# Phys6 - Light & Optics

 Constructive v. Destructive Interference constructive: waves are in phase (additive)destructive: waves are out of phase (if equal in amplitude waves cancel out perfectly) Interference w/ Light two light waves hit the same spot:if in phase → constructive overlapif out of phase → destructive overlap dsinθ = (m + 1/2)*λ If light wavelengths are different from each other by some variation of half a wavelength (eg. 1/2, 3/2, etc.) then the resulting light spot will appear DARK →*a dark light spot signifies DEstructive interferencem = any integer starting with 0 (1, 2, 3, etc.) dsinθ = m*λ if the “distance” a light wavelength travels is different from another by some variation of an integer of a wavelength (eg. 1, 2, 3, etc.) then the resulting light spot where the 2 wavelengths converge will appear BRIGHT → *a bright light spot signifies CONstructive interference Diffraction Grating a surface with slits equidistant all the way across itif light is shone through the surface there are many slits which it can pass throughd = distance between adjacent slitsequation for diffraction grating is the same as for just a double slit system: dsinθ = m*λ Speed of Light in a Vacuum cvac = 3 * 108 m/ssimilar value for the speed of light in air Index of Refraction (n) n = cvac / vmedcvac: speed of light in a vacuumvmed: velocity of light in whatever mediumvmed will NEVER be larger than cvac b/c light will never travel faster than it does in a vacuumtherefore the ratio for n will always be bigger than 1n ~ 1 for airn ~ 1.33 for water When light travels from a LOWER to HIGHER index of refraction, n, what happens to the wavelenght? it becomes INVERTED White Light smear of all wavelengths of light in the visible spectrumif you REMOVE a color from white light (eg. due to destructive interference?) what results is that color’s complimentary color across from it on the color wheel Mirrors if a mirror is concave → it’s convergingif a mirror is convex → it’s diverging Lens if a lens is convex → it’s convergingif a mirror is concave → it’s diverging Focal Distance always equal to exactly 1/2 the radius of curvaturef = 1/2r Lens Strength 1/f = dioptersfocal distance HAS to be in meters1 over the focal distance = Diopters (D) → which corresponds to the strength of a lens So if a questions gives you a mirror or lens’ radius of curvature, what values can you derive from r? 1. focal length (f) [f = 1/2r]2. strength of the lens in Diopters [D = 1/f] Upright v. Inverted if the image comes out below the principle axis → it’s inverted (upside down)if the image comes out above the principle axis → it’s upright Real v. Imaginary if light rays truly converge at di or a certain point → the image is Realif light rays do not truly converge at any certain point → the image is Virtual *a Real image is ALWAYS Inverted & a Virtual Image is ALWAYS Upright can remember using the mnemonics:IR spec: Inverted, RealUV light: Upright, Virtual(also remember just it’s counterintuitive - weird opposites) Equation for Focal Length 1/f = 1/do + 1/didi: distance from mirror to the imagedo: distance from mirror to the object Equation for Magnification m = hi / ho or m = – di / dohi: height of the imageho: height of the object Sign Rules for di & do object will ALWAYS be out in front of the mirror → do is positive (+do)if the image is REAL (i.e. the light rays that form the image actually converge) → di is positive (+di)however if the image is virtual (i.e. the light rays that form the image NEVER actually converge) → di is negative (–di) How to Interpret Magnification Equation if m is positive → then the image in question is uprightif m is negative → then the image is upside-down invertedif the absolute value of m is bigger than 1 → then the image appears bigger than the original objectif the absolute value of m is smaller than 1 (eg. 0.5) → then the image appears smaller than the object What is the focal distance for a diverging MIRROR? the focal distance for a diverging MIRROR is NEGATIVE (-f) For a Diverging Mirror the image is ALWAYS: 1. Upright 2. Virtual (UV) 3. smaller than the original object [regardless of whether the do is smaller or larger than the focal distance (inside or outside f)] A Diverging Lens: • always makes the resulting image look smaller • always results in a Upright, Virtual image Aberration when you’re looking at an image in a mirror or through a lens & it appears a little bit FUZZYthere are 2 explanations for why aberrations occur → both have to do with light rays not converging on a single point to form a clear focused image* Spherical Aberration light rays that pass through the outer edges of a lens (aka farther away from the principle axis) don’t quite converge exactly where the image forms? → results in a blurry image Chromatic Aberration as light of different colors passes through the medium of a lens, the light rays get refracted by the lensthe indices of refraction for different colors varies depending of the color → this causes some colors to be refracted more or less (angle of refraction differs) usually what results is the edge of the image appear fuzzily colored Authormse263 ID292614 Card SetPhys6 - Light & Optics DescriptionVideo Set 6 Updated2015-01-08T01:33:38Z Show Answers