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UVM PHYS 012 - Exam 3 Study Guide
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Physics 012 1st EditionExam # 3 Study Guide Lectures: 21 - 29Lecture 21 (March 16)Law of ReflectionQuestion: Two mirrors meet at an angle of 120 degrees. A beam of light hits the first mirror at 65 degrees relative to the normal and then reflects off of the second mirror. What is the angle between the final light beam and the normal of the second mirror?Answer:180 – 120 – 65 = 35 degrees (within triangle)90 – 35 = 55 degrees (complimentary angles)Lecture 22 (March 18) Curved MirrorsQuestion: An object is placed 25 cm in front of a concave mirror with a focal length of 10 cm. Find the distance between the mirror and the image and the magnification of the image.Answer: 1/di = 1/f + 1/do = 1/10 – 1/25 = 3/50 cm-1di = 50/3 cm (real image)m = -di/do = -(50/3)/(25) = 2/3 x (smaller, upright)Lecture 23 (March 20)Refraction of LightQuestion: A beam of light has a wavelength of 550 nm traveling in air is incident on a transparent material. The incident angle is 40 degrees to the normal. The refracted beam is 20 degrees to the normal. Assume n = 1 for air. Find n for and speed of light through the transparent material.Answer: n2 = (1)sinθ1/sinθ2 = sin(40)/sin(26) = 1.47n = c/vv = c/n = 2.04 x 10 8 m/sLecture 24 (March 23)Total Internal Reflection – Fiber OpticsQuestion: Find max value for θ1 to get total internal reflection in glass cylinder.Answer: ngsinθ = nasin90 = 1θ1 = sin-1(ngsinθ2/na) = 67.1 degreesLecture 25 (March 25)Thin LensesQuestion: An object is placed 20 cm in front of a diverging lens with a focal length of -32 cm. Find the distance from the lens to the image and the magnification of the image.Answer: 1/di = 1/f + 1/do = 1/-32 – 1/20di = -12.3 cm (virtual image, on same side of lens)m = -di/do = -(-12.3)/(20) = 0.615 x (smaller, upright)Lecture 26 (March 27)Laser Double Slit ExperimentQuestion: A laser is sent through a screen with two slits. The distance between the slits is 0.03 mm. The light is projected onto a second screen 1.2 m away from the first screen. The distance from the center of the screen to the second order fringe is 5.1 m. Find the wavelength of the light.Answer: tanθ = y/Ldsinθ = mλθ = tan-1(y/L)dsin(tan-1(y/L)) = mλλ = [dsin(tan-1(y/L))]/m = [(0.03)sin(tan-1(1.2/5.1))]/2 = 6.4 x 10 -7 mLecture 27 (March 30)Diffraction GratingsQuestion: Light of wavelength 406 nm (in vacuum) is incident on a diffraction grating that has a slit separation of 1.2 × 10-5 m. The distance between the grating and the viewing screen is 0.16m. A diffraction pattern is produced on the screen that consists of a central bright fringe and higher-order bright fringes (see the drawing). Determine the distance y from the central bright fringe to the second-order bright fringe.Answer: sinθ = mλ/dtanθ = y/Lsinθ ≈ tanθ (when d << L)mλ/d = y/Ly = mλL/d = (2)(410 x 10-9m)(0.15m)/(1.2 x 10-5m) = 0.01025 mLecture 28 (April 1)Thin Film Interference 27.18Question: A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in visible light with wavelength 643 nm in vacuum. Assuming that the visible spectrum extends from 380 to 750 nm, what is the longest visible wavelength (in vacuum) for which the film will appear bright due to constructive interference?Answer: t = λvac/2no2not = (m+ ½) λvac2no(λvac/2no) = (m+ ½) λvacλ = λred/(m+ ½)m = 0: λ = 2λred = 1280 nm (too big to see, outside of visible wavelength spectrum)m = 1: λ = ⅔ λred = 427 nm (visible)m = 2: λ = ⅖ λred =256 nm (too small to see, outside of visible wavelength spectrum)m>2: λ too small to see, outside of visible wavelength spectrumLecture 29 (April 3)ResolutionQuestion: A man is standing 1.6 km from two birds in a tree. Assuming the wavelength of light reflecting off the birds is 498 nm, determine the diameter of the pupils of his eyes that would be required for him to be able to resolve the birds as separate objects.Answer: sinθ = h/L ≈ θ (for small angles)θmin = 1.22 λ/D = θD = 1.22 λ/θmin = 1.22λL/h = 0.0097


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UVM PHYS 012 - Exam 3 Study Guide

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