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All Physics 26 Tips
| Question | Answer |
|---|---|
| Refraction | When light passes through a medium, it deviates from the original direction and bends towards the normal (refracted ray is across from incident + normal, underneath reflected). |
| Index of Refraction | Used to describe the difference between the speed of light in given material (speed of light in a vacuum). |
| Snell's Law | n1 equals medium where angle of incidence, n2 equals medium with angle of refraction is |
| When n2 is higher than n1... | the refracted ray bends towards the normal (towards right, side, lower half). Angle 2 will be less than angle 1. |
| When n2 is lower than n1... | the refracted ray bends away from the normal (left side, upper half). Angle 2 will be greater than angle 1. |
| Snell's Law Demo | when light passes through one medium to another, freq is unchanged but speed changes so wavelength has to change. The ray's don't continue on straight; instead, they bend towards normal if the initial index is higher. |
| Total Internal Reflection | When moving from high index to low index, some light gets reflected internally and other refracts. At some point, angle of refraction increases to 90 and θi = θc -> critical angle. |
| Critical Angle | Whenever θi > θc, then the light is totally reflected back -> total internal reflection. |
| Sin(90) = | 1 |
| How to solve for critical angle using index of refractions | Remember that n1 = the first medium you travel though the n2 - the second, then use the θc equation. |
| θc Relevance | Whatever angel you get for the critical angle -> ex: 24 -> means that this is the range where light will leave to the second medium, light only escapes to n2 if θ made with the normal equals 24 or less. TIR if LARGER! |
| Why is diamond shinier than glass? | Because the air-diamond relationship yields a small critical angle than air-glass, meaning the light has a shorter range to escape, keeping the diamond shiny. |
| Dispersion | When white light, like sunlight, enters a prison from one side, it leaves the other with colors separated from each other, forming a spectrum of colors. The reason for this is because index of refraction is wavelength/color dependent. |
| Red Vs. Violet | Red and violet are refracted at different angles. Because the prism surfaces are not parallel to each other, the rays will emerge diverging from each other and separate enough for the eye to resolve them. |
| The same index of refraction can be different from two different colors/wavelengths | Index of refraction for crowned glass is equal to 1.520 for the red color and 1.538 for the violet color. |
| Rainbow | Caused by dispersion of water droplets. Light enters the water droplets from your back and disperses into many different angles within the droplet cause the index of refraction is color-dependent. Many droplets are needed to see full spectrum. |
| Convex/Converging Lenses | Rays start parallel to the principal axis then all converge at the focal point. Distance between the converging lens and f is called focal length. Converging lens tend to be thicker. |
| Concave/Diverging Lenses | Rays are parallel to principal axis but diverge after passing through the lens. If the reflected rays are traced back in a straight line, they appear to converge at the focal point. Diverging lenses are thinner. |
| Thin Lenses | Made of transparent material that allows light to reflect. |
| Ray One (Converging) | Ray parallel to the axis of the lens reflects towards focal point on opposite side of the lens. |
| Ray Two (Converging) | Ray travels through the center of lens and continues without any bending. |
| Convex/Converging -> Object Placed Outside of Focal Point | Object comes first, then focal. Results in real, inverted image. |
| If you place an object 2F away from the lenses on one side... | An image will be formed 2F away from the lenses on the other side. The size of the image and object will be the same. |
| If you place an object between the focal point and the 2F point... | The image will be formed outside of 2F and it will be magnified. |
| If you can place an object outside of 2F... | The image will move between the focal point and 2F, becoming smaller. |
| Convex/Converging -> Object Placed Inside Focal Point | Focal comes first, then object. Results in upright, virtual, and magnified. |
| Memo: Convex Lenses -> | Converging Lenses |
| Memo: Convex Mirrors -> | Diverging |
| Memo: Concave Lenses -> | Diverging Lenses |
| Memo: Concave Mirrors -> | Converging |
| Tip about converging lenses/concave mirrors | Don't be confused by the fact that converging lenses are called convex; they share the same characteristics as converging mirrors in regards to focal/object pathways. |
