Refraction Chapter 14: Section 1.

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Presentation transcript:

Refraction Chapter 14: Section 1

Learning Targets Describe how the index of refraction determines the direction light will be refracted Calculate the index of refraction for various substances Use Snell’s Law to calculate the angle of incidence or reflection for a light ray traveling between transparent media

Refraction of Light When light reaches a boundary between water and air, the light rays are bent The bending of light as it travels from one medium to another is called refraction

If light travels from one transparent medium to another, the light ray changes directions at the boundary This occurs for every angle except those that hit exactly perpendicular to the surface

The angles between the normal and the refracted ray are called the angle of refraction (Θr) and the angle of incidence (Θi)

What Causes Refraction? Refraction occurs when light’s velocity changes Glass, water, ice, and diamonds are all examples of transparent media through which light can pass The speed of light in each of these materials is different Example: The speed of light in water is less than the speed of light in air

Direction of Refraction When light moves from a higher speed material to a lower speed material, the ray is bent toward the normal Example: air to glass

If light moves from a lower speed material to a higher speed material, the ray is bent away from the normal If the incident ray of light is parallel to the normal, then no refraction (bending) occurs

Law of Refraction An important property of transparent substances is the index of refraction The index of refraction for a substance is the ratio of speed of light in a vacuum to the speed of light in that substance

The index of refraction is a dimensionless number that is always greater than 1 This is because light always travels faster in a vacuum The larger the index of refraction, the slower the light travels in a substance and the more a light ray will bend

Index of Refraction The following formula is used to calculate the index of refraction

Sample Problem The speed of light in an unknown medium is measured to be 2.76 x 108 m/s. What is the index of refraction of the medium? 1.09

Effects of Refraction Objects appear to be in different positions due to refraction

When looking at a fish underwater, a cat sitting on a pier perceives the fish to be closer to the water’s surface than it actually is The fish also perceives the cat on the pier to be farther from the water’s surface than it actually is

The light ray that reaches the cat from the water’s surface is bent away from the normal This is because air has a lower index of refraction Extending the ray along a straight line makes the fish appear closer to the water’s surface than it actually is

On the other hand, the light ray that reaches the fish forms a smaller angle with respect to the normal This is because light is bent toward the normal when it travels from air to water Extending this ray along a straight line shows the cat’s image to be above the cat’s actual position

Wavelength and Refraction The indices of refraction are only valid for light that has a wavelength of 589 nm in a vacuum The amount that light bends depends on the wavelength of light as well as the speed This is why a spectrum is produced when white light enters through a prism Each color of light has a different wavelength and is refracted by different amounts

Snell’s Law Snell’s Law is used to determine the angle of refraction As mentioned earlier, the greater the index of refraction, the more refraction occurs Snell’s law can be used to find the angle of refraction for light traveling between two media for any wavelength

The equation for Snell’s Law is…

Sample Problem A light ray of wavelength 589 nm traveling through air strikes a smooth, flat slab of crown glass at an angle of 30° to the normal. Find the angle of refraction, Θr. 19.2 °