in this video we introduce converging lenses starting with a qualitative look at the
refractions that cause parallel rays to converge to a focal points how to determine the focal
length of the lens how to identify the principal rays in a lens analysis how to use ray diagrams to
find real and virtual images and how to calculate the location and magnification of an image
using the thin lens equation finally we wrap things up with quantitative examples of real
and virtual images formed by a convex lens so that's a lot to do in a single video so i divided
everything up into chapters so you can skip around and find what you need we start by qualitatively
discussing how a converging lens works on the left side we have a thin convex lens with parallel rays
of light coming into the lens parallel to its principal axis so that's the special name given
to the symmetry axis of the lens refractions in the lens bend all these rays toward the principal
axis bringing them together at a single point called the focal point because a convex lens
causes these parallel rays to converge it's also called a converging lens on the right we have a
picture of one particular ray showing why it gets refracted toward the principal axis of the lens
recall that when light passes from air into glass or plastic it will refract toward the
normal line because the index of refraction has increased so that's what happens at this
first contact you can see the refracted ray is closer to the normal line than the incident ray
then in the second refraction we're passing from the glass or plastic back into the air so
the index of refraction has decreased and the light ray refracts away from the normal line
so by the time that light ray emerges from the lens it's been bent toward the principal axis
if the lens has spherical surfaces then parallel rays coming into the lens will all refract to
that single focal points a little side note at this point is that we often talk about lenses
in terms of their power and the larger the power is the stronger the lens is in other words it's
going to cause all these rays to converge to a nearer focal point so power is defined as one
over the focal length the units of optical power are one over meters and this is given a special
name these are called diopters so we'll wrap up this chapter with a simple example
compute the power in diopters for a convex lens with a focal length of 10 centimeters
so all you do is just take 1 over f just put it in the right units so 0.1 meters
and i get 10 inverse meters or 10 diopters for the power suppose you have a converging lens and
you want to find the focal length one way to do this is to use the best set of parallel rays we
can find and those are rays coming from the sun because the sun is so far away rays of sunlight
are almost perfectly parallel so you might hold your lanes in the sunlight and move it up and down
over a target until these rays focus to a single point the distance above the target then tells
you the focal length sunlight is intense enough to start a fire if it's concentrated in this way
so be careful as you're experimenting so in reality what you've formed from the rays of
the sun is not really a single point it's an image of the sun and you can find the focal length of
a convex lens by focusing any image onto a target provided the object you're using for the image is
far enough away for the light rays to strike the lens nearly parallel note however in the second
picture that an object off the principal axis will form an image off the principal axis so i'm just
focusing on a tiny point on a tree here just a single leaf those rays are coming in parallel
to the lens and they're going to focus an image of the leaf off the axis so we define the focal plane
to be the plane on which distant objects will form an image and this plane is at the same distance
f from the lens so without risking a fire you can simply project an image on your wall by opening a
window and making an image of a distant tree when the image focuses the distance between the lens
and the wall is the focal length when we study lenses we usually want to find the image formed
as the light from an object passes through the lens we can use a ray diagram to draw the process
we focus on three principal rays that refract in a very simple way through the lens
first a ray coming into the lens parallel to the principal axis will be refracted through
the focus on the opposite side of the lens the second ray is the time reversal of the first
so light coming into the lens through the focus will be refracted parallel to the principal axis
finally a ray passing through the center of the lens will be unrefracted because the two surfaces
that it passes through on the lens on its way in and on its way out are parallel to each other
so to find an image formed by an object we start with the basic assumption that light is scattered
off the object in all directions all the time so we can choose any ray we like for the analysis
we represent the object with an arrow in order to keep track of the orientation
and we represent a converging lens as a line with two arrow heads that give you a hint as
to the curvature of the lens there are two basic cases for the image formed by a convex lens and
the first is when the object is outside the focus we choose principal rays emanating from the head
of the arrow and then we try to find out where those rays focus after passing through the lens
wherever those rays come together that's where the image of the head of the arrow will form
and we draw the rest of the arrow perpendicular to the principal axis and we've got our image
we start with the ray that's coming in parallel to the principal axis
and that's going to be refracted through the focus on the opposite side
next we pick a ray that's coming through the focus on the near side and that's going to be refracted
parallel to the principal axis finally we take a ray through the center of the lens just to verify
that it agrees with the first two and we see all three of those rays meeting at the same point
and that's the point at which the image of the head of the arrow will focus
and we obtain our image note first of all that the image is inverted
and second that the actual light rays are converging to form this image that means we could
project this image onto a piece of paper and this type of image is called a real image the second
case for a converging lens is when the object is inside the focus and we're going to see radically
different behavior in this case so we use the same sorts of principle rays i have one coming
in parallel to the principal axis and that gets refracted through the focus on the opposite side
i have one coming through the center of the lens and that's unrefracted and if a ray comes into the
lens from the direction of the nearside focus so like in this direction and there is no ray right
here so we're going to turn that into a dotted line then it will be refracted parallel to the
principal axis what we notice is that on the far side of the lens we have a bunch of rays that are
diverging from each other so these can't possibly focus to form a real image in the same way we got
for the first case well normally when we view an object we're viewing diverging rays that are
coming off in all directions from that object and our eye focuses those rays into an image
and where you perceive the object to be is where all these rays have a common
