Phys6  Light & Optics
Home > Flashcards > Print Preview
The flashcards below were created by user
mse263
on
FreezingBlue Flashcards. What would you like to do?

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 overlap
 if 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 interference
 m = 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 it
 if light is shone through the surface there are many slits which it can pass through
 d = distance between adjacent slits
 equation for diffraction grating is the same as for just a double slit system: dsinθ = m*λ

Speed of Light in a Vacuum
 c_{vac} = 3 * 10^{8} m/s
 similar value for the speed of light in air

Index of Refraction (n)
 n = c_{vac} / v_{med}
 c_{vac}: speed of light in a vacuum
 v_{med}: velocity of light in whatever medium
 v_{med} will NEVER be larger than c_{vac} b/c light will never travel faster than it does in a vacuum
 therefore the ratio for n will always be bigger than 1
 n ~ 1 for air
 n ~ 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 spectrum
 if 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 converging
 if a mirror is convex → it’s diverging

Lens
 if a lens is convex → it’s converging
 if a mirror is concave → it’s diverging

Focal Distance
 always equal to exactly 1/2 the radius of curvature
 f = 1/2r

Lens Strength
 1/f = diopters
 focal distance HAS to be in meters
 1 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 Real
 if 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, Real
 UV light: Upright, Virtual
 (also remember just it’s counterintuitive  weird opposites)

Equation for Focal Length
 1/f = 1/d_{o} + 1/d_{i}
 d_{i}: distance from mirror to the image
 d_{o}: distance from mirror to the object

Equation for Magnification
 m = h_{i} / h_{o} or m = – d_{i} / d_{o}
 h_{i}: height of the image
 h_{o}: height of the object

Sign Rules for d_{i} & d_{o}
 object will ALWAYS be out in front of the mirror → d_{o} is positive (+d_{o})
 if the image is REAL (i.e. the light rays that form the image actually converge) → d_{i} is positive (+d_{i})
 however if the image is virtual (i.e. the light rays that form the image NEVER actually converge) → d_{i} is negative (–d_{i})

How to Interpret Magnification Equation
 if m is positive → then the image in question is upright
 if m is negative → then the image is upsidedown inverted
 if the absolute value of m is bigger than 1 → then the image appears bigger than the original object
 if 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 FUZZY
 there 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 lens
 the 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