| Ray One (Diverging/Concave) | Ray starts parallel to principal axis, then diverges away but its extension appears to have originated from the focal on the left side. |
| Ray Two (Diverging/Concave) | Ray that travels through the center of the lenses continues directly without any bending. |
| Characteristics of Concave/Diverging Lenses | Don't be fooled by how the name is concave lens; the characteristics are the same as convex mirrors, regardless of where the object/f is placed. Always forms an upright, virtual image that is diminished and forms WITHIN f - not outside. |
| Real Images on Lenses | formed by lenses on the opposite side of where the light is coming from. Different than mirrors where the real images are formed on the same side as where the light is coming from. |
| Virtual Images on Lenses | formed by lenses on the same side as where the light is coming from. |
| Focal Length is pos for... | Converging lenses |
| Focal Length is neg for... | Diverging Lenses |
| do is pos if... | The object is on the same side as where the light is coming from (real object) |
| do is neg if... | The object is on the opposite side of where the light is coming from (virtual image) |
| di is pos if... | The image is on the opposite side of where the light is coming from (real image) |
| di is neg if... | The image is on the same side as where the light is coming from (virtual image) |
| m is pos... | If the image is upright with respect to the object. |
| m is neg... | If the image is inverted with respect to the object. |
| How to solve questions asking for characteristics | Determine the type of lenses, remember the sign conventions, draw a diagram as needed and use the necessary equations. |
| When light moves into a heavier medium... | Frequency remains unchanged while the wavelength decreases |
| When light moves into a lighter medium... | Frequency remains unchanged while the wavelength increases |
| Tip about focal length | Always remember to set your f as pos or neg. You may have to do this for the distances too — very important for di! |
| Cornea | Light enters the eye by passing through the cornea |
| Retina | Image is formed at the back of the eye on the retina |
| For distant objects, ciliary muscles... | Relax, then the lens become thinner and the focal length greater. |
| For nearby objects, ciliary muscles... | Contract, causing the curvature of the lenses to increase, therefore decreasing the focal length. |
| Pupil | Controls how much light enters the eye |
| Nearsightedness (myopia) | A person can see many objects clearly but objects at a distance at blurred — long eyeball. A diverging lenses converges the rays so the image comes to focus on the retina. |
| Farpoint and Myopia | The farthest part that an eye can use to focus images is called the farpoint. When the farpoint becomes too close to a given eye — because the eye is too long or any other reason — the person has myopia. |
| Farsightedness (hyperopia) | A person can see distant objects clearly but nearby objects are blurred (short eyeball). A converging lenses is used to converge the rays so the image come to focus on the retina. |
| Nearpoint and Hyperopia | Closest point that allows the eye to focus on an image is called the nearpoint, equal to 25 cm for a normal eye. The nearpoint distance can increase, causing hyperopia. |
| For both contacts and glasses... | Light entering the eye appears as if it is ongoing from the image formed by the lenses. To solve for the power of contacts or glasses, you'd use P = 1/f(m) |
| Are the images formed by glasses/contacts real or virtual? | They're always virtual, so the di will always be negative. |
| Reading Glasses | If you hear reading glasses, you'll immediately know we're dealing with farsightedness and there's a nearpoint involved. |
| How to solve a question that asks how far does the paper need to be so the print can be seen without glasses | The distance the paper is initially held at would equal do, the diopter amount can be used to solve for the focal point and then subbed into the di equation, di would be equal to your nearpoint. |
| Even though di is negative, the measurement of nearpoint should... | Still be a positive amount, you'll usually just take the absolute value. |
| Magnifying Glass | defined by the angular magnification, not ratio of heights. Main equation (NOT ON SHEET), M = θ'/θ = N/do. N = nearpoint, θ is the angle required to see the object with the naked eye, θ' is the angle required to see the object with the glass. |
| Gaillean Telescopes | concave eyepiece, produces upright images that are ideal for low power |
