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Refraction

In the previous sections we studied light reflecting off various surfaces. What happens when light passes through a medium? The speed of light, like that of all waves, is dependent on the medium in which it is travelling. When light moves from one medium into another (for example, from air to glass), the speed of light changes. The effect is that the light ray passing into a new medium is refracted, or bent. Refraction is therefore the bending of light as it moves from one optical medium to another.

Definition 1: Refraction

Refraction is the bending of light that occurs because light travels at different speeds in different materials.

When light travels from one medium to another, it will be bent away from its original path. When it travels from an optically dense medium like water or glass to a less dense medium like air, it will be refracted away from the normal (Figure 1). Whereas, if it travels from a less dense medium to a denser one, it will be refracted towards the normal (Figure 2).

Figure 1: Light is moving from an optically dense medium to an optically less dense medium. Light is refracted away from the normal.
Figure 1 (PG10C6_016.png)
Figure 2: Light is moving from an optically less dense medium to an optically denser medium. Light is refracted towards the normal.
Figure 2 (PG10C6_017.png)

Just as we defined an angle of reflection in the previous section, we can similarly define an angle of refraction as the angle between the surface normal and the refracted ray. This is shown in Figure 3.

Figure 3: Light moving from one medium to another bends towards or away from the surface normal. The angle of refraction θθ is shown.
(a)
Figure 3(a) (PG10C6_018.png)
(b)
Figure 3(b) (PG10C6_019.png)

Refractive Index

Which is easier to travel through, air or water? People usually travel faster through air. So does light! The speed of light and therefore the degree of bending of the light depends on the refractive index of material through which the light passes. The refractive index (symbol nn) is the ratio of the speed of light in a vacuum to its speed in the material.

Definition 2: Refractive Index

The refractive index of a material is the ratio of the speed of light in a vacuum to its speed in the medium.

Note: Interesting Fact :

The symbol cc is used to represent the speed of light in a vacuum.

c = 299 792 485 m · s - 1 c = 299 792 485 m · s - 1
(1)

For purposes of calculation, we use 3×108m·s-13×108m·s-1. A vacuum is a region with no matter in it, not even air. However, the speed of light in air is very close to that in a vacuum.

Definition 3: Refractive Index

The refractive index (symbol nn) of a material is the ratio of the speed of light in a vacuum to its speed in the material and gives an indication of how difficult it is for light to get through the material.

n = c v n = c v
(2)

where

Table 1
nn = refractive index (no unit)
cc = speed of light in a vacuum (3,00×108m·s-13,00×108m·s-1)
vv = speed of light in a given medium (m·s-1m·s-1)

Refractive Index and Speed of Light

Using

n = c v n = c v
(3)

we can also examine how the speed of light changes in different media, because the speed of light in a vacuum (cc) is constant.

If the refractive index nn increases, the speed of light in the material vv must decrease. Light therefore travels slowly through materials of high nn.

Table 2 shows refractive indices for various materials. Light travels slower in any material than it does in a vacuum, so all values for nn are greater than 1.

Table 2: Refractive indices of some materials. n air n air is calculated at STP.
Medium Refractive Index
Vacuum 1
Helium 1,000036
Air* 1,0002926
Carbon dioxide 1,00045
Water: Ice 1,31
Water: Liquid (2020C) 1,333
Acetone 1,36
Ethyl Alcohol (Ethanol) 1,36
Sugar solution (30%) 1,38
Fused quartz 1,46
Glycerine 1,4729
Sugar solution (80%) 1,49
Rock salt 1,516
Crown Glass 1,52
Sodium chloride 1,54
Polystyrene 1,55 to 1,59
Bromine 1,661
Sapphire 1,77
Glass (typical) 1,5 to 1,9
Cubic zirconia 2,15 to 2,18
Diamond 2,419
Silicon 4,01

Snell's Law

Now that we know that the degree of bending, or the angle of refraction, is dependent on the refractive index of a medium, how do we calculate the angle of refraction?

