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Chapter 35 The Nature of Light and the Laws of Geometric Optics EXAMPLES.

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Presentation on theme: "Chapter 35 The Nature of Light and the Laws of Geometric Optics EXAMPLES."— Presentation transcript:

1 Chapter 35 The Nature of Light and the Laws of Geometric Optics EXAMPLES

2 Chapter 35: Nature of Light and Laws of Geometric Optics; EXAMPLES

3 Example 35.1 Double Reflection Two mirrors make an angle of 120 o with each other. An incident ray strikes the mirror M 1 at an angle of 65 o to the normal. Find the direction of the ray after it is reflected from mirror M 2.  The reflected ray is directed toward the mirror M 2. Making an angle of 90 o  65 o = 25 o with the horizontal.  From the triangle made by the first reflection and the two mirrors: 180 o  120 o  25 o = 35 o (angle of the first reflected ray with M 2.  Therefore, this ray makes an angle of 55 o with the normal to M 2.  From the law of reflection the second reflected ray makes an angle of 55 o with the normal to M 2.

4 Example 35.2 Double Reflection (2 nd part) From Example 35.1 if the rays are extended behind the mirrors, they cross at 60 o, so that the overall change in direction of the light ray is 120 o (same angle between mirrors). If the angle between mirrors is changed, is the overall change in the direction of the light ray always equal to the angle between the mirrors?  NO! By using the law of reflection and the sum of the interior angles of a triangle:  = 180 o  (90 o   )   = 90 o +   .  From the highlighted triangle:  + 2  + 2 (90 o   ) =180 o   = 2(    )

5 Example 35.2 Double Reflection, final  The change in direction of the light ray is the angle  = 180 o     = 180 o  2(    )   = 180 o  2[   (90 o +    )]   = 360 o  2   If  = 120 o   = 120 o  If  = 90 o   = 180 o (RETROREFLECTION) Applications:

6 Quick Quiz 35.2 Following the Reflected and Refracted Rays Ray  is the incident ray Ray  is the reflected ray Ray  is refracted into the Lucite block Ray  is internally reflected in the block Ray  is refracted as it enters the air from the block

7 Example 35.3 Angle of Refraction for Glass Light of a wavelength 589 nm is refracted into a crown glass slab with  1 = 30.0 o. (A). Find the angle of refraction  2 = ? n 1 = 1.00 and n 2 = 1.52 (from Table 35.1) From eqn. (35.8) Then:  2 = sin -1 (n 1 / n 2 ) sin  1 = 19.2 o The ray bends toward the normal, as expected

8 Example 35.3 Angle of Refraction for Glass, 2 ( B). Find the speed of the light once enters the glass. From eqn. (35.4) (C). What is the wavelength of this light in the glass? From eqn. (35.7)

9 Example 35.4 Measuring n from a Prism The minimum angle of deviation (δ min ) for a prism occurs when the incident angle  1 is such that the refracted ray inside the prism makes the same angle with the normal to the two prism faces. Obtain an expression for the index of refraction of the prism material.

10 Example 35.4 Measuring n from a Prism, cont. From the geometry: From Snell’s Law (with n = 1 for air):

11 Example 35.5 Find Critical Angle  C Find  C for an Air-Water boundary. n 1 = 1.33 and n 2 = 1.00

12 Example 35.6 A view from the Fish’s Eye A fish in a still pond looks upward toward the water’s surface. What does it see? Using the result from Example 35.5  At an angle less than  C (48.8°) the fish can see out of the water  At 48.8°the light has to skim along the water’s surface before been refracted, so the fish can see the whole shore of the pond.  At an angle greater than  C (48.8°) the fish sees a reflection of the bottom of the pond.

13 Example 35.7 Light Passing Through a Slab ( A). A light beam passes from medium 1 to medium 2 with the latter medium being a thick slab of material whose index of refraction is. Show that the beam emerging into medium 1 from the other side is parallel to the incident beam. Apply Snell’s law to the upper surface: Apply Snell’s law to the lower surface: Therefore:

14 Example 35.7 Light Passing Through a Slab, 2 (B). If the thickness t of the slab is doubled, what does happen to the offset distance d? Find an expression for a from the yellow triangle: Find an expression for d from the red triangle: Combine these equations to find an expression for d in function of t: Conclusion between d and t: The angles are only related to index of refraction (constants), so. Therefore: If t doubles, so does d !

15 Material from the book to Study!!!  Objective Questions: 8-12  Conceptual Questions: 6-8  Problems: Material for the Midterm


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