in this video we introduced diverging lenses starting with a qualitative look
at the refractions that cause parallel rays to diverge from a focal point
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 the image formed by a diverging lens
and finally how to calculate the location and magnification of an image using the
thin lens equation so we start out by talking qualitatively about how a concave lens works
on the left side we have a thin concave lens it's also called a diverging lens
with parallel rays of light coming in from the left hand side parallel to the principal axis
that principal axis is the symmetry axis of the lens so refractions in the lens bend these rays
away from the principal axis so that they all appear to originate from a common point called
the focal point labeled with an f here so what you do is trace all these rays back to where
they appear to have a common origin that's the focus on the right we have a picture of
one particular ray showing why it gets refracted away from the principal axis of the lens
recall that when light passes from air into glass or plastic the index of refraction increases
and that causes the ray to refract closer to the normal line so in my first refraction at
this interface the light bends closer to the normal line and then i make it to the
second interface where the light is exiting the material and recall what happens when light exits
a material like glass or plastic and goes back into the air the index of refraction decreases and
the light refracts away from the normal line so both of these refractions contribute to the same
effect they're bending the light ray away from the principal axis and that's what's causing the light
rays to diverge provided the lens has spherical surfaces the divergence happens perfectly so
that if you trace all the rays back you get to one single common point that's the focus
as a side note here we talk about the strength of a diverging lens by talking
about power and this is different from the rate of change in energy power this is optical power
and the stronger the divergence that's caused by this lens in other words the
shorter the focal length the greater the power so it's defined as p equals one over f
the units of optical power are one over meters and these get a special name these are called diopters
so i put a simple example in here about diopters compute the power of a diverging lens with focal
length negative 10 centimeters so we're already introducing some of the sign conventions here
a diverging lens gets a minus sign on its focal length and that's true for computing its power
as well so i get one over f that's one over in the right units 0.1 meters with a minus sign
so i get a power of negative 10 diopters unlike a convex lens we can't find the focal length of a
concave lens by focusing an image of a distant object and that's because light rays coming
in parallel are going to fan out and diverge there's no way this is focusing a sharp image
instead it produces a virtual image that you could see with your eye but not project onto
anything however there's a neat little trick for computing the focal length of a diverging lens
so what i've done in this picture is i'm holding the lens exactly the right distance from a wall
to project a bright circle of light that has twice the diameter of the lens itself but this
creates two similar triangles the smaller pink triangle has a height equal to the focal length f
so what about the height of the larger green triangle well it has a base of 2d instead of
d and because all the dimensions of these triangles appear in proportion
it's going to have a height of 2f and that means this leftover piece here
the distance between the lens and the wall that we're projecting the light on
that must be a distance of f so to find the focal length of a diverging lens you just draw a circle
on a piece of paper with twice the diameter of the lens and tack that to the wall opposite an
open window you hold your lens away from the wall until the circle of light projected is the same
size as that circle twice the diameter of the lens and at that point the distance between the
lens and the wall is the focal length of the lens when we study lenses we usually want to find the
image formed as 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 very simple ways through the lens
first a ray coming into the lens parallel to the principal axis refracts away from the principal
axis so that it can be traced back to the focus on the near side the second principle rate is
the time reversal of the first if a ray is coming into the lens so that it's aiming at the opposite
focus it'll be refracted parallel to the principal axis and our third principle
array is if we take a ray aimed at the center of the lens it's going to pass through unrefracted
because the surface of the lens at the entry point and exit point those are parallel to each other
so these are the basic tools we're going to use to make ray diagrams when we try to find
an image 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 rays that we want
to for the analysis we represent the object with an arrow to keep track of orientation
and we represent the diverging lens as a line with inverted arrow tips
to hint at the shape of the concave lens unlike convex lenses the image formed by a concave lens
more or less works the same whether or not the object is inside the focus and in our example
we just happen to put it outside the focus so let's get our principal raised into the diagram
first we take a ray from the head of this arrow parallel to the principal axis
that's going to be refracted away from the principal axis so that it looks like it's
coming from the nearside focus so i used a light blue color there to indicate the tracing back to
where the focus is that's not an actual light ray coming from the object next why don't we just do
the easiest one the one that goes through the center of the lens that's unrefracted
and then the final principle array that we'll do is the one that's aimed at the opposite focus and
that's refracted out parallel to the principal axis so again i used a dashed and light blue line
to indicate where i'm trying to view a direction but the ray is not physically going that direction
it's aimed at the opposite focus the refraction happens out parallel to the principal axis
so on the right hand side of the lens we have three diverging rays and our eyes can gather
those diverging rays and see an image from them and we'll infer the image location is
where they have a common origin so we've got to trace all of these back to their common origin
and two of them are actually done already the one that's parallel to the principal axis is
the only one that still has to be done so we follow all three of these rays back to their
common origin and that's where we will see an image of the head of the arrow so first of all
this image is virtual because on the opposite side of the lens these rays are all diverging
in other words you can't project the image on a screen that's why it's called virtual
and we notice that it's upright and we notice that the image is smaller than the object so our
magnification would be less than one here's how we get quantitative about where the image forms
and how much it's magnified so we have a lot of sign conventions involved here and a couple simple
equations first by convention we're always going to put the object on the left side of the lens
second the object's distance is that's the distance between the object and the lens is
positive if the object is on the left side now we will rarely look at cases in a lens combination
where an image produced by the first lens lands on the wrong side of the second lens
in that case you have a negative object distance and it's called a virtual object and i'll post a
link to an example when it's ready but again the vast majority of the time we're talking about
positive object distances on the left side of the lens image distance positive if your image is on
the opposite side of the lens or the right side of the lens negative if it's on the same side as
the object so for diverging lenses our image distance is going to turn out to be negative
the focal length for a diverging lens by convention is a negative
finally the image distance object distance and focal length are related by the thin lens equation
1 over di plus 1 over do is equal to 1 over f that allows us to find the location of our image
then magnification that tells us how much bigger or smaller our image got and that's the image
height divided by the object height in other words if an object shrank to half its size the
magnification would be one half in addition the orientation of the image is taken care
of in the magnification equation if an image is inverted then the height would be called negative
if it's upright the height is positive and for a diverging lens it's going to be positive we
can express magnification in terms of the image distance and object distance as the negative of di
over do and because for diverging lenses our image distance is negative that means
the magnification is positive just like we saw in the ray diagram finally we'll work a
quantitative example so i have a diverging lens with a focal length of negative 20 centimeters
and i have an object sitting 27 centimeters from the lens i want to get the image distance
and the magnification now this ray diagram is basically the same as the one that we just did
hashing it out real quick array through the center is unrefracted a ray aiming at the opposite focus
is refracted parallel to the principal axis a ray coming in parallel to the lens is refracted
away from the principal axis so that it can be traced back to the focus on the near side
so we trace these rays back to a common point of origin and that's where the image of the
head of this arrow is going to form and what i'm looking for here is d i and the magnification
so we apply the thin lens equation taking care to put in a negative number
for the focal length and to get di we just combine negative 1 over 20 minus 1 over 27 and then take
1 over that answer to get the reciprocal three sig figs i get negative 11.5 centimeters for this
and my diagram is reasonably to scale so i can check for the reasonableness of my answer
by making decently careful ray diagrams and if this focal length is 20 centimeters i see 10
centimeters at about half that distance and di is a little bigger than that so that makes me happy
finally we get the magnification remember the minus sign keeps track of the orientation for us
so i plug in my negative image distance there and my magnification becomes positive
and i get a positive magnification of 0.43 so the image is smaller than the object by a little more
than a factor of two if you find the physics content on zack's lab helpful click on the
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