# TOTAL INTERNAL REFLECTION and the CRITICAL ANGLE

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TOTAL INTERNAL REFLECTION and the CRITICAL ANGLE

TOTAL INTERNAL REFLECTION
The inside surface of water or a glass block can act like a mirror.

Total Internal Reflection
Occurs when light reflects off of the inside wall of a denser medium (higher index of refraction)

Total Internal Reflection
Recall: When light passes from a dense material into a less dense medium, light reflects away from the normal

Total Internal Reflection
As the angle of incidence increases, the angle of refraction increases

Critical Angle At a certain angle, the refracted ray follows a path exactly along the surface of the dense medium When the angle of refraction is at 90o the incident angle is called the “critical angle”

Critical Angle Critical angle is the angle at which the refracted ray is at 90o

Critical Angle determines Total Internal Reflection
If the incident ray is increased beyond the critical angle, the light is no longer refracted Instead, it is reflected back inside the medium This is called “total internal reflection” When incident angle is greater than the critical angle (i>C), there is no refracted rays, all emergent rays are internally reflected

TOTAL INTERNAL REFLECTION
A light ray hits the inside face of a semicircular block.

Total Internal Reflection
A light ray hits the inside face of a semicircular block as follows. What will happen as the angle of incidence increases? air glass incident ray reflected ray refracted ray

Total Internal Reflection
air glass incident ray reflected ray refracted ray

Total Internal Reflection
air glass incident ray reflected ray refracted ray

Total Internal Reflection
air glass incident ray reflected ray refracted ray

Total Internal Reflection
air glass incident ray reflected ray

Total Internal Reflection
air glass incident ray reflected ray refracted ray Total Internal Reflection air glass incident ray reflected ray What happened?

Total Internal Reflection
At a small angle of incidence: Incident ray splits into refracted & reflected Angle of refraction < 90o Angle of reflection = angle of incidence air glass incident ray reflected ray refracted ray

Total Internal Reflection
As the angle of incidence increases, the angle of refraction increases until… air glass incident ray reflected ray refracted ray

Total Internal Reflection
Angle of refraction is at 90o, parallel to the surface of the medium At this point the angle of incidence = critical angle (C) Angle of incidence = C when angle of refraction = 90o air glass incident ray reflected ray refracted ray

Total Internal Reflection
As the angle of incidence increase beyond the critical angle (>C), there is no more refracted ray All emergent rays are totally reflected inside the medium = Total Internal Reflection (TIR) air glass incident ray reflected ray

Conditions for TIR & Critical Angle
Light is traveling slower in the first medium than in the second medium (v1 < v2) Thus light is moving from medium of higher refractive index to one of lower refractive index (n1 > n2) Critical angle (C) is defined when the angle of refraction is 90o to the normal (2 = 90o) TIR occurs when angle of incidence is larger than the critical angle (1 > C) air glass incident ray reflected ray

Calculating Critical Angle
Snell’s Law: n1 sin1 = n2 sin2 Critical angle C = 1 thus solve for 1 At the critical angle, the refracted ray is a 90o = 2 n1 sin1 = n2 sin90o (sin90o = 1) n1 sin1 = n2 1 n1 sin1 = n2 Rearrange equation to get: sin1 = n2 / n1 1 = sin-1 (n2 / n1) If medium 2 = air, then n2 = 1 1 = sin-1 (1 / n1) C = sin-1 (1 / n1) air glass C = 1 r refracted ray reflected ray incident ray R = 2

TOTAL INTERNAL REFLECTION
critical angles of different materials Medium Refractive index Critical angle Glass Water Perspex Diamond 1.50–1.70 30–42 1.33 49 1.5 42 2.42 24

TIR in Diamonds Sparkling is due to: Cut of diamond faces
High index of refraction which means a very small critical angle (n = 2.42, C = 24.4°) Incident rays can undergo multiple TIR inside a diamond before exiting the top of the diamond.

TIR in Fiber Optics Technology that uses light to transmit information along glass cables Fibre optics cable is made up of a bundle of glass fibres Sample materials: high-purity glass, Lucite

TIR in Fiber Optics Fiber optics cable has a small critical angle, thus a high refractive index Light entering will always have an angle of incidence greater than the critical angle

TIR in Fiber Optics Light does not escape as it travels along the fiber optics cable because it undergoes total internal reflection

TIR in Fiber Optics

TIR in Fiber Optics Advantages of Fiber Optics
Signals are not affected by electrical storms. Cable is smaller and lighter than copper cable. More signals can be carried over longer distances. 28

Fiber Optics in Endoscopes
An endoscope is a flexible fibre optic cable through which internal cavities can be viewed. Routinely used in the diagnosis of cancer and ulcers.

