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Unformatted text preview: REFLECTION AND REFRACTION E XERCISES Section 30.1 Reflection 12. Since 1 1 θ θ ′ = for specular reflection, (Equation 30.1) a reflected ray is deviated by 1 180 2 φ θ = °  from the incident direction. If rotating the mirror changes 1 θ by 1 , θ ∆ then the reflected ray is deviated by 1  2  φ θ ∆ =  ∆ or twice this amount. Thus, if 1 30 ,   15 . φ θ ∆ = ° ∆ = ° 13. I NTERPRET This problem involves sketching the path of the light ray reflected from the surfaces of two mirrors. D EVELOP The path of the reflected ray can be constructed using the law of reflection which states that the angle of incidence equals the angle of reflection: 1 1 . θ θ ′ = E VALUATE The first reflected ray leaves the upper mirror at a grazing angle of 30 ° , and therefore strikes the lower mirror normally. It is then reflected twice more in retracing its path in the opposite direction. A SSESS Our doublemirror arrangement is a retroreflector that sends light rays back to their point of origin. Retroreflection has many practical applications (e.g., taillights, stop signs, etc.). 14. A ray incident on the first mirror at a grazing angle α is deflected through an angle 2 α (this follows from the law of reflection). It strikes the second mirror at a grazing angle , β and is deflected by an additional angle 2 . β The total deflection is 2 2 2(180 ). α β θ + = °  If this is to be within 180 1 , °± ° then the angle between the mirrors, , θ must be within 1 2 90 . ° ± ° 15. I NTERPRET This problem is about the direction of the light ray after undergoing reflection from a twomirror arrangement. 30.1 30 30.2 Chapter 30 D EVELOP The path of the reflected ray can be constructed using the law of reflection: the angle of incidence equals the angle of reflection: 1 1 . θ θ ′ = E VALUATE Entering parallel to the top mirror, a ray makes an angle of incidence of 30 ° with the bottom mirror. It then strikes the top mirror also at 30 ° incidence, and is reflected out of the system parallel to the bottom mirror (see the figure). The total deflection is 240 ° CCW (or 120 ° CW) from the incident direction. A SSESS The reflected path follows from the law of reflection. Section 30.2 Refraction 16. Since the speed of light in a medium is 8 8 / , 3 10 /2.292 10 1.309. v c n n = = × × = This matches ice in Table 30.1. 17. I NTERPRET This problem is about the pit depth of a CD, given the wavelength of the laser light used and the index of refraction of the CD. D EVELOP Using Equation 30.2, / , n c v = and the reasoning in Example 30.3, the wavelength in the plastic can be written as air / . n λ λ = The pit depth is /4. λ E VALUATE Substituting the values given, we find the wavelength of the laser light to be air 780 nm 503 nm 1.55 n λ λ = = = The pit depth is one quarter of this, or 126 nm....
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 Spring '09
 Prof.Kim
 Physics, Reflection And Refraction, Angle of Incidence, Snell's Law, Total internal reflection, Geometrical optics, EVALUATEThus Snell

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