| Keplenan Telescopes | convex eyepiece, produces higher magnification and immediate focus, produces inverted images making them better for astronomy and high powered tasks. |
| Objective Lenses and Eyepiece | lens that is closer to the object is called the objective lens, and lens closer to the eye is the eyepiece. The eyepiece can be moved to force the image just inside the focal point, ultimately producing a virtual image that can be magnified multiple times |
| Formula for Telescopes (NOT ON SHEET) | m = θ'/θ = fo/fe |
| Tip for two lenses | Just because a certain set of characteristics can be applied to one of the lenses doesn't mean the same can be done for the other. If there's a IR on one side, there might be a UV on the other. |
| Find the Final Height of the Image | you can multiply the magnification received from the first lens m by the other m2 to get total magnification. Then you can use the height equation for a final height. |
| Approaching Microscope Problems with Multiple Lenses | set up everything normally for the first lens and evaluate do1. That will help you determine how far the second object is. Put in the focal length and evaluate the di2. Don't be fooled by the fact that it involves a microscope — it's same as lens problems |
| What causes light to form a circle | Light is leaving a higher medium and going into a lower one, triggering the critical angle and the total internal reflection so the circle is formed by only the ways that leave the medium. |
| Determine the depth of the pool | You can set up a right triangle with radius as the base, find θc with its equation, input into a tan relationship. |
| When the image is higher than the object... | You have converging/convex lenses, because diverging lens always form images that are smaller. |
| When the image is being projected onto a screen somewhere... | You have converging lenses, because diverging only forms virtual. |
| If you already know an object is going to produce an inverted image... | You can make its ho negative in the calculations. |
| Tip about di | If the problem says something about side-to-screen distance — or if you KNOW there's a lens separating your object — then your do will be from the object tot the lens and your di will be from the lens to the image. |
| Substituting | Remember that you can still make variables like di = 75do using the magnification and height equations, then substitute them into other things |
| When is do infinity? | When the object you're looking at with glasses is faraway/unspecified, you would consider do to be infinity. Don't forget to make your di negative because you're ALWAYS forming virtual images with glasses |
| When an image is formed to the left of the converging lens | it'll make a virtual image |
| If the line of sight makes a parallel line | Then there will be corresponding angles and the angle at the bottom equals the one formed by the eyes and upper body. |
| If there's a crick in the line of sight (not completely parallel) | Then look for alternate interior angles, those will also be equal to each other. |
| If you're solving for far point and the problem mentions there's distance between the eye and the glasses | Subtract the distance from the far point value to get di. |
| How to solve bifocal questions for the myopic half | Half the glasses are near-sighted, half are far-sighted. Your di will be the farpoint, remember to subtract distance between the eye and glasses from it. di will ultimately be neg for virtial images. Then cancel out do for infinity, solve for f and power. |
| How to solve bifocal questions for the hyperopic half | di will be nearpoint minus distance between eye and glasses. do is the object you'll be able to see ALSO minus the distance between eye and glasses. Solve for f and power. |
| How to solve lenses + mirror problems | The signs on the focal lengths can help you figure out what kind of lenses + mirrors you have. Solve for di with proper sign conventions, then subtract separation from that to get do2 (proper signs). Solve for di2, then m1 and m2 for total magnification |
| Fiber Optics | fiber optics use total internal reflection to keep light trapped inside the core, allowing it to bounce along the cables and transmit over long distances. |
| If you're solving for light leaving a medium out of a vertical side | Use Snell's Law normally to find θ2, then subtract that from 90 cause the normal is vertical, then input back into the equation with the indexes reversed. |
| When the image formed by a combination has the same size and orientation as the original one... | Solve for di like usual, but here, mtotal has to equa; 1, which you can use to find m2. Sub in di2 and do2, set up a relationship. Separation would be either added or subtracted. |
Created by:
smurtab