solution_pdf39

# solution_pdf39 - tovar(jdt436 homework 39 Turner(59070 This...

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tovar (jdt436) – homework 39 – Turner – (59070) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points A light ray is traveling in a medium with an index oF reFraction n 1 and it is re±ected at the boundary oF a second medium with an index oF reFraction n 2 . Considering the change oF the relative phases Δ φ due to their re±ections, which oF the Following conditions is correct? 1. IF n 1 > n 2 , then Δ φ = 0 and iF n 1 < n 2 , then Δ φ = π 2 . 2. IF n 1 > n 2 , then Δ φ = 0 and iF n 1 < n 2 , then Δ φ = 0. 3. IF n 1 > n 2 , then Δ φ = π 2 and iF n 1 < n 2 , then Δ φ = π . 4. IF n 1 > n 2 , then Δ φ = 0 and iF n 1 < n 2 , then Δ φ = π . correct 5. IF n 1 > n 2 , then Δ φ = π and iF n 1 < n 2 , then Δ φ = π . 6. IF n 1 > n 2 , then Δ φ = π 2 and iF n 1 < n 2 , then Δ φ = π 2 . Explanation: IF a light ray is traveling in a medium with an index oF reFraction n 1 and it is re±ected at the boundary oF a second medium with an index oF reFraction n 2 , the phase change is Δ φ = 0 when n 1 > n 2 and is π when n 1 < n 2 . 002 (part 2 oF 2) 10.0 points Consider the optical coating on a glass lens where the index oF reFraction oF the coating is n , where n is greater than the index oF reFraction oF the air. air lens t n θ θ 0 1 2 Assume: The index oF reFraction oF the lens is greater than that oF the coating. To minimize the re±ection oF a ray with a wavelength λ = 500 nm and incident angle θ 0, what is the minimum nonzero thickness t oF the coating? 1. t = λ 8 n 2. t = λ n 3. t = n λ 4. t = n λ 2 5. t = 3 λ 4 n 6. t = λ 2 n 7. t = n λ 4 8. t = n λ 8 9. t = λ 4 n correct 10. t = 3 n λ 4 Explanation: Destructive interFerence occurs when the di²erence between the phase angle oF the inci- dent ray re±ected From the outer surFace oF the coating (ray 1) and the phase angle oF the ray re±ected From the inner surFace oF the coating (ray 2) are at π, 3 5 π etc. The phase di²er- ences are due to the path di²erence oF the two rays and the change oF their relative phases due to re±ections. Putting them together, it

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solution_pdf39 - tovar(jdt436 homework 39 Turner(59070 This...

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