Sol-Ch35 - CHAPTER 35 E xercises 3. There are (360/ ) 1 = 5...

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Exercises 3. There are (360/ θ ) – 1 = 5 images. Use the idea of a “virtual” mirror that extends behind the real mirrors. 4. Each ray is deflected by Ф . 5. Let the mirrors lie in the xy, xz and yz planes. If the original direction is along (i + j   + k ), three reflections will result in (–i j   – k ). 6. (a) n 1 λ 1 = n 2 λ 2 , so n 2 = 1.5; (b) v = c/n = 2 × 10 8 m/s 7. sin θ i = 1.4sin32°, so θ i = 47.9° Angle required: (90 – 47.9) + (90 – 32) = 100.1˚ 8. sin θ i = n sin(90 – θ i ) = 1.52 cos θ i . Thus tan θ i = 1.52 and θ i = 56.7˚. 9. Within water ∆x 1 = 3tan30° = 1.73m, 1.33sin30° = sin θ r , θ r = 41.7˚, tan41.7° = ∆x 2 /(1 m), thus ∆x 2 = 0.89 m; ∆x 1 + ∆x 2 = 2.62 m 10. Slope = n = 1.33 11. d = t sin( θ – r). For small θ , sinx ≈ x, Snell’s law takes the form: θ = nr, thus d ≈ t( θ – r) = t θ (1 – 1/n) = t θ (n – 1)/n. 12. nsis68° = 4/3, so n = 1.43 and v = c/n = 2.1 × 10 8 m/s 13. (4/3) sin θ c = 1, so θ c = 48.5. r = 2tan48.5° = 2.27 m 14. (a) Refracted ray emerges along interface just prior to total internal reflection. (b) n 1 sin θ 1 = n 2 . Need n 2 < n 1 . 15. For i = 90°, sini = 1.5sinr = 1; thus maximum sinr = 2/3, or CHAPTER 35
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minimum cosr = (5) 1/2 /3. For TIR, 1.5 sin θ c = 1 = 1.5cosr, so we need cosr > 2/3. This condition is satisfied since (5) 1/2 > 2. 16. After reflection, the angle of incidence at the bottom surface is 30°. From nsin30° = sin θ , thus θ = 48.6°. 17. n = sin[(
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This note was uploaded on 11/09/2011 for the course EE EE taught by Professor Ee during the Spring '11 term at National Chiao Tung University.

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Sol-Ch35 - CHAPTER 35 E xercises 3. There are (360/ ) 1 = 5...

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