chap30 - REFLECTION AND REFRACTION 30 EXERCISES Section...

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30.1 REFLECTION AND REFRACTION 30 EXERCISES Section 30.1 Reflection 12. Since 11 θ = for specular reflection, (Equation 30.1) a reflected ray is deviated by 1 180 2 φ = °− from the incident direction. If rotating the mirror changes 1 by 1 , Δ then the reflected ray is deviated by 1 |2 | Δ=−Δ or twice this amount. Thus, if 1 30 , | | 15 . Δ= °Δ = ° 13. INTERPRET This problem involves sketching the path of the light ray reflected from the surfaces of two mirrors. DEVELOP 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: . ′ = EVALUATE 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. ASSESS Our double-mirror 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 (this follows from the law of reflection). It strikes the second mirror at a grazing angle , β and is deflected by an additional angle2. The total deflection is 2 2 2(180 ). += ° If this is to be within180 then the angle between the mirrors, 1 , °± ° , must be within 1 2 90 . 15. INTERPRET This problem is about the direction of the light ray after undergoing reflection from a two-mirror arrangement.
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30.2 Chapter 30 DEVELOP The path of the reflected ray can be constructed using the law of reflection: the angle of incidence equals the angle of reflection: 11 . θ ′ = EVALUATE 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. ASSESS The reflected path follows from the law of reflection. Section 30.2 Refraction 16. Since the speed of light in a medium is This matches ice in Table 30.1. 88 / , 3 10 /2.292 10 1.309. vc n n == × × = 17. INTERPRET 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. DEVELOP Using Equation 30.2, and the reasoning in Example 30.3, the wavelength in the plastic can be written as /, nc v = air /. n λ = The pit depth is /4. EVALUATE Substituting the values given, we find the wavelength of the laser light to be air 780 nm 503 nm 1.55 n = The pit depth is one quarter of this, or 126 nm. ASSESS A typical pit on a CD is about 100 nm deep and 500 nm wide. Our result in within this range. 18. Snell’s law (with gives air 1) n = 12 2 1 sin ( sin / ) sin (1.52sin 40 /1) 77.7 .
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.

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chap30 - REFLECTION AND REFRACTION 30 EXERCISES Section...

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