Physics Form 3 Notes : CHAPTER TWO – REFRACTION OF LIGHT

CHAPTER TWO

REFRACTION OF LIGHT

Introduction

Refraction is the change of direction of light rays as they pass at an angle from one medium to another of different optical densities.

Experiment: To investigate the path of light through a rectangular glass block

Apparatus: soft-board, white sheet of paper, drawing pins (optical), rectangular glass block.

Procedure

1. Fix the white plain paper on the soft board using pins.
2. Place the glass block on the paper and trace its outline, label it ABCD as shown below.
3. Draw a normal NON at point O.
4. Replace the glass block to its original position.
5. Stick two pins P₁ and P₂ on the line such that they are at least 6 cm apart and upright.
6. Viewing pins P₁ and P₂ from the opposite side, fix pins P₃ and P₄ such that they’re in a straight line.
7. Remove the pins and the glass block.
8. Draw a line joining P₃ and P₄ and produce it to meet the outline face AB at point O.

Explanation of refraction

Light travels at a velocity of 3.0×10⁸ m/s in a vacuum. Light travels with different velocities in different media. When a ray of light travels from an optically less dense medium to a more dense medium, it is refracted towards the normal. The glass block experiment gives rise to a very important law known as the law of reversibility which states that “if a ray of light is reversed, it always travels along its original path”. If the glass block is parallel-sided, the emergent ray will be parallel to the incident ray but displaced laterally as shown.

‘e’ is called the angle of emergence. The direction of the light is not altered but displaced sideways. This displacement is called lateral displacement and is denoted by ‘d’. Therefore:

XY = t / cos r

YZ = sin (i – r) × XY

So, lateral displacement, d = t sin (i – r) / cos r

Laws of refraction

1. The incident ray, the refracted ray, and the normal at the point of incidence all lie on the same plane.
2. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media.

   sin i / sin r = constant (k)

Refractive index

Refractive index (n) is the constant of proportionality in Snell’s law, hence:

sin i / sin r = n

Therefore, sin i / sin r = n = 1 / (sin r / sin i)

Examples

1. Calculate the refractive index for light travelling from glass to air given that _an_g = 1.5.

   Solution:

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   _gn_a = 1 / _an_g = 1 / 1.5 = 0.67

2. Calculate the angle of refraction for a ray of light from air striking an air-glass interface, making an angle of 60° with the interface. (_an_g = 1.5)

   Solution:

   Angle of incidence (i) = 90° – 60° = 30°

   1.5 = sin 30° / sin r, so sin r = sin 30° / 1.5 = 0.5 / 1.5 = 0.3333

   sin^-1 0.3333 = 19.5°

   Therefore, r = 19.5°

Refractive index in terms of velocity

Refractive index can be given in terms of velocity by the following equation:

n = velocity of light in medium 1 / velocity of light in medium 2

When a ray of light is travelling from vacuum to a medium, the refractive index is referred to as the absolute refractive index of the medium, denoted by ‘n’.

Refractive index of a material ‘n’ = velocity of light in vacuum / velocity of light in material

The absolute refractive indices of some common materials are given below:

Example

A ray of light is incident on a water-glass interface as shown. Calculate ‘r’. (Take the refractive index of glass and water as 3/2 and 4/3 respectively)

Solution:

Since _an_w sin θ_w = _an_g sin θ_g

(4/3) sin 30° = (3/2) sin r

(3/2) sin r = (4/3) × 0.5

sin r = (4/6) × (2/3) = 4/9 = 0.4444

r = 26.4°

1. The refractive index of water is 4/3 and that of glass is 3/2. Calculate the refractive index of glass with respect to water.

   Solution:

   _wn_g = _gn_a × _an_w, but _wn_a = 1 / _an_w = 3/4

   _wn_g = (3/4) × (3/2) = 9/8 = 1.13

Real and apparent depth

Consider the following diagram:

The depth of the water OM is the real depth, and the distance IM is known as the apparent depth. OI is the distance through which the coin has been displaced and is known as the vertical displacement. The relationship between refractive index and the apparent depth is given by:

Refractive index of a material = real depth / apparent depth

Note: This is true only if the object is viewed normally.

Example

A glass block of thickness 12 cm is placed on a mark drawn on plain paper. The mark is viewed normally through the glass. Calculate the apparent depth of the mark and hence the vertical displacement. (Refractive index of glass = 3/2)

Solution:

_an_g = real depth / apparent depth

Apparent depth = real depth / _an_g = (12 × 2) / 3 = 8 cm

Vertical displacement = 12 – 8 = 4 cm

Applications of refractive index

Total internal reflection

This occurs when light travels from a denser optical medium to a less dense medium. The refracted ray moves away from the normal until a critical angle is reached, usually 90°, where the refracted ray is parallel to the boundary between the two media. If this critical angle is exceeded, total internal reflection occurs and at this point no refraction occurs but the ray is reflected internally within the denser medium.

Relationship between the critical angle and refractive index

Consider the following diagram:

From Snell’s law:

_gn_w = sin C / sin 90°, but _an_g = 1 / _gn_a since sin 90° = 1

Therefore, _an_g = 1 / sin C, hence sin C = 1 / n or n = 1 / sin C

Example

Calculate the critical angle of diamond given that its refractive index is 2.42.

Solution:

sin C = 1 / n = 1 / 2.42 = 0.4132

C = sin^-1 0.4132 = 24.4°

Effects of total internal reflection

1. Mirage: These are ‘pools of water’ seen on a tarmac road during a hot day. They are also observed in very cold regions but the light curves in the opposite direction such that a polar bear seems to be upside down in the sky.
2. Atmospheric refraction: The Earth’s atmosphere refracts light rays so that the sun can be seen even when it has set. Similarly, the sun is seen before it actually rises.

Applications of total internal reflection

1. Periscope: A prism periscope consists of two right-angled glass prisms of angles 45°, 90°, and 45° arranged as shown below. They are used to observe distant objects.

2. Prism binoculars: The arrangement of lenses and prisms is as shown below. Binoculars reduce the distance of objects such that they seem to be nearer.

3. Pentaprism: Used in cameras to change the inverted images formed into erect and actual images in front of the photographer.
4. Optical fibre: This is a flexible glass rod of small diameter. Light entering through them undergoes repeated internal reflections. They are used in medicine to observe or view internal organs of the body.

Dispersion of white light

The splitting of light into its constituent colours is known as dispersion. Each colour represents a different wavelength as they strike the prism and therefore are refracted differently as shown.