#Waves *these mirrors are infinitely thin for the most part. get rid of any issues w/ refraction NOW!* # Your Myopia and Lenses - Intro Shortsightedness isn't the rarest thing you see here - most people are shortsighted to some degree, and those who claim to have 'perfect' vision are probably faking it. Hell, tell them that there's something tall far away - see if they believe you. Most of the time, people wear glasses or contact lenses to correct this - converging lenses that help increase the clarity of the images our mind creates. %%diagram once you're done with spherical mirrors!%% # Mirrors - The Plane Typically, mirrors are flat - we use them to see ourselves in a desperate bid for reassurance. However, let's go through some of the 'maths' behind it - if you can even call it maths. A mirror is typically made of a reflective surface - usually something with a high albedo like a metal. When you stand in front of a mirror, what happens is that light rays from other sources (usually a LED light) bounce off you and hit the mirror, forming an *image* within it. One way we can express this is through ray diagrams. These give us general approximations for the directions of EM waves, much like the wavefronts you see in [[Waves Foundations#More Transverse Waves|transverse waves]]. Here's a diagram! ![[Pasted image 20231215193157.png]] The distance O corresponds to the same linear distance as the point I from the mirror. Let's go generalise this into an identity! $i = -p$ Where $p$ takes the distance to the mirror as a vector from the imaginary point $I$. This point is known as a *focal point*, where the light rays from that point converge behind the mirror. # Curving the Contact Lens You could call this part a case study - but it goes a lot, lot further than just contact lenses! Here, we'll be delving into spherical lenses - lenses that take the shape of *any part* of a sphere. ![[Pasted image 20231222152228.png]] By curving these mirrors, the sizes of the images we'll create will change. Let's start off by drawing some ray diagrams! ![[Pasted image 20231227163331.png]] *Mirror - Concave* ![[Pasted image 20231227175236.png]] *Mirror - Convex* Look closely at the mirrors. When you orient them different ways, how does the end result image look? >[!Success]- Answer >A concave mirror creates a larger image and pushes it further from its original spot. A convex mirror does the opposite! When a mirror takes a full surface area, we can ignore any reference points and instead take generalised wave diagrams. We can find the foci of these mirrors using these! ![[Pasted image 20231227173651.png]] The length from the lens to the center of curvature $C$ is the *radius of curvature.* In the case of a mirror, the *focal length* is the length from the edge of the mirror to the focal point. Basically this means that an ideal mirror like this would give you pretty good aspect ratios! >[!Success]- Focal Length rules >Let's look at our convex mirror again. Imagine each ray's displacement from the centre is the same (meaning that if the baseline were the X axis, they'd have equal absolute values!) > >Since the mirror is part of a sphere, the incident angles of the light rays will be the same. > >Therefore, **as long as this distance from the center is always the same, the focal length will always be half the radius of curvature!** >$f = \frac{1}{2}r$ >[!Example]- Sample Problem - Spherical Optics # The Mirror Equation and Images Now that we've clamped our foot down on the 'mirror lingo', let's take a look at how moving an object along the radius of curvature will change the image! First, let's set up our mirror and its focal length. Let's pretend you're at the Chicago [Cloud Gate](https://www.chicago.gov/city/en/depts/dca/supp_info/chicago_s_publicartcloudgateinmillenniumpark.html), and you're staring at one of its faces. Where's the dirt, though? ![[Pasted image 20231229110542.png]] We'll start off far away from the bean and move closer. Let's treat the object as an ellipsoid - we'll use preexisting data (listed down below) to find the focal length! - [[Conics - Circles, Parabolas & More! (Maths)#Ellipses|Semi-minor axis]]: 10m - Approximate height under arch = 1.8m (average human height) + 1.2m = 3m Taking the bottom face of the bean we can get a value of 3m for the radius of curvature. For now, you'll be crouching down, then. ![[Pasted image 20231229113001.png]] Time to complete our ray diagram! In our case, we're looking to draw rays from the top of our 'face' - as a reference for the rest of our body. ![[Pasted image 20231229113308.png]] Just to reiterate, this is an example of a **real image** - where the rays converge in the same plane as the initial rays. Alright, now stand up so you're 1.5 meters tall - the focal length of the mirror. You should see that the parts of you at that height just don't show up - there is no image: ![[Pasted image 20231229113538.png]] Now jump up and down for a while! I want you to tell me what you see - will you appear bigger, or smaller? ![[Pasted image 20231229113909.png]] Since the reflected light rays diverge, what you're actually seeing on a mirror is a [[Refraction (Waves)|mirage]] - a virtual image, so to say. There is an equation that dictates this, which we can write as: $\frac{1}{p} + \frac{1}{i} = \frac{1}{f}$ Where: - $p$ is the object distance from the mirror - $i$ is the image distance from the mirror - $f$ is the focal length >[!Example]- Problem Taster - Module 34-2 Problem 9 >A concave shaving mirror has a radius of curvature of 35.0 cm. It is positioned so that the (upright) image of a man’s face is 2.50 times the size of the face. How far is the mirror from the face? > >**Solution:** >Let's start off with our obligatory diagram! > # Magnification All things considered, I really did overthink this chapter. Simply put, magnification is just the ratio of the size of the image against the size of the object. Most of the time, we just take one dimension as a reference since the other dimensions will just follow the same scaling rule. $|m| = \frac{h'}{h}$ Where: - $h'$ is the height of the image - $h$ is the height of the object - $m$ is the magnification We can also give the magnification in terms of the object and image distances: $|m| = -\frac{i}{p}$ This shows us that the distance of the object from the lens is >[!Success]- Proof # Important Words! - *Center of Curvature:* The center of the sphere which the lens is part of once extended. # What's Next?