Presentation on theme: "TOTAL INTERNAL REFLECTION and the CRITICAL ANGLE"— Presentation transcript:
1 TOTAL INTERNAL REFLECTION and the CRITICAL ANGLE
2 TOTAL INTERNAL REFLECTION The inside surface of water or a glass block can act like a mirror.
3 Total Internal Reflection Occurs when light reflects off of the inside wall of a denser medium (higher index of refraction)
4 Total Internal Reflection Recall:When light passes from a dense material into a less dense medium, light reflects away from the normal
5 Total Internal Reflection As the angle of incidence increases,the angle of refraction increases
6 Critical AngleAt a certain angle, the refracted ray follows a path exactly along the surface of the dense mediumWhen the angle of refraction is at 90o the incident angle is called the “critical angle”
7 Critical AngleCritical angle is the angle at which the refracted ray is at 90o
8 Critical Angle determines Total Internal Reflection If the incident ray is increased beyond the critical angle, the light is no longer refractedInstead, it is reflected back inside the mediumThis 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
9 TOTAL INTERNAL REFLECTION A light ray hits the inside face of a semicircular block.
10 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?airglassincident rayreflected rayrefracted ray
11 Total Internal Reflection airglassincident rayreflected rayrefracted ray
12 Total Internal Reflection airglassincident rayreflected rayrefracted ray
13 Total Internal Reflection airglassincident rayreflected rayrefracted ray
14 Total Internal Reflection airglassincident rayreflected ray
16 Total Internal Reflection At a small angle of incidence:Incident ray splits into refracted & reflectedAngle of refraction < 90oAngle of reflection = angle of incidenceairglassincident rayreflected rayrefracted ray
17 Total Internal Reflection As the angle of incidence increases, the angle of refraction increases until…airglassincident rayreflected rayrefracted ray
18 Total Internal Reflection Angle of refraction is at 90o, parallel to the surface of the mediumAt this point the angle of incidence = critical angle (C)Angle of incidence = C when angle of refraction = 90oairglassincident rayreflected rayrefracted ray
19 Total Internal Reflection As the angle of incidence increase beyond the critical angle (>C), there is no more refracted rayAll emergent rays are totally reflected inside the medium = Total Internal Reflection (TIR)airglassincident rayreflected ray
20 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)airglassincident rayreflected ray
21 Calculating Critical Angle Snell’s Law: n1 sin1 = n2 sin2Critical angle C = 1 thus solve for 1At the critical angle, the refracted ray is a 90o = 2n1 sin1 = n2 sin90o (sin90o = 1)n1 sin1 = n2 1n1 sin1 = n2Rearrange equation to get:sin1 = n2 / n11 = sin-1 (n2 / n1)If medium 2 = air, then n2 = 11 = sin-1 (1 / n1)C = sin-1 (1 / n1)airglassC = 1rrefracted rayreflected rayincident rayR = 2
22 TOTAL INTERNAL REFLECTION critical angles of different materialsMediumRefractive indexCritical angleGlassWaterPerspexDiamond1.50–1.7030–421.33491.5422.4224
23 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.
24 TIR in Fiber OpticsTechnology that uses light to transmit information along glass cablesFibre optics cable is made up of a bundle of glass fibresSample materials: high-purity glass, Lucite
25 TIR in Fiber OpticsFiber optics cable has a small critical angle, thus a high refractive indexLight entering will always have an angle of incidence greater than the critical angle
26 TIR in Fiber OpticsLight does not escape as it travels along the fiber optics cable because it undergoes total internal reflection
28 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
29 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.
30 Gastroscopy Endoscopy examination of a stomach Endoscope inserted through the patient's mouth and fed down through throatImage obtained by endoscope is projected onto a screenA surgical instrument for obtaining a biopsy has been fed through the endoscope cable and controlled by the doctorGastroscopy
32 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 CancerPhotographed in Belgium.
33 TIR in Prisms I1 I2 I3 Object Plane mirror = glass + silvered surface multiple reflection inside the glassmultiple images formednuisance in optical instrumentsglass sheetsilvered surfaceI I I3Object33
34 45PrismsIf 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.4534
35 Prisms in PeriscopesInstrument for observation from a concealed positionUses two triangular prisms (or mirrors) to change direction of light by 90oUsed in war and in submarines
36 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.
38 SubmarinesAmerican submarine commander inspects the horizon through the periscope (1942)
39 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.
40 Prisms in BinocularsUses 2 prisms to change direction of light by 180o40
41 Prisms in Single-lens Camera A five-sided ‘pentaprism’ reflects light from the mirror into the eye.filmmirror41
42 Prisms and Retro-reflectors 45Device that returns incident light back in exactly the same direction from which it cameApplications in bike reflectors, reflective strip on clothing, road signs
43 Practice Problems3018EOA 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.
44 Practice Problems E 30 O 18 Given: 1 = 30, 2 = 18, n1 (air) = 1 Required: n2 (block)Analysis: n1 sin1 = n2 sin2Solution:1 x sin 30 = n2 sin 18n2 = sin 30 / sin 18 = 1.62Phrase:The index of refraction of the block is 1.623018EO
45 Practice Problems E 30 O 18 b. Find the critical angle C for the block.3018EORecall: If medium 2 = air, thenC = 1 = sin-1 (1 / n1)
46 Practice Problems E 30 O 18 b. Find the critical angle C for the block.Given: 2 = 90, n2 (air) = 1, n1 (block) = 1.62Required: 1Analysis: n1 sin1 = n2 sin2Solution:1.62 x sin 1 = 1 x sin 90sin 1 = 1 / 1.621 = sin-1 (1 / 1.62) = 38.1Phrase:The critical angle of the block is 38.13018EORecall: If medium 2 = air, thenC = 1 = sin-1 (1 / n1)
47 Practice Problemsc. If the ray is incident on surface BC, from which surface and at what angle will the ray leave the block?30ABCD
48 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 = 60n1 (air) = 1n2 (block) = 1.62Required: 2Analysis:n1 sin1 = n2 sin2Solution:1 x sin 60 = 1.62 x sin 2sin 2 = 1 x sin 60 / 1.622 = sin-1 (1 x sin 60 / 1.62) = 32.330ABCD
49 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 = 60oAngle of refraction = 32.3oCritical angle = 38.1o30ABCD32.357.760Draw 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.
50 Practice ProblemsWhich of the following angles is the critical angle of glass?ABCD
51 Practice ProblemsWhich of the following angles is the critical angle of glass?ABCD
52 Practice Problems A horizontal light ray hits a prism as shown. 45A horizontal light ray hits a prism as shown.What happens to the light ray?BCA
53 Practice Problems A horizontal light ray hits a prism as shown. 45A horizontal light ray hits a prism as shown.What happens to the light ray?BCA