MCAT Physics: Light and Optics Chapter Summary – Electromagnetic Spectrum & Geometrical Optics

hello everyone my name is Iman welcome back to my YouTube channel today we're covering chapter8 for mcap physics and

math playlist and this chapter is titled light and Optics in this chapter we're going to cover the following objectives

we're going to start with the electromagnetic spectrum here we're going to discuss the nature of

electromagnetic waves and focus on the visible spectrum to understand how energy travels through space next we're

going to dive into geometrical Optics this is a dense section where we're going to explore reflection refraction

and lenses examining how light behaves at different surfaces and media third we're going to discuss

defraction here we're going to go over the concept of interference and we're going to analyze how light spreads

through a single slit a slit lens system and multiple slits finally we'll review polarization

by defining plain polarized and circularly polarized light ultimately learning how light's orientation can be

manipulated with that introduction let's go ahead and get started with our first objective we're going to begin with

discussing the electromagnetic spectrum now one of the ways that energy travels through space is by electromagnetic

radiation electromagnetic radiation or EMR is a form of energy that moves as oscillating electric and magnetic fields

propagating at the speed of light these oscillations are perpendicular to each other and to the direction of the waves

travel and that makes electromagnetic radiation unique in its ability to carry energy across vast distances even

through a vacuum to fully understand electromagnetic radiation it's important to recognize its three defining

characteristics wavelength frequency and speed wavelength frequency and speed are all related through this equation right

here this equation says that the speed of electromagnetic radiation in a vacuum is equal to wavelength multiplied by

frequency now the speed of electromagnetic radiation in a vacuum is approximately 3 * 10 8

m/s wavelength is defined as the distance between two consecutive Peaks or two consecutive troughs in a

weight frequency is defined as the number of wave Cycles passing a specific point in space per second now the

relationship between wavelength and frequency is inversely proportional so as wavelength decreases frequency

increases and vice versa the electromagnetic spectrum itself it encompasses a vast range of

wavelengths and frequencies each corresponding to different types of radiation with unique

characteristics starting from the longest wavelengths we have radio waves which range from about 10 to the 6 M

down to 1 M these waves are widely used for communication such as in radio and television

broadcasts next we encounter microwaves with wavelengths between 1 M and 1 mm which are commonly used in radar and for

heating food in microwave ovens moving into shorter wavelengths we reach infrared region spanning from 1 mm to

700 nanom infrared radiation is primarily associated with heat and it's used in applications like thermal

imaging next we have our visible light range this includes wavelengths detectable by the he human eye and that

ranges from 700 nanometers which we perceive as red down to 400 nanometers which we perceive as Violet this visible

spectrum is a major focus of this chapter because it directly relates to human vision and practical Optical

applications Beyond visible light we have ultraviolet radiation this covers wavelengths from about 400 nanom down to

about 50 nanom ultraviolet light has higher energy and it can cause ionization which is why it can lead to

Sunburn and other biological effects next we have x-rays with wavelengths between 50 NM and about 10 -2 nanm

they're known for their medical imaging uses because x-rays have the ability to penetrate soft tissues allowing us to

view structures inside the body finally at the shortest end of the spectrum we find gamma rays with wavelengths less

than 10 -2 nanm these extremely energetic waves are produced in nuclear reactions and

certain Cosmic events it becomes clear to us looking at this that each region of the electromagnetic spectrum

has distinct properties and applications that are influenced by its wavelength and frequency and this diversity allows

electromagnetic radiation to interact with matter in various ways from transmitting radio signals to

penetrating the body for medical imaging to tie this back to what we said earlier as electromagnetic radiation moves

through space it generates oscillating electric and magnetic fields along its path this interrelationship between

electricity and magnetism reflects the intrinsic Unity of these two forces a changing electric field induces a

magnetic field and vice versa and this oscillation Behavior allows electromagnetic radiation to propagate

as discrete pack of energy or photons which interact with matter in a diverse manner with that we've completed

objective one and we can go ahead and move into objective two in this objective we're going to cover

geometrical Optics to start let's look at how light behaves when it travels through a medium with a uniform

composition a homogeneous medium in this medium light moves in a straight straight line which is a behavior we

call rectilinear propagation however when the Light reaches a boundary between two different

materials say air and water its path can change this shift in direction is explained by the principles of

geometrical Optics which help us understand how light interacts with surfaces geometrical Optics they cover

key phenomena like reflection and refraction which form the foundation for understanding mirrors and lenses

let's begin with reflection reflection happens when light strikes a surface and it bounces back

into its original medium instead of passing through this behavior is governed by the law of reflection which

states that the angle at which light hits the surface called the angle of incidence

