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Chapter 25: Optics

25.B: Refraction and Lenses

Refraction
The change in direction of light as it enters a new medium at a nonzero angle due to the light’s speed changing in the new medium.
Lens
A disk of transparent material with one or both sides machined into a curved surface.
Converging Lens
A lens that causes light rays to come together, or to converge; it is thicker at the middle than at the edges.
Diverging Lens
A lens that causes light to spread apart, or diverge; it is thinner at the middle than at the edges.
Real Image
An image formed by converging light rays that can be displayed on a screen.
Virtual Image
An image formed by diverging light rays that cannot be displayed on a screen.

Snell’s Law

  • Snell’s law provides a relationship between the angle of refraction (θr) and the angle of incidence (θi): ni sin θi = nr sin θr.
  • θi: angle of incidence
  • θr: angle of refraction
  • ni: index of refraction of the medium containing the incident ray
  • nr: index of refraction of the medium containing the refracted ray

Using Ray Diagrams: Converging Lenses

  • A ray strikes the lens and bends; then it leaves the lens and bends again.
  • Both refractions bend the ray toward the principal axis and toward the focus.
  • Converging lenses form real images on the opposite side of the lens from its object.
  • There are six cases for converging lenses, including when the object is at infinity, beyond the radius of curvature, at the radius of curvature, between the radius of curvature and the focal point, at the focal point, and between the focal point and the lens.

Using Ray Diagrams: Diverging Lenses

  • A ray strikes the lens and bends; then it leaves the lens and bends again.
  • Both refractions bend the ray away from the principal axis.
  • Diverging lenses produce only one kind of image: virtual, upright, and reduced.
  • The virtual image appears to be on the same side of the lens as the object.

Example: Lens Problem—Ray Diagrams

A 2.50 cm object is 12.0 cm from a converging lens whose focal length is 5.00 cm. Use ray diagrams to determine the position and height of the image, then determine the magnification of the image. Finally, describe the image.

di = 8.6 cm

hi = –1.8 cm

M = hi / ho = –1.8 cm / 2.50 cm = –0.72

The image is real, reduced (72%), and inverted with a height of 1.8 cm. The image is 8.6 cm from the lens on the opposite side.

Questions for Students

  1. Define refraction and its cause.
  2. Explain Snell’s law and provide an example calculation.
  3. Describe the difference between converging and diverging lenses.
  4. Explain how to use ray diagrams to determine the position and height of an image formed by a lens.
  5. Calculate the magnification of an image given the object and image heights.