origin so we're going to trace them back to a common origin and we found a single point
that acts as a common origin for all these diverging rays so what you're going to see if
you look through this lens is an enlarged upright image of the object now because these rays are not
actually converging and we can't actually project the image onto a screen it's called a virtual
image so there's our two cases when the object is outside the focus of the converging lens we end up
with an inverted real image on the opposite side when the object is inside the focal point of the
converging lens we end up with an upright virtual image on the same side of the lens as the object
finally we want to get quantitative with the image formed by a lens so what we're doing here
is defining all the sign conventions for where the object is where the image is for the focal length
and then a couple simple equations to help us get the image distance and the magnification
for the object so by convention i'm always going to put the object on the left side of the lens
the object distance will be positive as long as the object is on the left side of the lens and
you should be aware there are rare cases where we have a virtual object in a lens combination
where the object appears on the opposite side of the lens and i'll post a link to
an example of that once i've got it done the image distance the convention here is that
it will be positive if it's on the right side of the lens in other words opposite the object
negative if it's on the left side or the same side as the objects the focal length by convention is
positive for convex lenses and finally using all these sign considerations
the image distance the object distance and the focal length are all related by the thin
lens equation so provided these lenses are thin enough this will give you the
approximate location of the image finally a little note on magnification so magnification is a way of
talking about how much bigger or smaller the image is compared to the objects and that's given by the
image height divided by the object height where the sine convention is that the image height is
negative if the image is inverted positive if it's upright and as it turns out magnification
can be written in terms of the image distance and object distance has negative image distance
over object distance so this means the images on the opposite side of the lens from the object
those have positive image distances remember those are going to have negative magnifications
in other words they are inverted and we already saw an example of this where for an object outside
the focal length of a converging lens the image is formed on the opposite side of the lens but
the magnification is negative because the object is flipped upside down finally we look at a couple
typical examples of image formation by convex lens and we're getting quantitative with it now
so our instructions here are to solve for the image distance and magnification in each of
the following cases and the ray diagrams have been provided so just reviewing the
ray diagram real quick for the first one i have a focal length of 10 centimeters
i have an object distance of 16 centimeters by convention those are both positive values
and then my ray diagram is constructed by taking a parallel ray to the principal axis
that's refracted through the focal point on the opposite side array through the center of the
lens that's unrefracted an array coming through the near side focus and that's refracted parallel
to the principal axis again you only really need two of these rays but the third one is there to
confirm that the first two were done correctly and what we're trying to get out of this is the
image distance and once we have that we'll be able to plug that in and get a magnification as well
so we write down the thin lens equation one over d i plus one over d o is equal to one over f
so one over d i is equal to one over f minus one over d o we can just work in centimeters here
we get one over ten because the focal length is ten minus 1 over 16. so to solve for di i'm going
to take 1 over 10 minus 1 over 16 and then take 1 divided by that answer to get the reciprocal
and i get an image distance of 26.7 centimeters out of this now if you're very careful in your
ray diagram you can verify by looking at your picture that your answer is approximately correct
and i tried to scale this reasonably remember the focal point was 10 centimeters out
so twice that would be about 20 centimeters right about there does it look plausible that we're
looking at 27 centimeters approximately yeah it looks pretty good next we get the magnification
and that's negative d i over d o it has that minus sign to keep track of the fact that the image is
inverted and i have negative 26.7 over my original object distance of 16 and i get a magnification
of negative 1.67 meaning the scale of this thing has been blown up by a factor of 1.67
and it's been inverted next we look at the converging lens same focal length
but our object is inside the focus d o is six centimeters focal length is tan positive by
convention and the ray diagrams are definitely trickier when you're looking at a virtual image
i take a ray coming in parallel to the lens that's refracted through the opposite focus
but because i end up with a bunch of diverging rays coming out the far side of the lens i have
to trace back and find a common origin for those so i'm going to trace back a ray that is coming
from the direction of the near side focus is refracted parallel to the principal axis
and again i've got a trace back to where the common origin is for these diverging rays
and then finally a ray that goes through the center of the lens is unrefracted
but i've got to trace back to find this common origin and this should all meet in the same place
and that's where we'll perceive that this object is now it looks like it has a rather large
magnification and it's upright and virtual let's plug into the thin lens equation i'm going to skip
one of these steps this time and just solve for one over d i that's
one over f minus one over d o and that's one over ten minus one over six one over six is bigger
than 1 over 10 that means we're going to get a negative out of this
and i get negative .0667 then i take 1 over that answer and i've got d i
and d i is negative 15 centimeters remember that convention that a negative image distance is on
the same side of the lens as the object started which for us will always be the left hand side
and then i can ask whether or not that looks reasonable in the picture and roughly yeah it's
not perfect but my focal length is 10 i should have 10 plus half of 10 to get to negative 15 over
here it looks a little bit short but because i didn't draw my diagram quite perfectly and finally
i'm going to get the magnification for this and that's negative d i which was negative 15 over d
o which was six and this gives me a positive magnification confirming that my image is upright
and my magnification is positive two and a half so this thing will be blown up by a factor of two
and a half and will appear upright just a final note before we wrap things up this second example
is exactly what a magnifying glass is you just take a converging lens
and you put it close enough to what you want to look at so that the object is inside the focus
and you're going to see a magnified upright image of whatever the object is if you find
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