is equal to the angle at which it reflects off the surface known as the angle of

reflection both of these angles are measured in relation to a line perpendicular to the surface called the

normal to reiterate the law of reflection states that the angle of incidence is equal to the angle of

reflection reflection is Central to how mirrors create images in general images produced are

classified as either real or virtual a real image is one where light rays actually meet at a point meaning that if

you placed a screen there you would see the image on the screen in contrast a virtual image forms when the light rays

only appear to come from a point behind the mirror even though they don't actually

converge there now for plain mirrors the images are typically virtual and upright because the reflected Rays stay parallel

and do not converge to build on this a little more one important property of plain mirrors is that they don't cause

the reflected light rays to converge or diverge so when parallel light rays hit a plain mirror they remain parallel

after reflection since a plain mirror has no curvature it doesn't bring the Rays together or spread them apart and

this lack of convergence means that plain mirrors always produce virtual images the reflected light stays in

front of the mirror but the image appears to be the same distance behind the mirror as as the object is in front

and this creates the illusion that the light rays originate from a point behind the mirror again even though no actual

Rays pass that point interestingly enough we can think of plain mirrors as a special type of spherical mirror with

an infinite radius of curvature and this is going to make more sense after we cover spherical mirrors and we Define

the radius of curvature so let's do that next spherical mirrors come in two main varities concave and convex a concave

mirror has an inwardly curved reflective surface like the inside of a sphere which causes light rays to converge or

come together in contrast a convex mirror has an outwardly curved surface similar to the outside of a sphere and

it causes light rays to diverge or spread apart to understand how these mirrors work we have to Define some key

terms related to their geometry let's go ahead and get started with Center of curvature this is the center of the

imaginary sphere from which the mirror segment is taken it's located on the optical axis at a distance from the

mirror surface equal to the mirror's radius of curvature the radius of curvature

denoted as lowercase R this is the distance between the mirror surface and the center of

curvature then we have the focal point denoted as capital F this is a crucial point on the optical

axis where parallel rays of light upon reflection either converge or appear to

diverge the focal point is positioned halfway between the mirror and the center of

curvature the focal length denoted as lowercase f this is the distance from the mirror to the focal

point then we have the principal axis this is the straight line that passes through through the center of curvature

and the midpoint of the mirror surface the object distance denoted lowercase o this is the distance between

the object and the mirror and this is in contrast to the image distance denoted lowercase i which is the distance

between the resulting image and the mirror lastly we have magnification this is the ratio of the height of the image

to the height of the object it also corresponds to the ratio of the image distance to the object distance and it

helps us understand the relative size and orientation of the image now there's an important

relationship that links these distances together and it's known as the mirror equation this equation says 1 over the

focal length is equal to 1 over the object distance plus 1 over the image distance and that is equal to 2 / R

where R is the radius of curvature this equation allows us to calculate the position of an image

formed by a spherical mirror based on the object distance and the focal length by convention if the image distance I is

positive then the image is real and it's located in front of the mirror if I is negative then the image is virtual and

it appears behind the mirror we can also determine the size and orientation of the image using the magnification

formula the magnification formula says that the magnification is equal to the negative of I which is the image

distance divided by o which is the object distance the negative sign here

indicates that a positive magnification results in an upright image while a negative magnification results in an

inverted image so keep in mind that if the absolute value of magnification is less than one

then the image is smaller than the object if the absolute value of magnification is greater than one then

the image is larger than the object and if it's just equal to one then the image and the object are the same

size understanding these principles helps us predict the behavior of light as it reflects off of spherical mirrors

so for example example concave mirrors can produce both real and virtual images depending on the

object's position relative to the focal point convex mirrors however always produce virtual images that are upright

and reduced in size we definitely want to elaborate on that some more now one thing we could

use to help us visualize how these images form when we're discussing convex and concave mirrors for example is by

using Ray diagrams Ray diagrams are a valuable tool for approximating where an image will appear its size and whether

it's going to be upright or inverted and you can do this by tracing the path of the light rays as they reflect off of

the mirror surface now this is a great tool but sometimes it's not the most practical tool when you're taking a

timed exam and an exam that is specifically very high stress so I'm going to take a bit of a different

approach and just cover some of the different scenarios you can encounter let's start with convex mirrors a convex

mirror has an outwardly curved reflective surface which causes incoming light rays to diverge after reflection

because these Rays diverge they spread apart they don't actually meet in front of the mirror instead they appear to

come from a point behind the mirror forming a virtual image this virtual image is always upright and reduced in

size compared to the object and these convex mirrors are commonly used in applications like vehicle side mirrors

