CIE iGCSE Co-ordinated Sciences-P3.2.2 Refraction of light- Study Notes- New Syllabus
CIE iGCSE Co-ordinated Sciences-P3.2.2 Refraction of light – Study Notes
CIE iGCSE Co-ordinated Sciences-P3.2.2 Refraction of light – Study Notes -CIE iGCSE Co-ordinated Sciences – per latest Syllabus.
Key Concepts:
CIE iGCSE Co-Ordinated Sciences-Concise Summary Notes- All Topics
Refraction
(a) Definition of Refraction
Refraction is the change in direction of a light ray when it passes from one medium into another due to a change in speed of light.
- If light passes from a less dense to a more dense medium (e.g. air → glass), it slows down and bends towards the normal.
- If light passes from a more dense to a less dense medium (e.g. glass → air), it speeds up and bends away from the normal.
- The frequency remains constant, but the wavelength changes since: $ v = f \lambda $
(b) Normal, Angle of Incidence, and Angle of Refraction
- Normal: An imaginary line drawn perpendicular to the boundary at the point where the ray strikes.
- Angle of incidence (\(\theta_i\)): The angle between the incident ray and the normal.
- Angle of refraction (\(\theta_r\)): The angle between the refracted ray and the normal.
These definitions are used when applying Snell’s Law:
$ n_1 \sin \theta_i = n_2 \sin \theta_r $
Example
A light ray enters water from air at an angle of incidence of \( 45^\circ \). The refractive index of water is \( 1.33 \). Calculate the angle of refraction inside the water.
▶️Answer/Explanation
Step (1) – Write Snell’s Law:
$ n_1 \sin \theta_i = n_2 \sin \theta_r $
Here, \( n_1 = 1.0 \) (air), \( \theta_i = 45^\circ \), \( n_2 = 1.33 \).
Step (2) – Solve for \(\sin \theta_r\):
$ \sin \theta_r = \dfrac{n_1 \sin \theta_i}{n_2} = \dfrac{1.0 \times \sin 45^\circ}{1.33} = \dfrac{0.707}{1.33} \approx 0.532 $
Step (3) – Find \(\theta_r\):
$ \theta_r = \sin^{-1}(0.532) \approx 32.1^\circ $
Final Answer:
The angle of refraction inside water is approximately \( 32^\circ \).
Refractive Index
(a) Definition of Refractive Index
The refractive index, \( n \), of a medium is defined as the ratio of the speed of light in one medium to the speed of light in another medium:
$ n = \dfrac{c}{v} $
- \( c \) = speed of light in vacuum (or air, approximately).
- \( v \) = speed of light in the medium.
Alternatively, when comparing two different media:
$ n_{21} = \dfrac{v_1}{v_2} $
Where
- \( v_1 \) = speed of light in medium 1,
- \( v_2 \) = speed of light in medium 2.
(b) Passage of Light Through a Transparent Material
When light passes through a transparent material (e.g. glass block, water tank), the behaviour is limited to what happens at the boundaries:
- At the first boundary (air → material): the light ray slows down and bends towards the normal.
- Inside the material: the ray travels in a straight line at constant (reduced) speed.
- At the second boundary (material → air): the light ray speeds up and bends away from the normal.
- If the two boundaries are parallel (e.g. rectangular glass block), the emergent ray is parallel to the incident ray but laterally shifted.
Example
The speed of light in air is \( 3.0 \times 10^8~\text{m/s} \). In a certain transparent material, the speed of light is \( 2.25 \times 10^8~\text{m/s} \). Calculate the refractive index of the material.
▶️Answer/Explanation
Step (1) – Use definition of refractive index:
$ n = \dfrac{c}{v} $
Step (2) – Substitute values:
$ n = \dfrac{3.0 \times 10^8}{2.25 \times 10^8} = \dfrac{3.0}{2.25} = 1.33 $
Final Answer:
The refractive index of the material is \( 1.33 \).
Total Internal Reflection
Total internal reflection (TIR) occurs when light attempts to pass from a denser medium (higher refractive index) into a less dense medium (lower refractive index), such as glass → air or water → air.
- At small angles of incidence, part of the light is refracted out and part is reflected inside.
- At a certain angle, called the critical angle, the refracted ray emerges along the boundary (\( 90^\circ \) to the normal).
- For incidence angles greater than the critical angle, no refraction occurs; instead, all the light is reflected back into the denser medium. This is total internal reflection.
(b) Critical Angle
The critical angle, \( C \), is defined as the angle of incidence in the denser medium at which the angle of refraction in the less dense medium is \( 90^\circ \).
From Snell’s Law:
\( n_1 \sin C = n_2 \sin 90^\circ \)
\( \sin C = \dfrac{n_2}{n_1} \)
Where
- \( n_1 \) is the refractive index of the denser medium.
- \( n_2 \) is the refractive index of the less dense medium.
Example
The refractive index of glass is \( 1.50 \), and that of air is \( 1.00 \). Calculate the critical angle for total internal reflection at a glass–air boundary.
▶️Answer/Explanation
Step (1) – Use formula for critical angle:
\( \sin C = \dfrac{n_2}{n_1} \)
Step (2) – Substitute values:
\( \sin C = \dfrac{1.00}{1.50} = 0.667 \)
Step (3) – Find \( C \):
\( C = \sin^{-1}(0.667) \approx 41.8^\circ \)
Final Answer:
The critical angle for glass–air is approximately \( 42^\circ \).
Total Internal Reflection in Optical Fibres
(a) Structure of an Optical Fibre
An optical fibre is a very thin, flexible strand of transparent material (usually glass or plastic) with a high refractive index core and a lower refractive index cladding.
- Core: The central region where light travels.
- Cladding: A surrounding layer with slightly lower refractive index, ensuring total internal reflection occurs at the boundary.
- Protective coating: Outer layer providing mechanical strength.
(b) Total Internal Reflection in Optical Fibres
- Light enters the fibre at one end and strikes the core–cladding boundary.
- Since the core has a higher refractive index, if the angle of incidence is greater than the critical angle, the light undergoes total internal reflection.
- This process repeats many times along the fibre, guiding the light efficiently without significant loss.
- The light ray emerges at the other end of the fibre, even if the fibre is bent or curved.
(c) Applications of Optical Fibres
- Telecommunications: Transmit digital data (internet, telephone, TV) as light pulses over long distances.
- Medical endoscopes: Carry light into the body and transmit images back for internal examination.
- Decorative lighting: Fibre optics used in lamps, signs, and displays.
- Military and aerospace: Secure, lightweight communication systems.
Example
The refractive index of the core of an optical fibre is \( 1.55 \), while the cladding has refractive index \( 1.50 \). Calculate the critical angle at the core–cladding boundary.
▶️Answer/Explanation
Step (1) – Formula for critical angle:
\( \sin C = \dfrac{n_{\text{cladding}}}{n_{\text{core}}} \)
Step (2) – Substitute values:
\( \sin C = \dfrac{1.50}{1.55} = 0.968 \)
Step (3) – Find \( C \):
\( C = \sin^{-1}(0.968) \approx 75.5^\circ \)
Final Answer:
The critical angle at the core–cladding boundary is approximately \( 75.5^\circ \).