Physics 012 1st Edition Lecture 31Outline of Last Lecture I. Particle Wave Dualitya. Particles can behave like waves and exhibit interference effects.i. Electrons and other particles can exhibit wave-like properties.II. Black Body Radiationa. Black body: an ideal system that absorbs all radiation incident on it, thus appearing blacki. Common model: hollow spherical container with small opening, increasestemperature as it absorbs radiationb. For black body radiation, there is a characteristic distribution of wavelengths, dependent on the temperature of the object.c. Wien’s Law: λmaxT = 2.898 x 10-3i. As T increases, λmax decreases.ii. Ultraviolent catastrophe: prediction of classical physics that an ideal blackbodyat thermal equilibriumwill emit radiationwith infinite powerd. Total power of radiation emitted depends on temperature.i. P = σAeT41. σ = Stefan-Boltzmann constant = 5.7 x 10-8 W/(m2K4)These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.2. A = surface area of object3. e = emissivity (0 to 1), how easily an object emits/absorbs radiation, perfect black bodies have emissivity of exactly 14. T = temperature (must be in Kelvins, °C + 273)III. Max Planck – proposed solution to ultraviolet catastrophe problem: energy is quantized in discrete “packets” or valuesa. En = nhfi. h = Planck’s contant = 6.63 x 10-34 J*sii. n = integer multipleiii. f = frequency of oscillating modelOutline of Current Lecture IV. Problem: A diverging lens with focal length of -20 cm is placed 50 cm to the left of a converging lens with focal length of 15 cm. An object is placed 10 cm to the left of the diverging lens. How far is the final image from the second lens?a. 1/di1 = 1/f1 – 1/do1 = 1/-20 – 1/10 = -3/20b. di1 = -20/3 cmc. do2 = di1 + 50 cmd. 1/di2 = 1/f2 – 1/do2 = 1/15 – 1/(50 +di1)V. Problem 47 from chapter 26 in text booka. nred = 1.662b. nairsinθ1 = nredsinθ2c. θ2 = sin-1(sin60/nred) = 31.4 degreesd. θ3 = 90 – (180-60-58.6) = 90 – 61.4 = 28.6 degreese. nredsinθ3 = nairsinθ4f. θ4 = sin-1(nredsin(28.6)) = 52.7 degreesVI. Problem 37 from chapter 27 in text booka. λ = 498 nm, L = 1.6 kmb. θmin = 1.22λ/Dc. sinθ ≈ h/L ≈ θd. D = 1.22λ/θmin = 1.22λL/h = 0.0097
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