Download Quiz 3: Refraction and Optics and more Quizzes Health sciences in PDF only on Docsity! Version PREVIEW – quiz 3 (sample) – li – (58025) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Refraction 03 001 16.66 points In the figure, a ray of light enters the liquid from air and is bent toward the normal as shown. 7.6 10 10 7.6 n What is the index of refraction n for the liquid? 1. @@@ 2. @@@ 3. @@@ 4. @@@ 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 1.31579. Explanation: Let : a = 10 , and b = 7.6 . b a a b c c θ1 θ2 By Snell’s Law, n1 sin θ1 = n2 sin θ2 1 a c = n2 b c n2 = a c · c b = a b = 10 7.6 = 1.31579 . Double Slits 04 002 16.66 points In the double slit arrangement shown in the figure, the slits S1 and S2 are illuminated with light whose wavelength is 643 nm . y 2800 cm 0 .7 5 m m S1 S2 θ v ie w in g s cr e e n What is the spacing between the adjacent bright fringes on the viewing screen? 1. @@@ 2. @@@ 3. @@@ 4. @@@ 5. @@@ 6. @@@ 7. @@@ Version PREVIEW – quiz 3 (sample) – li – (58025) 2 8. @@@ 9. @@@ 10. @@@ Correct answer: 2.40053 cm. Explanation: Let : L = 2800 cm d = 0.75 mm λ = 643 nm r2 r1 y L d S1 S2 θ = tan −1 ( y L ) v ie w in g s cr e e n δ ≈ d sin θ ≈ r2 − r1 P O 6 S2 Q S1 ≈ 90 ◦ Q Solution: The bright fringes are the inten- sity maxima resulting from the constructive interference, which requires the path differ- ence to be integer multiples of the wavelength of the light d sin θ = mλ , where m is an integer starting from 0,1,2....... r2 r1 d S1 S2 θ = tan −1 ( y L ) θ δ ≈ d si n θ ≈ r2 − r1 6 S2 QS1 ≈ 9 0 ◦ Q But (assuming θ is small, sin θ ≈ tan θ) sin θ = y L , so the distances of these bright fringes from the center of the screen can be calculated as ym = m λ L d . The spacing ∆y between the adjacent bright fringes is ∆y = λ L d = (643 nm) (2800 cm) (0.75 mm) = 2.40053 cm . Tipler PSE5 32 76 003 16.66 points Suppose the eye were designed like a camera with a lens of fixed focal length 2.5 cm that could move toward or away from the retina. Approximately how far would the lens have to move to focus the image of an object 39 cm from the eye onto the retina? 1. @@@ 2. @@@ 3. @@@ 4. @@@ 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 0.171233 cm. Explanation: Let : f = 2.5 cm , and s = 39 cm . Using the thin-lens equation, the location of the image is 1 s′ = 1 f − 1 s = s − f f s s′ = f s s − f = (2.5 cm) (39 cm) 39 cm − 2.5 cm = 2.67123 cm ,
