Objective Questions: 8-12 Conceptual Questions: 6-8 Problems: 6-8-15-16-20-37-39 Material as long as the Midterm

Objective Questions: 8-12 Conceptual Questions: 6-8 Problems: 6-8-15-16-20-37-39 Material as long as the Midterm www.phwiki.com

Objective Questions: 8-12 Conceptual Questions: 6-8 Problems: 6-8-15-16-20-37-39 Material as long as the Midterm

Brennan, Dan, Morning Host has reference to this Academic Journal, PHwiki organized this Journal Chapter 35 The Nature of Light in addition to the Laws of Geometric Optics EXAMPLES Chapter 35: Nature of Light in addition to Laws of Geometric Optics; EXAMPLES Example 35.1 Double Reflection Two mirrors make an angle of 120o with each other. An incident ray strikes the mirror M1 at an angle of 65o to the normal. Find the direction of the ray after it is reflected from mirror M2. The reflected ray is directed toward the mirror M2. Making an angle of 90o 65o = 25o with the horizontal. From the triangle made by the first reflection in addition to the two mirrors: 180o 120o 25o = 35o (angle of the first reflected ray with M2. There as long as e, this ray makes an angle of 55o with the normal to M2. From the law of reflection the second reflected ray makes an angle of 55o with the normal to M2.

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Example 35.2 Double Reflection (2nd part) From Example 35.1 if the rays are extended behind the mirrors, they cross at 60o, so that the overall change in direction of the light ray is 120o (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 in addition to the sum of the interior angles of a triangle: = 180o (90o ) = 90o + . From the highlighted triangle: + 2 + 2 (90o ) =180o = 2( ) Example 35.2 Double Reflection, final The change in direction of the light ray is the angle = 180o = 180o 2( ) = 180o 2[ (90o + )] = 360o 2 If = 120o = 120o If = 90o = 180o (RETROREFLECTION) Applications: Quick Quiz 35.2 Following the Reflected in addition to 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

Example 35.3 Angle of Refraction as long as Glass Light of a wavelength 589 nm is refracted into a crown glass slab with 1 = 30.0o. (A). Find the angle of refraction 2 = n1 = 1.00 in addition to n2 = 1.52 (from Table 35.1) From eqn. (35.8) Then: 2 = sin-1(n1 / n2) sin 1 = 19.2o The ray bends toward the normal, as expected Example 35.3 Angle of Refraction as long as 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) Example 35.4 Measuring n from a Prism The minimum angle of deviation (min) as long as 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 as long as the index of refraction of the prism material.

Example 35.4 Measuring n from a Prism, cont. From the geometry: From Snell’s Law (with n = 1 as long as air): Example 35.5 Find Critical Angle C Find C as long as an Air-Water boundary. n1 = 1.33 in addition to n2 = 1.00 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 be as long as e 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.

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: There as long as e: 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 as long as a from the yellow triangle: Find an expression as long as d from the red triangle: Combine these equations to find an expression as long as d in function of t: Conclusion between d in addition to t: The angles are only related to index of refraction (constants), so . There as long as e: If t doubles, so does d ! Material from the book to Study!!! Objective Questions: 8-12 Conceptual Questions: 6-8 Problems: 6-8-15-16-20-37-39 Material as long as the Midterm

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Brennan, Dan Morning Host

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