Refraction of light

The normal is drawn perpendicular to the surface of the glass block at the point where the ray of light hits the surface. (2 marks)

Diagram: A rectangle representing the glass block, an incident ray entering the block, a refracted ray within the block, and an emergent ray exiting the block. Label the incident ray, refracted ray, emergent ray, normal, angle of incidence (i), and angle of refraction (r).

Procedure:
1. Place the glass block on a piece of white paper.
2. Shine a ray of light at an angle onto one side of the block.
3. Mark the path of the incident ray before it enters the block, the refracted ray inside the block, and the emergent ray after it leaves the block.
4. Remove the block and draw straight lines to connect the points, showing the complete path of the light ray.
5. Draw a normal (a line perpendicular to the surface) at the point where the light ray enters the block.
6. Measure the angle of incidence (i) between the incident ray and the normal, and the angle of refraction (r) between the refracted ray and the normal, using a protractor. Repeat for different angles of incidence.

Explanation: Refraction occurs because light changes speed when it passes from air to glass (and vice-versa). This change in speed causes the light to bend.

When a ray of light passes through the curved surface of a semi-circular glass block towards the center, it enters the block perpendicularly to the surface (along the normal). Therefore, there is no refraction at this first surface. Only at the exit point (the flat side) will refraction occur as the light exits the glass block. With a rectangular block, there is refraction at both the entry and exit points.

Snell's Law: n1 * sin(θ1) = n2 * sin(θ2)
1. 0 * sin(60°) = 1.5 * sin(θ2)
sin(θ2) = (1.0 * sin(60°))/1.5
sin(θ2) = 0.577
θ2 = arcsin(0.577)
θ2 = 35.3 degrees

Explanation: Snell's Law relates the refractive indices of two mediums to the angles of incidence and refraction. We used it to calculate the angle of refraction after the light ray entered the glass.

The speed of light decreases as it enters the glass from air. Since the frequency of light remains constant, and the speed of light is equal to frequency times wavelength (v = fλ), the wavelength of the light also decreases as it enters the glass.

Explanation: Because glass has a higher refractive index than air, the light slows down. Constant frequency leads to a corresponding decrease in wavelength.

When the angle of incidence in the glass is equal to the critical angle, the ray of light is refracted along the glass-air boundary. The angle of refraction in the air is 90 degrees.

The ray of light is completely reflected back into the glass. This phenomenon is called total internal reflection.

1. Optical fibres used for communication, endoscopes and decoration
2. Reflectors e.g. in bicycle lights.

Refractive index, *n* = speed of light in air / speed of light in glass
*n* = (3.0 x 10⁸ m/s) / (2.0 x 10⁸ m/s)
*n* = 1.5

The refractive index is a dimensionless quantity; it has no units. It indicates how much slower light travels in the glass compared to air.

n = sin i / sin r
n = sin(60°) / sin(35°)
n = 0.866 / 0.574
n = 1.51

The refractive index is a ratio of the sine of the angles and therefore has no units.

n = sin i / sin r
Rearrange to make sin i the subject:
sin i = n * sin r
sin i = 2.42 * sin(20°)
sin i = 2.42 * 0.342
sin i = 0.828
i = sin⁻¹(0.828)
i = 55.9°

The angle of incidence is the angle between the incident ray and the normal.

Formula: n = 1/sin(c)
Calculation: n = 1/sin(41.8°) = 1/0.667 = 1.50
Answer: n = 1.50
Explanation: The refractive index is the inverse of the sine of the critical angle.

When light travels from a higher refractive index to a lower refractive index medium, it bends away from the normal. As the angle of incidence increases, the angle of refraction approaches 90°. At the critical angle, *c*, the angle of refraction is 90°. If the angle of incidence exceeds the critical angle, the light is entirely reflected back into the higher refractive index medium; this is total internal reflection. The critical angle *c* and refractive index *n* are related by the equation n = 1/sin(c). Thus, a higher refractive index is related to a lower critical angle, making total internal reflection more likely to occur.

1. Light signals enter the optical fibre.
2. Due to the high refractive index of the fibre core and low refractive index of the cladding, light undergoes total internal reflection at the core-cladding boundary.
3. The light repeatedly reflects along the fibre, transmitting the signal with minimal loss.
4. Optical fibres allow for the transmission of signals as light over great distances.

1. Higher bandwidth: Optical fibres can carry significantly more data than copper cables, allowing for faster communication speeds.
2. Lower signal loss: Signals transmitted through optical fibres experience less attenuation (loss of signal strength) compared to copper cables, enabling longer transmission distances without the need for repeaters.