HW3solution08

HW3solution08 - Version PREVIEW – HW 3 – Savrasov...

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Unformatted text preview: Version PREVIEW – HW 3 – Savrasov – (39824) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Wavelength Comparison 001 10.0 points The speed of light is 2 . 998 × 10 8 m / s. The ratio of the wavelength of visible light of wavelength 600 nm to that of an FM radio wave of frequency 100 MHz is about 1. 10 − 1 2. 10 − 7 correct 3. 10 − 3 4. 10 − 14 5. 10 − 10 Explanation: The wavelength of the FM radio wave is λ FM = c f FM = 2 . 998 × 10 8 m / s 100 MHz = 2 . 998 m . Then the ratio of the wavelength of visible light to that of the FM radio wave is about 1 × 10 − 7 . keywords: Perpendicular Mirrors 002 (part 1 of 2) 10.0 points Consider the case in which light ray A is in- cident on mirror 1 , as shown in the figure. The reflected ray is incident on mirror 2 and subsequently reflected as ray B. Let the an- gle of incidence (with respect to the normal) on mirror 1 be θ A = 50 ◦ and the point of incidence be located 20 cm from the edge of contact between the two mirrors. 90 ◦ x A B mirror 1 mirror 2 θ A θ B What is the angle of the reflection of ray B (with respect to the normal) on mirror 2 ? Correct answer: 40 ◦ . Explanation: If the angle of the incident ray A is θ A , the angle of reflection must also be θ A . Since the mirrors are perpendicular to each other, angle θ B is equal to 90 ◦- θ A θ B = 90 ◦- θ A = 90 ◦- 50 ◦ = 40 ◦ . 003 (part 2 of 2) 10.0 points Determine the angle between the rays A and B. 1. 50 ◦ 2. ◦ 3. 80 ◦ 4. 40 ◦ 5. 180 ◦ correct 6. 100 ◦ 7. 90 ◦ Explanation: Refer to the figure below. Consider the angle each ray makes with mirror 1 . Ray A Version PREVIEW – HW 3 – Savrasov – (39824) 2 makes an angle of...
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This note was uploaded on 01/10/2010 for the course PHY na taught by Professor Richardscalettar during the Fall '09 term at UC Davis.

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HW3solution08 - Version PREVIEW – HW 3 – Savrasov...

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