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All Physics 27 Tips
| Question | Answer |
|---|---|
| Interference: | two identical coherent sources producing waves of the same frequency and amplitude produce an interference pattern. According to the principle of linear superposition, the waves combine constructively or destructively. |
| Constructive Interference: | when two waves at point P are both oscillating the same direction and combine to create an amplified signal with more intense light. |
| Destructive Interference: | when two waves at point P are oscillating out of phase and combine to destroy each other, forming no signal and light with no intensity. |
| Young’s Double Slit Experiment: | an experiment where light passes through two very close slits, creating two coherent sources (waves with the same frequency and constant phase difference) that interfere with each other to produce a pattern of bright and dark fringes on a screen. |
| Distance (d) | the distance (d) between the two slits is very small and in the order of millimeters. |
| distance (L) | The distance (L) between the slits and the screen is in the order of meters. |
| d & L | When the lights coming from the two slits interfere on a screen located L distance away, an interference pattern is formed. The interference pattern consists of bright and dark alternating fringes, all of which are equal in size. |
| If light is emerging from two different slits and both heading to the same fringe, then... | The light emerging from the lower slit would have the greater L value -> longer distance to pass. |
| Path Difference (Core Idea) | When light from two different slits reaches a point on the screen, the waves travel slightly different distances. ΔL = dsinθ. ΔL = path difference d = distance between slits θ = angle to a point on the screen. |
| Constructive Interference (Bright Fringes) | dsinθ = mλ m = 0,1,2,3 (fringe order) |
| Destructive Interference (Dark Fringes) | sinθ = (m + 1/2)(λ/d) m = 0,1,2,3 (fringe order) |
| Small Angle Approximation (NOT INCLUDED, SUPER IMPORTANT) | sinθ = tanθ = ym/L ym = distance from main axis to specific fringe orders L = distance from slits to screen Valid as long as the angle is small |
| Small Angle Approximation for Constructive Interference (NOT INCLUDED, SUPER IMPORTANT) | ym/L = mλ/d |
| Small Angle Approximation for Destructive Interference (NOT INCLUDED, SUPER IMPORTANT) | ym/L = (m+1/2)λ/d |
| How to solve a Young's Double Slit Experiment asking you to find the wavelength of the light | First take stock of what values you have (d, L, y1) then set up the dsinθ = mλ equation. Solve for θ, the angle formed from d to y1 using tan. Input it into the equation and get your wavelength, make sure everything's in the right units! |
| Bright Fringes & Dark Fringes | The bright fringes formed by an interference pattern are constructive interference while the dark spots in between them are destructive interference. |
| Central Maximum | The bright fringe right in the center, connected to the central line. m = 0 |
| When the problem is referring to "the first bright fringe..." | You have your y1 value. |
| Converting m to nm | Multiply by 10^9 |
| Converting nm to m | Dividing by 10^9 |
| Alternate Way to solve for wavelength of the light if you have a small angle | You can use the small angle approximation in the form of mλ/d = ym/L. You wouldn't even have to solve for the angle if you rearrange the terms correctly. |
| Acronym for Small Angle Approximation | MLD = YML |
| Phase shift of 180° (flip) | Phase shift of 180° (flip) -> λ/2 Happens when reflecting off a higher refractive index medium Light: going from air → glass |
| No phase shift | Happens when reflecting off a lower refractive index medium Light: glass → air |
| No Net Phase Shift Situation | Overall result for phase shifts. Ex: 3 mediums, n = 1.0, 1.4, and 1.5. Ray comes through 1.0, touches 1.4 and then reflects back up -> λ/2 phase shift. Does the same between 1.4 & 1.5, but because there's 2 phase shifts, they cancel each other out. |
| Net Phase Shift Situation | Overall result for phase shifts. Ex: 3 mediums, n = 1.0, 1.3, and 1.0. Ray comes through 1.0, touches 1.3 and then reflects back up. -> λ/2 phase shift. Does the same between 1.3 & 1.0, but because it's high to low, no cancellation. |
| Soap film has an index of refraction of | 1.33 |
| Trick for Net Phase Shift | Essentially, if there's a straight pattern -> like lower to lower to lower or higher to higher to higher -> you can expect the phase shifts to cancel out. |
| Why does the topmost band of a soap film appear dark? | Because when you mount the film vertically, the soap solution starts to drain towards the bottom of the ring and the thickness at the top becomes thinner. When it eventually reaches zero, destructive interference occurs to make the band dark. |
