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Refraction of Light at Interfaces. Scott P. Drexler O.D. Objectives. Snell’s Law Critical Angle Total Internal Reflection Chromatic Aberration. Interface. The boundary between two media with different indices. n. n’. n. n’. n. n’. air. water. air. glass. glass. water.
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Refraction of Light at Interfaces Scott P. Drexler O.D.
Objectives • Snell’s Law • Critical Angle • Total Internal Reflection • Chromatic Aberration
Interface • The boundary between two media with different indices n n’ n n’ n n’ air water air glass glass water
Index of Refraction • Recall that in media other than a vacuum, light waves slow down and the wavelength also decreases v=fl • n= Speed of light in a vacuum (C) Speed of light in material • n= 3.0x108m/s Speed of light in material
Index of Refraction • Since c is always the greatest, n is always greater than 1. • It is convention to treat the nair as 1.0 • Vacuum =1 • Air (nonpolluted) =1 • Water= 1.33 • PMMA =1.49 • Crown glass =1.52 • Diamond =2.417 • Cornea =1.376 • Zeiss hi-index =1.8 • Crystalline lens= 1.42
Index of Refraction • For certain prescriptions we will recommend a “high index” eyeglass lens. This means that the lens material is more optically dense so the eyeglass lens will be thinner because light is bent more per unit thickness of lens. • When the velocity of light, wavelength, and frequency are altered by the increased index of refraction the “nature “of light changes which can cause vague adaption problems
Snell’s Law n1sinq1 = n2sinq2 • n1= index of material before refraction • n2= index of material after refraction • 1= incident angle q 2= refracted angle
Snell’s Law • Light travelling from a less dense to a denser material will be refracted TOWARDS the normal. • Light travelling from a more dense to a less dense material will be refracted AWAY from the normal.
Refraction problem A light ray strikes air/ water interface at 30 degrees. What is the angle of refraction? nsinq = n’sinq’ 1.0 sin 30 = 1.33 sin X Sin X= (1.0)(.5)/1.33 = .3759 X= sin-1(.3759) = 22.08 degrees
Critical Angle • The largest possible angle of incidence which still results in a refracted ray • The refracted ray at the critical angle travels along the boundary of the 2 media • Only occurs when the light passes from higher index to lower index • Angle greater than the critical angle gives total internal reflection
Critical Angle- (n>n’) • For a high-low (water to air ,etc) interface, there is a physical limit of 90 degrees for the refracted angle- Law of Reflection • The critical angle is the incident angle that yields a refracted angle of 90 degrees • Ni•sinCA = nr•sin(90) CA=sin-1(n’/n)
Critical Angle Problem • Find the critical angle at an interface between crown glass (n=1.52) and air CA=sin-1(n’/n) = sin -1 (1.0/1.52) = .657 = sin-1(.667) = 41.10
Critical Angle and Total Internal Reflection N • If the angle of refraction were greater than 90 degrees, then the refracted ray would lie on the incident side of the medium • The light must be moving from the more dense to the less dense material • The angel of incidence must be greater than the CA air n n’ water cornea n n’ air
TIR and CA in Gonioscopy • Light going out of the eye is not visible because of TIR at the cornea ( n= 1.376) to air ( n= 1.0 ) interface
The cornea (n= 1.376 ) to gonioscopy lens (n=1.49) interface is not subject to TIR due to the increase in n value.
Chromatic Aberration • Speed of light = wavelength x frequency • n= velocity in vacuum velocity in material • A single material will have slightly different n values for different colors (wavelengths) • Blue( shorter wavelength) will refract more than red ( longer wavelength) • Eye has 3D spread of focus used clinically for Red-Green Test.
Chromatic Aberration Minimize by : • Changing lens shape-aspheric better • Changing Refractive Index- lower better • Changing Aperture- smaller better, central better
Use Snell’s law repeatedly to figure out all angles Snell’s Law for Refraction through multiple interfaces 1 2 3 4 air air 1.33 n3 n2 n4 n1 n5 1.5 1.5 25
n1sinq1 = n2sinq2 • 1: • 2: • 3: • 4: 1.0sin25=1.5sinX=16.3 1.5sin 16.3=1.33sinX=18.45 1.33sin 18.45=1.5sinX=16.3 1.5sin16.3=1.0sinX = 25