the critical angle.

(c)

1. State the formula for the critical angle. $n = \frac{1}{\sin c}$ [Correct formula stated]
2. Rearrange the equation to solve for critical angle. $\sin c = \frac{1}{n} = \frac{1}{1.60} = 0.625$ [Correct substitution and rearrangement]
3. Calculate the critical angle. $c = \sin^{-1}(0.625) = 38.7^\circ$ $\boxed{c = 38.7^\circ}$ [Correct value for the critical angle, with units]

How to earn full marks:

• State the formula for the critical angle.
• Correct substitution of $n$ into the formula.
• Calculate the correct critical angle with units.

Common Pitfall: Remember that total internal reflection only happens when light goes from a denser to a less dense medium. Also, make sure you use the correct formula relating refractive index and critical angle – it's easy to mix up!

Exam-Style Question 3 — Short Answer [6 marks]

Question:

A student investigates the refraction of light through a rectangular glass block. They shine a ray of light at different angles of incidence onto the block and measure the corresponding angles of refraction.

(a) Describe how the student could accurately measure the angle of incidence and the angle of refraction. [3]

(b) State two precautions the student should take to ensure accurate results. [2]

(c) The student calculates the refractive index of the glass to be 1.52. State what this value represents. [1]

Worked Solution:

(a)

1. Shine a ray of light from a ray box towards the glass block. [Correct starting point]
2. Place the glass block on a piece of paper and trace around it. Also, mark the incident and refracted rays with crosses. [Good description of the setup]
3. Remove the block and draw the incident and refracted rays with a ruler. Draw the normal at the point of incidence. Use a protractor to measure the angle of incidence and the angle of refraction. [Clear description of the measurement process]

How to earn full marks:

• Shine a ray of light towards the glass block.
• Trace the block and mark the rays.
• Remove the block, draw the rays and normal, and measure angles with a protractor.

(b)

1. Ensure the ray box produces a narrow, well-defined ray of light. $\boxed{\text{Use a narrow ray of light}}$ [Correct precaution stated]
2. View the ray from directly above to avoid parallax error when marking the rays. $\boxed{\text{View the ray from directly above}}$ [Correct precaution stated]

How to earn full marks:

• Use a narrow ray of light.
• View the ray from directly above to avoid parallax error.

(c)

1. This value represents the ratio of the speed of light in a vacuum (or air) to the speed of light in the glass. $\boxed{\text{Ratio of the speed of light in vacuum to the speed of light in glass}}$ [Correct statement of the refractive index definition]

How to earn full marks:

• State that the refractive index is the ratio of the speed of light in a vacuum (or air) to the speed of light in the glass.

Common Pitfall: When describing the experiment, be really specific about how to measure the angles accurately. Don't just say "use a protractor" – explain where to draw the normal and how to align the protractor. Also, remember that parallax error can affect your readings, so viewing from directly above is important.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A waterproof LED torch is used underwater. The LED emits light with a wavelength of 580 nm in air. The refractive index of the torch lens is 1.50 and the refractive index of water is 1.33.

(a) Calculate the speed of light in the torch lens. (Speed of light in air = $3.0 \times 10^8$ m/s) [3]

(b) Calculate the wavelength of the light inside the water. [3]

(c) Explain why the torch lens is designed to have a small curvature (i.e., almost flat) where the light exits the lens and enters the water. [3]

Worked Solution:

(a)

1. State the refractive index formula for speed of light. $n = \frac{\text{speed of light in air}}{\text{speed of light in lens}}$ [Correct formula stated]
2. Rearrange to find the speed of light in the lens. $\text{speed of light in lens} = \frac{\text{speed of light in air}}{n} = \frac{3.0 \times 10^8 \text{ m/s}}{1.50}$ [Correct substitution and rearrangement]
3. Calculate the speed of light in the lens. $\text{speed of light in lens} = 2.0 \times 10^8 \text{ m/s}$ $\boxed{\text{speed of light in lens} = 2.0 \times 10^8 \text{ m/s}}$ [Correct value with units]

How to earn full marks:

• State the refractive index formula.
• Substitute the values correctly.
• Calculate the correct speed of light in the lens with units.

(b)

1. Recognize that the frequency of light remains constant when it enters the water. $v = f\lambda$ [Correct idea]
2. Calculate the frequency of the light in air (and therefore in water). $f = \frac{v}{\lambda} = \frac{3.0 \times 10^8 \text{ m/s}}{580 \times 10^{-9} \text{ m}} = 5.17 \times 10^{14} \text{ Hz}$ [Correct use of formula and unit conversion]
3. Calculate the speed of light in water. $v_{water} = \frac{3.0 \times 10^8 \text{ m/s}}{1.33} = 2.26 \times 10^8 \text{ m/s}$ [Correct use of refractive index formula]
4. Calculate the wavelength of the light in water. $\lambda_{water} = \frac{v_{water}}{f} = \frac{2.26 \times 10^8 \text{ m/s}}{5.17 \times 10^{14} \text{ Hz}} = 4.37 \times 10^{-7} \text{ m} = 437 \text{ nm}$ $\boxed{\lambda_{water} = 437 \text{ nm}}$ [Correct value with units]

How to earn full marks:

• Calculate the frequency of light in air.
• Calculate the speed of light in water using the refractive index formula.
• Calculate the correct wavelength of light in water with units.

(c)

1. A small curvature means the angle of incidence of the light rays will be small. [Correct effect of small curvature]
2. With a small angle of incidence, the angle of refraction will also be small. [Correct reasoning about angle of refraction]
3. This minimizes refraction and reflection at the boundary, allowing more light to exit the lens and travel into the water. This makes the torch brighter. [Correct explanation of the effect on light output]

How to earn full marks:

• Small curvature leads to a small angle of incidence.
• Small angle of incidence leads to a small angle of refraction.
• Minimizing refraction/reflection increases the light output.

Common Pitfall: Part (b) is tricky because you need to remember that the frequency of the light stays the same when it enters the water, but the speed and wavelength change. Don't forget to convert nanometers (nm) to meters (m) in your calculations! Also, make sure you understand how a small curvature affects the angle of incidence.