| If the question asks you to determine the thickness of the film at the first three bright bands... | You would start counting from m = 0, meaning you're including 0,1, and 2. |
| Air Tip | Usually, your first index of refraction is going to be air (1.00). In soap bands specifically, you might go from 1.0 to 1.33 to 1.0 alternating. |
| How to solve a question that asks for the first three bright bands of a soap film that is drained from the top. | Determine what the first three bright bands are (m=0,1,2) then determine whether you have a net phase shift, use constructive interference cause you have bright bands + you need thickness, get λf using equation, solve for the 3 bands' thickness. |
| Thin Film Interference | Light waves traveling from one medium to another undergo partial reflection and partial transmission at the interface of the two mediums. |
| Single Slit Diffraction | due to the combined effects of diffraction and interference, monochromatic light passing through a single slit of width (W) produces an interference pattern of alternating bright and dark fringes. Made of central bright fringe + smaller ones |
| Why are there dark spots formed when we're using a single slit | The light, after passing through a single slit, starts to diffract in all different directions. |
| Tip about m for Single Slits | make sure that m does not equal zero for the single slit diffraction experiment. The central slit doesn't get labeled, but the ones next to it would be m = 1 on both sides, then 2, then so on. |
| W & d | Rather than using d (distance between the slits) for a single-slit problem, you would just use W, which is the width of the slit. |
| Destructive Interference Equation for Single Slits | Wsinθ = mλ There is no mathematical expression for constructive interference — only destructive. |
| Does Small Angle Approximation Work for Single Slits | Yes, you can just swap out the d for W. It would still be MLD = YML, just the d and the W would be swapped. |
| If the problem asks you to calculate the distance from the central maximum to the second-order dark fringe, what will you do? | Set up a tan relationship between the theta, y2, and L. Use the sinθ general equation to retrieve the value of theta, then input it into the tan. You can ALSO just use the small angle approximation. |
| Instant way to tell you've done something wrong | Only diffraction grating is supposed to have large angles. So if you have large angles anywhere else, there was probably a calculation error. |
| Diffraction Grating | a diffraction grating consists of a large number of closely spaced parallel slits which diffract light incident on the grating. |
| m for Diffraction Grating | The central maximum would be m = 0, while the two bright fringes beside it would be m = 1 on both sides, then m = 2, so on. |
| Can you use Small Angle Approximation? | No, you cannot because now the angle has become too large. However, you can still use tanθ = ym/L to solve for the theta. |
| Diffraction Grating Demo | When light passes through the gradient, the light diffracts in all directions and forms an interference pattern that consists of bright spots that are very well localized with large spaces in between. As such, the diffraction angles are large. |
| d & the lines per milimeter | When the problem gives you a value like 600 lines per milimeter, that would be your d value. You would convert that to meters by doing (1/600)/1000. |
| How to solve problem that asks for angular deflection of bright fringes for a diffraction grating situation | Determine your d value by doing (1mm/lines)/1000, then inputting that into the bright spots equation for theta. If it's asking for first-order bright fringe, you're using m = 1. If second, m =2. Make sure everything is in meters! |
| How to solve problem that asks what is the maximum number of bright fringes that can be observed for a diffraction grating situation | You would set theta equal to 90, then use the rest of the equation normally and solve for m. You must round your value DOWN. There's fringes on both sides for gratings, plus the central, your answer might look like this if you get 2 -> 2 + 2 + 1 = 5. |
| Resolving power | the ability of a lens, microscope, telescope, or other optical instrument to distinguish two very close objects as separate. Even if two objects are extremely close, a system with high resolving power will show them as distinct, not blurred together. |
| Two principal factors limit the resolution of a lens | Aberrations – imperfections in the lens shape or material, causing distortions. Diffraction – the wave nature of light, which spreads out at the edges of apertures. |
| Rayleigh criterion | defines the minimum resolvable separation between two point sources, occurring when the center of one diffraction pattern overlaps with the first minimum of another |