The angles of incidence and refraction when light travels from one medium to another can be calculated using Snell's Law.

Definition 4: Snell's Law
n 1 sin θ 1 = n 2 sin θ 2 n 1 sin θ 1 = n 2 sin θ 2
(4)

where

Table 3
n1n1 = Refractive index of material 1
n2n2 = Refractive index of material 2
θ1θ1 = Angle of incidence
θ2θ2 = Angle of refraction

Remember that angles of incidence and refraction are measured from the normal, which is an imaginary line perpendicular to the surface.

Suppose we have two media with refractive indices n1n1 and n2n2. A light ray is incident on the surface between these materials with an angle of incidence θ1θ1. The refracted ray that passes through the second medium will have an angle of refraction θ2θ2.

Exercise 1: Using Snell's Law

A light ray with an angle of incidence of 35 passes from water to air. Find the angle of refraction using Snell's Law and Table 2. Discuss the meaning of your answer.

Solution
  1. Step 1. Determine the refractive indices of water and air :

    From Table 2, the refractive index is 1,333 for water and about 1 for air. We know the angle of incidence, so we are ready to use Snell's Law.

  2. Step 2. Substitute values :

    According to Snell's Law:

    n 1 sin θ 1 = n 2 sin θ 2 1 , 33 sin 35 = 1 sin θ 2 sin θ 2 = 0 , 763 θ 2 = 49 , 7 or 130 , 3 n 1 sin θ 1 = n 2 sin θ 2 1 , 33 sin 35 = 1 sin θ 2 sin θ 2 = 0 , 763 θ 2 = 49 , 7 or 130 , 3
    (5)

    Since 130,3130,3 is larger than 9090, the solution is:

    θ 2 = 49 , 7 θ 2 = 49 , 7
    (6)
  3. Step 3. Discuss the answer :

    The light ray passes from a medium of high refractive index to one of low refractive index. Therefore, the light ray is bent away from the normal.

Exercise 2: Using Snell's Law

A light ray passes from water to diamond with an angle of incidence of 7575. Calculate the angle of refraction. Discuss the meaning of your answer.

Solution
  1. Step 1. Determine the refractive indices of water and air :

    From Table 2, the refractive index is 1,333 for water and 2,42 for diamond. We know the angle of incidence, so we are ready to use Snell's Law.

  2. Step 2. Substitute values and solve :

    According to Snell's Law:

    n 1 sin θ 1 = n 2 sin θ 2 1 , 33 sin 75 = 2 , 42 sin θ 2 sin θ 2 = 0 , 531 θ 2 = 32 , 1 . n 1 sin θ 1 = n 2 sin θ 2 1 , 33 sin 75 = 2 , 42 sin θ 2 sin θ 2 = 0 , 531 θ 2 = 32 , 1 .
    (7)
  3. Step 3. Discuss the answer :

    The light ray passes from a medium of low refractive index to one of high refractive index. Therefore, the light ray is bent towards the normal.

If

n 2 > n 1 n 2 > n 1
(8)

then from Snell's Law,

sin θ 1 > sin θ 2 . sin θ 1 > sin θ 2 .
(9)

For angles smaller than 9090, sinθsinθ increases as θθ increases. Therefore,

θ 1 > θ 2 . θ 1 > θ 2 .
(10)

This means that the angle of incidence is greater than the angle of refraction and the light ray is bent toward the normal.

Similarly, if

n 2 < n 1 n 2 < n 1
(11)

then from Snell's Law,

sin θ 1 < sin θ 2 . sin θ 1 < sin θ 2 .
(12)

For angles smaller than 9090, sinθsinθ increases as θθ increases. Therefore,

θ 1 < θ 2 . θ 1 < θ 2 .
(13)

This means that the angle of incidence is less than the angle of refraction and the light ray is away toward the normal.

