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Learning objectives

By the end of this section, you will be able to:

  • Explain the phenomenon of total internal reflection.
  • Describe the workings and uses of fiber optics.
  • Analyze the reason for the sparkle of diamonds.

The information presented in this section supports the following AP® learning objectives and science practices:

  • 6.E.1.1 The student is able to make claims using connections across concepts about the behavior of light as the wave travels from one medium into another, as some is transmitted, some is reflected, and some is absorbed. (S.P. 6.4, 7.2)

A good-quality mirror may reflect more than 90% of the light that falls on it, absorbing the rest. But it would be useful to have a mirror that reflects all of the light that falls on it. Interestingly, we can produce total reflection using an aspect of refraction .

Consider what happens when a ray of light strikes the surface between two materials, such as is shown in [link] (a). Part of the light crosses the boundary and is refracted; the rest is reflected. If, as shown in the figure, the index of refraction for the second medium is less than for the first, the ray bends away from the perpendicular. (Since n 1 > n 2 size 12{n rSub { size 8{1} }>n rSub { size 8{2} } } {} , the angle of refraction is greater than the angle of incidence—that is, θ 2 > θ 1 size 12{θ rSub { size 8{1} }>θ rSub { size 8{2} } } {} .) Now imagine what happens as the incident angle is increased. This causes θ 2 size 12{θ rSub { size 8{2} } } {} to increase also. The largest the angle of refraction θ 2 size 12{θ rSub { size 8{2} } } {} can be is 90º size 12{"90"°} {} , as shown in [link] (b).The critical angle     θ c size 12{θ rSub { size 8{c} } } {} for a combination of materials is defined to be the incident angle θ 1 size 12{θ rSub { size 8{1} } } {} that produces an angle of refraction of 90º . That is, θ c is the incident angle for which θ 2 = 90º . If the incident angle θ 1 is greater than the critical angle, as shown in [link] (c), then all of the light is reflected back into medium 1, a condition called total internal reflection .

Critical angle

The incident angle θ 1 size 12{θ rSub { size 8{1} } } {} that produces an angle of refraction of 90º size 12{"90"°} {} is called the critical angle, θ c size 12{θ rSub { size 8{1} } } {} .

In the first figure, an incident ray at an angle theta 1 with a perpendicular line drawn at the point of incidence travels from n1 to n2. The incident ray suffers both refraction and reflection. The angle of refraction is theta 2. In the second figure, as theta 1 is increased, the angle of refraction theta 2 becomes 90 degrees and the angle of reflection corresponding to 90 degrees is theta c. In the third figure, theta c greater than theta i, total internal reflection takes place and instead of refraction, reflection takes place and the light ray travels back into medium n1.
(a) A ray of light crosses a boundary where the speed of light increases and the index of refraction decreases. That is, n 2 < n 1 size 12{n rSub { size 8{2} }<n rSub { size 8{1} } } {} . The ray bends away from the perpendicular. (b) The critical angle θ c size 12{θ rSub { size 8{c} } } {} is the one for which the angle of refraction is . (c) Total internal reflection occurs when the incident angle is greater than the critical angle.

Snell’s law states the relationship between angles and indices of refraction. It is given by

n 1 sin θ 1 = n 2 sin θ 2 . size 12{n rSub { size 8{1} } "sin"θ rSub { size 8{1} } =n rSub { size 8{2} } "sin"θ rSub { size 8{2} } } {}

When the incident angle equals the critical angle ( θ 1 = θ c size 12{θ rSub { size 8{1} } =`θ rSub { size 8{c} } } {} ), the angle of refraction is 90º size 12{"90"°} {} ( θ 2 = 90º size 12{θ rSub { size 8{2} } =`"90" rSup { size 8{0} } } {} ). Noting that sin 90º =1 size 12{"sin"" 90"°"=1"} {} , Snell’s law in this case becomes

n 1 sin θ 1 = n 2 . size 12{n rSub { size 8{1} } "sin"θ rSub { size 8{1} } =n rSub { size 8{2} } } {}

The critical angle θ c size 12{q rSub { size 8{c} } } {} for a given combination of materials is thus

θ c = sin 1 n 2 / n 1 for size 12{θ rSub { size 8{c} } ="sin" rSup { size 8{ - 1} } left ( {n rSub { size 8{2} } } slash {n rSub { size 8{1} } } right )} {} n 1 > n 2 . size 12{n rSub { size 8{1} }>n rSub { size 8{2} } } {}

Total internal reflection occurs for any incident angle greater than the critical angle θ c size 12{q rSub { size 8{c} } } {} , and it can only occur when the second medium has an index of refraction less than the first. Note the above equation is written for a light ray that travels in medium 1 and reflects from medium 2, as shown in the figure.

How big is the critical angle here?

What is the critical angle for light traveling in a polystyrene (a type of plastic) pipe surrounded by air?

Strategy

The index of refraction for polystyrene is found to be 1.49 in [link] , and the index of refraction of air can be taken to be 1.00, as before. Thus, the condition that the second medium (air) has an index of refraction less than the first (plastic) is satisfied, and the equation θ c = sin 1 n 2 / n 1 size 12{θ rSub { size 8{c} } ="sin" rSup { size 8{ - 1} } left ( {n rSub { size 8{2} } } slash {n rSub { size 8{1} } } right )} {} can be used to find the critical angle θ c size 12{θ rSub { size 8{c} } } {} . Here, then, n 2 = 1 . 00 size 12{n rSub { size 8{2} } =1 "." "00"} {} and n 1 = 1 . 49 size 12{n rSub { size 8{1} } =1 "." "49"} {} .

Solution

The critical angle is given by

θ c = sin 1 n 2 / n 1 . size 12{θ rSub { size 8{c} } ="sin" rSup { size 8{ - 1} } left ( {n rSub { size 8{2} } } slash {n rSub { size 8{1} } } right )} {}

Substituting the identified values gives

θ c = sin 1 1 . 00 / 1 . 49 = sin 1 0 . 671 42.2º. alignl { stack { size 12{θ rSub { size 8{c} } ="sin" rSup { size 8{ - 1} } left ( {1 "." "00"} slash {1 "." "49"} right )="sin" rSup { size 8{ - 1} } left (0 "." "671" right )} {} #="42" "." 2° "." {} } } {}

Discussion

This means that any ray of light inside the plastic that strikes the surface at an angle greater than 42.2º will be totally reflected. This will make the inside surface of the clear plastic a perfect mirror for such rays without any need for the silvering used on common mirrors. Different combinations of materials have different critical angles, but any combination with n 1 > n 2 size 12{n rSub { size 8{1} }>n rSub { size 8{2} } } {} can produce total internal reflection. The same calculation as made here shows that the critical angle for a ray going from water to air is 48 . size 12{"48" "." 6°} {} , while that from diamond to air is 24 . size 12{"24" "." 4°} {} , and that from flint glass to crown glass is 66 . size 12{"66" "." 3°} {} . There is no total reflection for rays going in the other direction—for example, from air to water—since the condition that the second medium must have a smaller index of refraction is not satisfied. A number of interesting applications of total internal reflection follow.

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Questions & Answers

Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly.
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Practice Key Terms 4

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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