where a wider field of view and smaller upright images are beneficial now let's turn to concave

mirrors which have an inwardly curved reflective surface the type of image a concave mirror produces really depends

on the object's position relative to two key points the focal point and the center of curvature so here we're going

to cover a couple of scenarios our first case is when the object is beyond the center of curvature

here the mirror forms an image that is real inverted and reduced in size this image appears between the focal point

and the center of curvature the second case is when the object is at the center of curvature

here the image formed is real inverted and the same size as the object in this case the image appears at the center of

curve cature the third case is when the object is between the center of curvature and

the focal point the mirror produces an image that is real inverted and enlarged and it's going to be located beyond the

center of curvature our fourth case is when the object is at the focal point here no

image is formed because the reflected Rays become parallel and they do not converge or diverge to create an image

and finally we're going to consider when the object is between the focal point and the mirror in this case the image is

virtual upright and enlarged and it appears behind the mirror giving the impression that the light rays are

originating from a point behind the mirror now we're going to see more of this later in our problem sets but

before we get to that let's look at this table that summarizes our main takeaways for sign

conventions these conventions are going to help us determine the characteristics of the image based on the values in our

equations 4 o which is the object distance when this is positive it indicates that the object is

in front of the mirror if this is negative then the object is considered to be behind the mirror which is

uncommon but it can apply in certain Optical setups then we have our image distance

when that is positive all right that means that the image is formed in front of the mirror

indicating a real image on the other hand if we have a negative value all right then this is suggesting that the

image appears behind the mirror and that image is virtual now for radius of curvature and

for the focal length for concave mirrors both the radius of curvature and the focal length are positive because these

mirrors have an inward curvature for convex mirrors both values are negative due to the outward curvature

then finally we can discuss magnification a positive magnification indicates an upright image while a

negative magnification indicates an inverted image the absolute value of magnification also tells us the relative

size of the image compared to the object and with that we're going to jump into a practice

problem this problem States an object is Place 6 cm in front of a concave mirror that has a 10 cm radius of curvature

determine the image distance magnification whether the image is real or virtual and whether it is inverted or

upright to solve this problem we're going to use our Optics equation the equation says that 1 over the focal

length is equal to one over the object's distance plus one over the image distance which is all also equal to 2 /

the radius of curvature in the problem we're given the radius of curvature and we're also given

the object distance we want to figure out the image distance so we can use this portion of the equation to solve

for the image distance we're going to go ahead and isolate this variable right here 1/ I is equal to 2 / Rus 1/ o and

we have both of these variabl so we can go ahead and plug them in our radius of curvature is 10 cm and our object

distance is 6 cm now we can go ahead and simplify this to be 1

over5 now we have 1 over 5 - 1/ 6 we're going to adjust the denominator so that they have a common denominator we're

going to go ahead and do 30 so to get five to be 30 we have to multiply it by six and what we do to the denominator we

have to do to the numerator so 1 over 5 is equivalent to 6 over 30 and we want to do the same thing to 1 over 6 we have

to multiply 6 by 5 to get 30 what we do to the denominator we have to do to the numerator so 1/ 6 is equivalent to 5

over 30 if we go ahead and subtract 6 over 30 and 5 over 30 we get 1 over 30 this is equal to 1 over I we want

just I we want the image distance so we're going to have to do the reciprocal of both sides and we get that I is equal

to 30 cm a positive value for i signifies that the image is in front of the mirror and

it's therefore real so we figured out the image distance 30 cm and we figured out that the image is real now how do we

figure out if it's going to be inverted or upright here we have to look at the value of O the object distance if o is

greater than zero which is the case here then the real image will always be inverted so that is how we figure out

whether the image is inverted or upright we have to look at the value of the object distance if it's greater than

zero and I know they look very similar to each other here this is an O and this is a zero if the object distance is

greater than zero then the real image will always be inverted okay now for the last thing that we have

to calculate and figure out magnification the magnification is equal to I / o i is image distance o is object

distance we know the image distance is 30 cm the we know the object distance is 6 cm 30 / 6 is equal to 5 and we cannot

forget our negative sign so the magnification is equal to -5 now here the negative sign on the

magnification confirms that the image is inverted and then we want to look at the absolute value of the magnification the

absolute value of -5 is five 5 is greater than 1 when the magnification is greater than one when the absolute value

of the magnification is greater than one that indicates that the image is enlarged

and with that we've completed this practice problem and we're going to go ahead and stop the lecture video here in

the next part we're going to go ahead and finish this chapter I hope this was helpful thus far please let me know if

you have any questions comments concerns down below other than that good luck happy studying and have a beautiful

beautiful day future doctors