(Chap 24) Lecture #19 Electromagnetic Waves Change in magnetic field → electric fields & vice versa Combo of magnetic & electric fields [Diagram of electromagnetic wave] [C = \frac{1}{\sqrt{\epsilon_0 \mu_0}} = 3 \times 10^8 \text{ m/s}] E is the electric field in wave & B is magnetic field. E = c(B) [Diagram of a wave] fλ = v a wave w/ λ = 3m has f = (3 \times 10^8 \text{ m/s} / 3m = 1 \times 10^8 \text{ Hz} = 100 \text{ MHz } (1 \text{ MHz } = 10^6 \text{ Hz})) Light has a shorter λ, between 400 nm - 750 nm. X-rays have λ on the order of the size of an atom, useful to determine molecular structure. Electromagnetic waves carries energy. Since its electric & magnetic field, energy density =: [u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2 \mu_0} B^2] & E = cB & [c = \sqrt{\frac{1}{\epsilon_0 \mu_0}}] so, [\frac{1}{\mu_0} B^2 = \frac{1}{\mu_0 c^2} E^2 = \frac{1}{2} \epsilon_0 E^2] , thus [u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \epsilon_0 E^2] [u = \epsilon_0 E^2] [Diagram of a surface] How much energy flows through the surface? AcAtu = AcAtε₀E² to find energy through a surface per area & per time is S = cε₀E² But E is always changing thus take average Sav = ½ cε₀E₀² units: Watts per m² (W/m²) Ex) Laser puts an avg. power of 1 mW in a beam w/ d = 1 mm. What's E₀? A = (0.5 x 10⁻³ m)² π = 7.85 x 10⁻⁷ m² Sav = (1 x 10⁻³ W) / (7.85 x 10⁻⁷ m²) = 1.27 x 10³ W/m² Then solve for E₀ which = 9.8 x 10³ V/m [Note on the left side: "linked power by 1000"] Ex) Lasers can operate in a continuous beam or pulsed mode. Let's assume laser is on pulses of T = 2 x 10⁻⁵ s in a beam area of 4 mm². The max E₀ in each pulse = 4 x 10⁸ V/m. What's energy in each pulse? Sav = ½ cε₀E₀² = 2.12 x 10¹⁰ W/m² 2.12 x 10¹⁰ W/m² (4 x 10⁻⁶ m²) = 4.48 x 10⁴ W Energy = Power (time) = (4.48 x 10⁴ W) (2 x 10⁻⁵ s) = 1.7 x 10⁰ J Chapter 23: Geometric Optics Geometric Optics have 2 rules 1st: describes what happens when a ray hits a reflecting surface such as a mirror. 2nd: how a ray bends as it goes from 1 medium to another (air -> water) Law of Reflection [\theta_i = \theta_r] [Diagram of reflection] C = location of the image of the point source @A blc the reflected rays seem to come from a point source @C. Blc no light @C, it's called a virtual image. Triangle ADB is congruent to ACDB, thus do = di Lecture #20 do = di & ho = hi [Diagram of object and image] Ex) Show that any ray that is reflected from 2 mirrors that make a 90° angle w/ each other will reverse direction. [Diagram of two mirrors at 90 degrees] Both coming & out going rays move here an angle of 90° w/ the horizontal. Corner reflector

Electricity and Magnetism Lecture Note #19

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