Snell’s law of refraction

Table of the indices of refraction for various materialsThe angle by which a light ray is refracted toward or away from the normal depends on the index of refraction on both sides of the boundary. The higher the difference between the two indices of refraction on either side of the boundary, the greater the refraction. Among common materials, the index of refraction varies from 1 (air) to 2.42 (diamond). The table on the right lists some typical values. The greatest difference we would normally see occurs between air and diamond. In part, this explains the characteristic “sparkle” of a polished diamond. Read the text aloud
Snell’s law of refraction relates the angles of the incident and refracted rays to the index of refraction on both sides of the refracting boundary. The formula says that the product of the index of refraction multiplied by the sine of the angle is the same on both the incident and refracting sides of the boundary. In equation (21.1) for Snell’s law of refraction, , ni and θi refer to the medium containing the incident ray. Similarly, nr and θr refer to the medium containing the refracted ray. Read the text aloud
(21.1) n i sin θ i = n r sin θ r
ni  = index of refraction, incident medium
nr  = index of refraction, refracted medium
θi  = angle of incidence (degrees)
θr  = angle of refraction (degrees)
Snell’s law
of refraction
When using Snell’s law, always remember to reference the angles of incidence and refraction to the normal, not to the boundary interface between the materials! The solved problem below illustrates what we mean. The sine of an angle is a mathematical function that varies from zero to one as the angle changes from 0° to 90°. In standard mathematical notation, sin(0°) = 0 and sin(90°) = 1. Read the text aloud
What is the index of refraction of the water?A light ray traveling from air (nair = 1.0) into the ocean is deflected, according to the figure at right. What is the index of refraction of the water?
Asked: index of refraction of water, nw
Given: index of refraction of air, na = 1.0;
figure showing incident and refracted angles
Relationships: Snell’s law: n a sin θ i = n w sin θ r
Solution: From the figure, recognize that the incident angle is
θi = 90° − 50° = 40°
and the refracted angle is
θr = 90° − 61.3° = 28.7°
Solve for the index of refraction of water: n w = n a sin θ i sin θ r = ( 1.0 )sin 40° sin 28.7° =1.34
Answer: nw = 1.34.
(Note that this is the value for saltwater!)
Read the text aloud
For light traveling through air and then entering another medium, which material bends light more, water or diamond? Show
What would happen to a light ray that enters a parallelogram prism with parallel sides at an angle? Show

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