Gastroscopy Endoscopy examination of a stomach
Endoscope inserted through the patient's mouth and fed down through throat Image obtained by endoscope is projected onto a screen A surgical instrument for obtaining a biopsy has been fed through the endoscope cable and controlled by the doctor Gastroscopy

Digestive Endoscopy

Doctors using a fibroscope to investigate suspected lung cancer in a patient's bronchi (airways). A fibroscope is a flexible fibre optic cable with a camera on the end, similar to an endoscope. Diagnosing Cancer Photographed in Belgium.

TIR in Prisms I1 I2 I3 Object Plane mirror = glass + silvered surface
multiple reflection inside the glass multiple images formed nuisance in optical instruments glass sheet silvered surface I I I3 Object 33

45 Prisms If light rays strike the inside face at an angle > 42, glass prism behaves like a perfect mirror. Prisms are more useful than mirrors because it reflects almost 100% of light internally. Mirrors lose some light through absorption. Emergent can be 90o or 180o relative to incident ray. 45 34

Prisms in Periscopes Instrument for observation from a concealed position Uses two triangular prisms (or mirrors) to change direction of light by 90o Used in war and in submarines

In War Land Periscope used by a German Staff Officer during 1914
Lens was sixteen feet above the ground, protecting the officer from enemy observation.

Submarines American submarine commander inspects the horizon through the periscope (1942)

Submarines carry all kinds of extendable devices in their sail which allow them to sense above the ocean's surface. This shows a deployed periscope (on left) and other electronic surveillance and communications probes.

Prisms in Binoculars Uses 2 prisms to change direction of light by 180o 40

Prisms in Single-lens Camera
A five-sided ‘pentaprism’ reflects light from the mirror into the eye. film mirror 41

Prisms and Retro-reflectors
45 Device that returns incident light back in exactly the same direction from which it came Applications in bike reflectors, reflective strip on clothing, road signs

Practice Problems 30 18 E O A ray of light traveling in the direction EO in air enters a rectangular block at an angle of incidence = 30. The resulting angle of refraction = 18. a. Find the refractive index n of the block.

Practice Problems E 30 O 18 Given: 1 = 30, 2 = 18, n1 (air) = 1
Required: n2 (block) Analysis: n1 sin1 = n2 sin2 Solution: 1 x sin 30 = n2 sin 18 n2 = sin 30 / sin 18 = 1.62 Phrase: The index of refraction of the block is 1.62 30 18 E O

Practice Problems E 30 O 18
b. Find the critical angle C for the block. 30 18 E O Recall: If medium 2 = air, then C = 1 = sin-1 (1 / n1)

Practice Problems E 30 O 18
b. Find the critical angle C for the block. Given: 2 = 90, n2 (air) = 1, n1 (block) = 1.62 Required: 1 Analysis: n1 sin1 = n2 sin2 Solution: 1.62 x sin 1 = 1 x sin 90 sin 1 = 1 / 1.62 1 = sin-1 (1 / 1.62) = 38.1 Phrase: The critical angle of the block is 38.1 30 18 E O Recall: If medium 2 = air, then C = 1 = sin-1 (1 / n1)

Practice Problems c. If the ray is incident on surface BC, from which surface and at what angle will the ray leave the block? 30 A B C D

Practice Problems C D B A
c. If the ray is incident on surface BC, from which surface and at what angle will the ray leave the block? Given: 1 = 60 n1 (air) = 1 n2 (block) = 1.62 Required: 2 Analysis: n1 sin1 = n2 sin2 Solution: 1 x sin 60 = 1.62 x sin 2 sin 2 = 1 x sin 60 / 1.62 2 = sin-1 (1 x sin 60 / 1.62) = 32.3 30 A B C D

Practice Problems A B C D 32.3 57.7 60
c. If the ray is incident on surface BC, from which surface and at what angle will the ray leave the block? Recall: Angle of incidence = 60o Angle of refraction = 32.3o Critical angle = 38.1o 30 A B C D 32.3 57.7 60 Draw refracted ray2. Measure angle of ray2 hitting block. Angle is greater than critical angle of 38.1o thus ray3 will reflect. Draw ray3 following Law of Reflection. Measure angle of ray3. Since it is the same angle as ray2, it will refract out at the same angle. Thus ray4 refracts at 60o from surface AD.

Practice Problems Which of the following angles is the critical angle of glass? A B C D

Practice Problems Which of the following angles is the critical angle of glass? A B C D

Practice Problems A horizontal light ray hits a prism as shown.
45 A horizontal light ray hits a prism as shown. What happens to the light ray? B C A

Practice Problems A horizontal light ray hits a prism as shown.
45 A horizontal light ray hits a prism as shown. What happens to the light ray? B C A