Ch33Word - Ch. 33 p. 1 CHAPTER 33 Light: Reflection and...

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Ch. 33 p. 1 CHAPTER 33 – Light: Reflection and Refraction 1. ( a ) The speed in ethyl alcohol is v = c / n = (3.00 × 10 8 m/s)/(1.36) = 2.21 × 10 8 m/s . ( b ) The speed in lucite is v = c / n = (3.00 × 10 8 m/s)/(1.51) = 1.99 × 10 8 m/s . 2. We find the index of refraction from v = c / n ; 2.29 × 10 8 m/s = (3.00 × 10 8 m/s)/ n , which gives n = 1.31 . 3. The time for light to travel from the Sun to the Earth is ? t = L / c = (1.50 × 10 11 m)/(3.00 × 10 8 m/s) = 5.00 × 10 2 s = 8.33 min . 4. We convert the units: d = (4.2 ly)(3.00 × 10 8 m/s)(3.16 × 10 7 s/yr) = 4.0 × 10 16 m . 5. The length in space of a burst is ? s = c ? t = (3.00 × 10 8 m/s)(10 –8 s) = 3 m . 6. The speed in water is v water = c / n water . We find the index of refraction in the substance from v = c / n ; 0.92 v water = 0.92 c / n water = c / n , which gives n = 1.33/0.92 = 1.45 . 7. The eight-sided mirror would have to rotate 1/8 of a revolution for the succeeding mirror to be in position to reflect the light in the proper direction. During this time the light must travel to the opposite mirror and back. Thus the angular speed required is ω = ? θ /? t = (2p rad/8)/(2 L / c ) = (p rad) c /8 L = (p rad)(3.00 × 10 8 m/s)/8(35 × 10 3 m) = 3.4 × 10 3 rad/s (3.2 × 10 4 rev/min). 8. For a flat mirror the image is as far behind the mirror as the object is in front, so the distance from object to image is d o + d i = 1.8 m + 1.8 m = 3.6 m . 9. We show some of the rays from the tip of the arrow that form the three images. Single reflections form the two side images. Double reflections form the third image. The two reflections have reversed the orientation of the image. I 1 I 2 I 3 Object
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Ch. 33 p. 2 10. The angle of incidence equals the angle of reflection. Thus we have t a n θ = ( H h )/ L = h / x ; (1.54 m – 0.40 m)/(2.30 m) = (0.40 m)/ x , which gives x = 0.81 m = 81 cm . 11. From the triangle formed by the mirrors and the first reflected ray, we have + α + φ = 180°; 40° + 135° + = 180°, which gives = 5°. 12. Because the rays entering your eye are diverging from the image position behind the mirror, the diameter of the area on the mirror and the diameter of your pupil must subtend the same angle at the image: D mirror / d i = D pupil /( d o + d i ); D mirror /(80 cm) = (5.0 mm)/(80 cm + 80 cm), which gives D mirror = 2.5 mm. Thus the area on the mirror is A mirror = # p D mirror 2 = # p(2.5 × 10 –3 m) 2 = 4.9 × 10 –6 m 2 . 13. For the first reflection at A the angle of incidence 1 is the angle of reflection. For the second reflection at B the angle of incidence 2 is the angle of reflection. We can relate these angles to the angle at which the mirrors meet, , by using the sum of the angles of the triangle ABC: + (90° – 1 ) + (90° – 2 ) = 180°, or = 1 + 2 . In the same way, for the triangle ABD, we have + 2 1 + 2 2 = 180°, or = 180° – 2( 1 + 2 ) = 180° – 2 . At point D we see that the deflection is β = 180° – = 180° – (180° – 2 ) = 2 . 14. ( a ) The velocity of the wave, which specifies the direction of the light wave, is in the direction of the ray. If we consider a single reflection, because the angle of incidence is equal to the angle of reflection, the component of the velocity perpendicular to the mirror is reversed, while the component parallel to the mirror is unchanged. When we have three mirrors at right angles, we can choose the orientation of the mirrors for our axes. In each of the three reflections, the component of velocity perpendicular to the mirror will reverse.
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch33Word - Ch. 33 p. 1 CHAPTER 33 Light: Reflection and...

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