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Chapter 24 Geometric Optics

Chapter 24 Geometric Optics. Outline of Chapter. Ray Model of Light Reflection : Image Formation by Plane Mirrors Image formation by Spherical Mirrors Index of Refraction Refraction : Snell’s Law Visible Spectrum and Dispersion Total Internal Reflection Fiber Optics

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Chapter 24 Geometric Optics

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  1. Chapter 24Geometric Optics

  2. Outline of Chapter • Ray Model of Light • Reflection: Image Formation by Plane Mirrors • Image formation bySpherical Mirrors • Index of Refraction • Refraction: Snell’s Law • Visible Spectrum and Dispersion • Total Internal Reflection • Fiber Optics • Refraction at a Spherical Surface

  3. The Ray Model of Light • Ray Optics (sometimes called geometric optics) • involves the study of the propagation of light. • Uses the assumption that light travels in a • straight-line path in a uniform medium & • changes its direction when it meets the • surface of a different medium or if the optical • properties of the medium are nonuniform. • The ray approximation is used to • represent beams of light.

  4. The Ray Model of Light • Often, light travels in straight lines, so we can • represent it using Rays. • Rays are straight lines coming from an object. • This is a model or an idealization, but it is very • useful for geometric optics. Light Rays

  5. Reflection:Image Formation-Plane Mirrors • Experiments show that rays always obey • The Law of Reflection: • Angle of reflection (angle that the ray makes WITH THE NORMAL to the mirror) = angle of incidence. • θi = θr

  6. Image Formation by a Plane Mirror • When light reflects from a rough surface, the law of • reflection still holds, but the angle of incidence • varies. This is called diffuse reflection.

  7. With diffuse reflection, • the eye sees reflected • light at all angles. • With specular • reflection • (from a mirror), the • eye must be in the • correct position.

  8. Image Formation by a Plane MirrorExample: Reflection from Flat Mirrors • Two flat mirrors are perpendicular to each other. • An incoming beam of light makes an angle of 15° • with the first mirror as shown. • What angle θ5will the outgoing beam make with the • second mirror? Note that the angles in the Law of • Reflection are with respect to mirror normal,NOT • with respect to the mirror itself!)

  9. Example: Reflection from Flat Mirrors • Solution:Geometry θ1 + 15º = 90º, so • θ1 = 75º. • Law of reflection: θ2 = 75º. Also, θ2 + θ3 + 90º = 180º. • So θ3 = 15º. Law of reflection: θ4 = 15º. • So, θ4 = 75º. So, the outgoing ray is parallel to the • incoming ray!

  10. When you look into a plane (flat) mirror, you see an • image appearing to be behind the mirror. This is called a • Virtual Image, because the light does not go through it. • The distance of the image from the mirror is equal to the distance of the object from the mirror.

  11. Example: How tall must a full-length mirror be? • A woman, 1.60 m tall, is in • front of a vertical plane mirror. • Find the minimum mirror height • & the distance its lower edge • must be from the floor, so that • she can see her whole body. • Assume that her eyes are 0.1 m • below the top of her head.

  12. Example: How tall must a full-length mirror be? • A woman, 1.60 m tall, is in front • of a vertical plane mirror. • Find the minimum mirror height & • the distance its lower edge must be • from the floor, so that she can see • her whole body. • Assume that her eyes are 0.1 m • below the top of her head. • Solution: Consider the ray that leaves her foot at A. If it comes to her • eye at E, this means a reflection at B, with θi = θr. So, in the figure, • BD = (½)AE. We know that AE = 1.6 – 0.1 = 1.50 m. So, BD = 0.75 m. • Similar analyses can be made for other rays leaving the woman, reflecting • & going to her eye. For example, from the top of her head at G to get to her eye • at E, the top edge of the mirror needs to be at F, which is 0.05 m below G. • So, DF = 1.55 m& the mirror needs to have height FB = 1.55 – 0.75 = 0.8 m. • Also, distance BD = 0.75 m = distance of the lower mirror edge above the floor.

  13. Conceptual Example:Is the photo upside down? Close examination of this photograph reveals that in the top portion, the image of the Sun is seen clearly, whereas in the lower portion, the image of the Sun is partially blocked by the tree branches. Show why the reflection is not the same as the real scene by drawing a sketch of this situation, showing the Sun, the camera, the branch, and two rays going from the Sun to the camera (one direct and one reflected). Is the photograph right side up?

  14. Solution:

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