| The minimum resolvable angle (Rayleigh criterion): | Two objects can be considered resolved if they are separated by the following angle. θ = 1.22(λ/D) λ = wavelength D = diameter of aperture (or slit width) |
| Arc length formula | Resolving power gives you the smallest angle you can distinguish θ=s/R lets you convert that into a real distance s = arc length (or separation distance along a circle) R = radius (distance from eye to object) θ = angle in radians |
| How to solve for a question that asks what is the minimum angular separation an eye can resolve? | You would set your values equal to meters and use the θ = 1.22(λ/D) equation. |
| How to solve for a question that asks how far away can a human eye distinguish certain objects located blank meters apart? | You would use the θ value taken from the 1.22(λ/D) equation and s, then solve for R using the θ=s/R equation. |
| How to solve problems with rectangular plates stacked next to each other? | First determine whether they are experiencing a net phase shift, then use constructive or destructive interference (depending on whether you're looking for dark or light bands), then set your wavelength f = wavelength air (NO CHANGES). Then solve. |
| Rounding down tip | Whenever the problem asks for maximum number of fringes, or total number of bands etc., remember to round your m answer! |
| How to calculate size of central maximum fringe in single slit problems? | You can use the small angle approximation to single out y1, then multiply y1 times 2 for your answer. Don't forget that this would be the version of MLD = YML where you swap out d for W. |
| How to solve a diffraction grating asking you for the width of the first order spectrum over a multitude of wavelengths? | You would use the tanθ = y1/L equation but first you'd find the theta for both wavelengths using the dsinθ = mλ. Answering finding the two angles, you plug them each into the tan. Then find the difference between the y's for your width. |
| When the problem tells you the location of two fringes are coinciding and to find the wavelength of the second fringe based on the information you have for the first | Figure out if you're doing con or des interference. Ex: for con, do the dsinθ = mλ and then set the mλ portions of the equation equivalent to each other, solving for the wavelength. You wouldn't need to solve for the theta, just set the halves equal! |
| A thin film of oil is located on smooth wet pavement. How would you set up a diagram for this? | Air index -> oil index -> water index |
| If the problem gave you a scenario where there was one color completely blocked and the other revealed... | The blocked color is equivalent to destructive interference and the other would be constructive. You can set up a relationship using that and find the values of m1 and m2. Those values can also be used to find the minimum thickness. |
| Asking for the width of the slit on a single slit diffraction | If they give you a y value, you would take half of that and form a right triangle. Remember the central bright spot does not equal m=0, you would start from the m=1 next to it. You can set up the small angle approx without solving for theta, solve. |
| Tip about MAX | Whenever the question asks for maximum number of bright fringes or highest order of visible maximum (size of maximum fringe is an exception), you will set theta equal to 90. |
| As the wavelength decreases, the value of m increases... | So when the question asks for the highest-order visible maximum, you're going to use the lowest wavelength available to you. |
| What would increase if the wavelength of light in a single-slit diffraction increased? | The angular separation between the bright fringes |
| Do NOT forget that even if you're using the small angle approximation... | Depending on whether you have constructive or destructive interference in a thin film interference, you might have to put the 1/2 next to your m |
| Which of the interferences uses m = 0? | All of them tend to use m = 0 save for the single slit which uses a central maximum, meaning that if they ask you for the third fringe, etc. you would have to start from 0 and go 0,1,2 |
| Key word for s variable | Apart |
| Key word for R variable | Away/above |
| Useful formula for pupil & resolution formulas | Rtanθ = y. |
| Major tip for resolution problems | Set up an equation with the variables you have and use the result to solve for variables you don't. |
| Double Slits and Brights (m) | 2m + 1 |
| Double Slits and Darks (m) | 2m + 2 |
| Single Slit and Dark (m) | 2m |
| Why do you not factor in the 1/2 for an equation? | When you are finding a difference in thickness between two fringes |
| If they're asking you to find the wavelength regarding the distance between dashed lines for a single slit, then... | Count how many fringes there are, divide the m and the y by 2. Then solve using small angle approximation. |
Created by:
smurtab