Exam4-solutions

# Exam4-solutions - Version 058 – Exam4 – Hoffmann...

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Unformatted text preview: Version 058 – Exam4 – Hoffmann – (56745) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The unit for the answer choices is centimeter (cm). A concave mirror has a focal length of 36 cm. What is the position of the resulting image if the image is inverted and 4 times smaller than the object? 1. 40.5 2. 64.1667 3. 48.0 4. 45.0 5. 12.25 6. 24.0 7. 90.0 8. 18.2857 9. 61.2 10. 47.5 Correct answer: 45 cm. Explanation: 1 p + 1 q = 1 f = 2 R m = h ′ h =- q p Concave Mirror f > ∞ >p> f f <q < ∞ >m>-∞ f >p>-∞ <q < ∞ >m> 1 Let : f = 36 cm and n = 4 . Since the image is inverted, n =- 1 m . From the mirror equation 1 p + 1 q = 1 f and p q = n , so 1 nq + 1 q = 1 f q = ( n + 1) f n = (4 + 1) (36 cm) 4 = 45 cm . 002 10.0 points A beam of light of wavelength 530 nm travel- ing in air is incident on a slab of transparent material. Given: The incident beam makes an angle of 40 . 6 ◦ with the normal, and the refracted beam makes an angle of 24 . 7 ◦ with the nor- mal. Find the index of refraction of the material. 1. 1.62899 2. 2.32801 3. 1.90638 4. 1.58386 5. 1.93073 6. 2.1496 7. 1.55737 8. 1.79652 9. 1.69718 10. 2.07826 Correct answer: 1 . 55737. Explanation: Snell’s law of refraction n 1 sin θ 1 = n 2 sin θ 2 with these data gives n 2 = n 1 sin θ 1 sin θ 2 = 1 sin(40 . 6 ◦ ) sin(24 . 7 ◦ ) = 1 . 55737 . 003 10.0 points Two parallel polarizer disks are aligned with the transmission axis of the first oriented at an angle θ 1 clockwise from the vertical, and that of the second at an angle θ 2 clockwise from the vertical, with 0 < θ 1 < θ 2 . A horizontal beam of polarized light with intensity I and the polarization direction pointing vertically passes through both polarizers. The intensity I of the light is 1. I = I cos 2 θ 2 . Version 058 – Exam4 – Hoffmann – (56745) 2 2. I = I cos 2 θ 1 cos 2 ( θ 2- θ 1 ) . correct 3. I = I cos 2 θ 2 . 4. I = I cos 2 θ 1 . 5. I = I cos 2 θ 1 cos 2 θ 2 . 6. I = I cos 2 θ 2 cos 2 θ 1 . 7. I = I cos 2 θ 1 cos 2 θ 2 . 8. I = I cos 2 θ 1 . 9. I = I . 10. None of these. Explanation: When polarized light passes through a polarizer, the transmitted intensity is I = I cos 2 θ , where θ is the angle between the di- rection of polarization of the light and the axis of the polarizer. Now we have two successive polarizers, so the resulting intensity will be I = I cos 2 θ 1 cos 2 ( θ 2- θ 1 ) . 004 10.0 points The unit for the answer choices is millimeter (mm). Consider the setup of a single slit experi- ment. The wavelength of the incident light is λ = 500 nm . The slit width and the distance between the slit and the screen is specified in the figure....
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## This note was uploaded on 01/27/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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Exam4-solutions - Version 058 – Exam4 – Hoffmann...

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