Both these situations can be seen in Figure 4.

Figure 4: Refraction of two light rays. (a) A ray travels from a medium of low refractive index to one of high refractive index. The ray is bent towards the normal. (b) A ray travels from a medium with a high refractive index to one with a low refractive index. The ray is bent away from the normal.
(a)
Figure 4(a) (PG10C6_020.png)
(b)
Figure 4(b) (PG10C6_021.png)

What happens to a ray that lies along the normal line? In this case, the angle of incidence is 00 and

sin θ 2 = n 1 n 2 sin θ 1 = 0 θ 2 = 0 . sin θ 2 = n 1 n 2 sin θ 1 = 0 θ 2 = 0 .
(14)

This shows that if the light ray is incident at 00, then the angle of refraction is also 00. The ray passes through the surface unchanged, i.e. no refraction occurs.

Investigation : Snell's Law 1

The angles of incidence and refraction were measured in five unknown media and recorded in the table below. Use your knowledge about Snell's Law to identify each of the unknown media A - E. Use Table 2 to help you.

Table 4
Medium 1 n 1 n 1 θ 1 θ 1 θ 2 θ 2 n 2 n 2 Unknown Medium
Air 1,000036 38 11,6 ? A
Air 1,000036 65 38,4 ? B
Vacuum 1 44 0,419 ? C
Air 1,000036 15 29,3 ? D
Vacuum 1 20 36,9 ? E

Investigation : Snell's Law 2

Zingi and Tumi performed an investigation to identify an unknown liquid. They shone a beam of light into the unknown liquid, varying the angle of incidence and recording the angle of refraction. Their results are recorded in the following table:

Table 5
Angle of Incidence Angle of Refraction
0,0 0,00
5,0 3,76
10,0 7,50
15,0 11,2
20,0 14,9
25,0 18,5
30,0 22,1
35,0 25,5
40,0 28,9
45,0 32,1
50,0 35,2
55,0 38,0
60,0 40,6
65,0 43,0
70,0 ?
75,0 ?
80,0 ?
85,0 ?
  1. Write down an aim for the investigation.
  2. Make a list of all the apparatus they used.
  3. Identify the unknown liquid.
  4. Predict what the angle of refraction will be for 70, 75, 80 and 85.

Figure 5
Khan academy video on Snell's Law - 1

Apparent Depth

Imagine a coin on the bottom of a shallow pool of water. If you reach for the coin, you will miss it because the light rays from the coin are refracted at the water's surface.

Consider a light ray that travels from an underwater object to your eye. The ray is refracted at the water surface and then reaches your eye. Your eye does not know Snell's Law; it assumes light rays travel in straight lines. Your eye therefore sees the image of the at coin shallower location. This shallower location is known as the apparent depth.

The refractive index of a medium can also be expressed as

n = real depth apparent depth . n = real depth apparent depth .
(15)

Figure 6
Figure 6 (PG10C6_022.png)

Exercise 3: Apparent Depth 1

A coin is placed at the bottom of a 40 cm deep pond. The refractive index for water is 1,33. How deep does the coin appear to be?

Solution
  1. Step 1. Identify what is given and what is asked :

    n = 1,33

    real depth = 40 cm

    apparent depth = ?

  2. Step 2. Substitute values and find answer :
    n = real depth apparent depth 1 , 33 = 40 x x = 40 1 , 33 = 30 , 08 c m n = real depth apparent depth 1 , 33 = 40 x x = 40 1 , 33 = 30 , 08 c m
    (16)

    The coin appears to be 30,08 cm deep.

Exercise 4: Apparent Depth 2

A R1 coin appears to be 7 cm deep in a colourless liquid known to be listed in Table 2. The depth of the liquid is 10,43 cm.

  1. Determine the refractive index of the liquid.
  2. Identify the liquid.
Solution
  1. Step 1. Identify what is given and what is asked :

    real depth = 7 cm

    apparent depth = 10,43 cm

    nn = ?

    Identify the liquid.

  2. Step 2. Calculate refractive index :
    n = real depth apparent depth = 10 , 43 7 = 1 , 49 n = real depth apparent depth = 10 , 43 7 = 1 , 49
    (17)
  3. Step 3. Identify the liquid :

    Use Table 2. The liquid is an 80% sugar solution.

Refraction

  1. Explain refraction in terms of a change of wave speed in different media.
    Click here for the solution.
  2. In the diagram, label the following:
    1. angle of incidence
    2. angle of refraction
    3. incident ray
    4. refracted ray
    5. normal
    Figure 7
    Figure 7 (PG10C6_023.png)
    Click here for the solution.
  3. What is refraction?
    Click here for the solution.
  4. Describe what is meant by the refractive index of a medium.
    Click here for the solution.
  5. State Snell's Law.
    Click here for the solution.
  6. In the diagram, a ray of light strikes the interface between two media.
    Figure 8
    Figure 8 (PG10C6_024.png)
    Draw what the refracted ray would look like if:
    1. medium 1 had a higher refractive index than medium 2.
    2. medium 1 had a lower refractive index than medium 2.
    Click here for the solution.
  7. Light travels from a region of glass into a region of glycerine, making an angle of incidence of 40.
    1. Describe the path of the light as it moves into the glycerine.
    2. Calculate the angle of refraction.
    Click here for the solution.
  8. A ray of light travels from silicon to water. If the ray of light in the water makes an angle of 69 to the surface normal, what is the angle of incidence in the silicon?
    Click here for the solution.
  9. Light travels from a medium with n=1,25n=1,25 into a medium of n=1,34n=1,34, at an angle of 27 from the interface normal.
    1. What happens to the speed of the light? Does it increase, decrease, or remain the same?
    2. What happens to the wavelength of the light? Does it increase, decrease, or remain the same?
    3. Does the light bend towards the normal, away from the normal, or not at all?
    Click here for the solution.
  10. Light travels from a medium with n=1,63n=1,63 into a medium of n=1,42n=1,42.
    1. What happens to the speed of the light? Does it increase, decrease, or remain the same?
    2. What happens to the wavelength of the light? Does it increase, decrease, or remain the same?
    3. Does the light bend towards the normal, away from the normal, or not at all?
    Click here for the solution.
  11. Light is incident on a glass prism. The prism is surrounded by air. The angle of incidence is 23. Calculate the angle of reflection and the angle of refraction.
    Click here for the solution.
  12. Light is refracted at the interface between air and an unknown medium. If the angle of incidence is 53 and the angle of refraction is 37, calculate the refractive index of the unknown, second medium.
    Click here for the solution.
  13. A coin is placed in a bowl of acetone (nn = 1,36). The coin appears to be 10 cm deep. What is the depth of the acetone?
    Click here for the solution.
  14. A dot is drawn on a piece of paper and a glass prism placed on the dot according to the diagram.
    Figure 9
    Figure 9 (PG10C6_025.png)
    Use the information supplied to determine the refractive index of glass.
    Click here for the solution.
  15. Light is refracted at the interface between a medium of refractive index 1,5 and a second medium of refractive index 2,1. If the angle of incidence is 45, calculate the angle of refraction.
    Click here for the solution.
  16. A ray of light strikes the interface between air and diamond. If the incident ray makes an angle of 30 with the interface, calculate the angle made by the refracted ray with the interface.
    Click here for the solution.
  17. Challenge Question: What values of nn are physically impossible to achieve? Explain your answer. The values provide the limits of possible refractive indices.
    Click here for the solution.
  18. Challenge Question: You have been given a glass beaker full of an unknown liquid. How would you identify what the liquid is? You have the following pieces of equipment available for the experiment: a laser, a protractor, a ruler, a pencil, and a reference guide containing optical properties of various liquids.
    Click here for the solution.

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