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>> In this video, we're going to
take a closer look at lenses.
In our closer look on mirrors,
we learned strategies
to help answer questions
about mirrors.
Much of that content
applies to lenses too.
In this segment,
we'll highlight those areas
and get practice using
those approaches with lenses.
As you answer questions
about lenses and mirrors,
be careful to keep in mind
which kind you are working with.
Though they may look
similar in a picture,
mirrors reflect light,
while lenses refract light.
That means the ray diagram
for a mirror will look
really different from
the ray diagram of a lens.
Speaking of ray diagrams,
they are always
part of our approach to
answering lens questions.
Keep the good habit
you began with mirrors
by drawing ray diagrams
as you begin a lens question.
Before we try a few examples,
I will show you a way
of drawing lens diagrams
that clarify the situation,
and give you confidence
in your answer.
In a lens diagram,
there are three rays we can draw
that are called principal rays.
These are lines that
connect the object, image,
and focal points
in the scenario.
Here's a picture of an object
in front of a converging lens.
Since the lens spins light
coming from either side equally,
there are two focus points.
We can call them F1 and F2.
The first principal ray
begins at the object
and runs parallel
to the optical axis
until it reaches the lens.
The lens then bends the ray,
so it passes through F2,
the focal point on
the opposite side.
Notice that just by
drawing one principal ray,
we can't tell where
the image forms.
It could be anywhere
along the ray.
The second principal ray
runs from the object
straight through the center
of the lens to the other side.
That way, it doesn't
bend at all, but keeps on
going straight through.
The third principal ray
begins at the object
and passes through
the first focal point, F1,
then onto the lens.
The lens refracts the ray
so it moves parallel
to the optical axis
on the other side.
These are the rays to include
whenever we draw a ray diagram.
With diverging lenses,
you will use the same principal
rays as you can see here.
These lenses make an image
that is virtual.
It can't be projected
on a surface.
Wherever all three meet,
that's where the image forms.
When the lines diverge
from one another,
we trace dotted lines
from the lens
to where they converge
on the same of the lens
as the object.
If two of the lines meet up,
but the third doesn't cross
at that same point,
that tells us
we need to go back
and double check our work.
In general, if our
lens diagram is correct,
our answer to the question
is going to be on target too.
Just like with mirrors,
the focal length,
object distance
and image distance of lenses
are related by this equation.
One over the lens focal length
equals one over
the object distance,
plus one over
the image distance.
Along with where the image
is located is the question
of how big it is.
This is called
the image magnification.
Capital M.
Image magnification
equals the image height,
divided by the object height,
or negative the image distance,
divided by the
object distance.
When magnification
is less that one,
the image is smaller
than the object.
And when it's greater than one,
the image is larger
than the object.
If the answer is negative,
we know it is inverted.
Solving lens exercises requires
carefully keeping track of
positive and negative signs.
This table shows that
when the lens focal length
is positive,
the lens is converging.
When the focal length
is negative, that means
the lens is diverging.
If the image distance
is positive,
the image is real.
And on the far side of
the lens as the object,
while a negative image distance
means the image is virtual,
it exists on the same side
of the lens as the object.
Positive object distance
indicates the object is real.
And negative object distance
tells us the object is virtual.
When image magnification
is positive,
the image is upright.
And when it's negative,
the image is upside down,
inverted.
Remember these conventions
when you're solving
lens exercises.
They will help you find
the correct result.
All right, let's try out
some lens exercises.
Using a converging lens,
with a focal length
of 22 centimeters,
an object is placed
45 centimeters from the lens.
Where does the image form,
and is it real or virtual?
And what is the
image magnification?
Our sketch of the
information given,
shows the object further
from the lens than
the focal point, F1.
Let's trace in
the principal rays passing
through this converging lens.
Remember, the principal rays
begin at the tip of the object.
One passes through the
distant focal point, F2,
one goes through
the center of the lens,
and one through the nearby
focal point, F1.
Now you give it a try.
Pause the video
and take a minute
with a ruler and pencil
to draw in the three
principal rays.
Then come back
and check your work.
Okay, we're back.
And here's a diagram showing
the principal rays involved.
In this case,
the image is real.
It appears on the opposite
side of the lens as the image
and it's inverted.
Now let's find out
its position.
We'll rely on an equation
we use in the mirrors segment.
One over the focal length
equals one over
the object distance,
plus one over
the image distance.
To rearrange to solve
for d sub i,
we'll multiply both sides
by the focal length,
image distance
and object distance.
Then isolate the terms
with d sub i.
Factor that term out,
and finally divide
both sides of the equation
by d sub o minus f.
That distance equals the focal
length of 22 centimeters
times the object distance
of 45 centimeters.
Divided by 45 centimeters
minus 22 centimeters.
Which, altogether,
is 43 centimeters.
In a minute, we're going
to check this answer
to see if it makes sense
with our ray diagram.
Before we do that,
let's figure out what
the image magnification is.
Since we don't know
the object or image height,
we're going to need to rely
on object and image distance.
Our magnification equation
is a ratio of the negative
of image distance
to the object distance.
That means the image
magnification in our case
is negative 43 centimeters,
divided by 45 centimeters,
which is negative 0.96.
Now we've figured out what
the question asked about,
but do our answers
make sense together?
Are they consistent?
Let's bring together
our ray diagram,
the image distance
and the object distance.
The ray diagram indicated
our image distance
will be positive.
Since it's the opposite side
of the lens as the object.
It also looks like
the image distance
and object distance
are about the same,
and that the image is inverted.
The results we find
for image distance and
magnification confirms this.
The object and image distances
are about the same.
And the image magnification
is nearly negative one,
meaning the image and object
are nearly the same size,
but that one is inverted
to compare it to the other.
This is what we expected.
So we can be confident
in our solutions.
There's more than one sign
they're correct.
Let's move the object
closer to the lens
between the focal point
and the lens.
The object distance
is 15 centimeters and
the focal length of the lens
is 22 centimeters.
Where does the image form,
and what is its magnification?
Let's draw a diagram
that shows our set up
and the information given.
We'll carefully trace
the three principal rays.
If you're doing this on paper,
you'll want to use a ruler
and graph paper.
The principal rays
diverge from one another
as they move to the right.
They converge to a point only
if we trace them backward.
That means the image
that forms is virtual,
and that we expect the image
distance to be negative.
We also see the image is upright
and bigger than the object.
That tells us we can expect
the magnification
to be more than one.
See all the information
our drawing can give us.
A good ray diagram
is worth the effort.
Let's move ahead and solve for
the image distance, d sub i.
Our lens equation
will again be useful.
We're given the focal length
and the object distance,
and we want to find the third
variable, image distance.
We'll rearrange the equation
because we want to find d sub i,
the distance from the
object's image to the lens.
Now we plug in our values
for focal length
and object distance,
and calculate an image distance
of negative 47 centimeters.
Notice a couple of things
about this answer.
Number one, it's negative,
meaning the image is virtual.
That's consistent
with our ray diagram.
Number two,
it's bigger in magnitude
than our focal length
or object distance.
That also checks
with the diagram.
Now, let's look at
the image magnification.
Magnification is the ratio
of the image height
to the object's height.
It also can be found by
taking the negative ratio
of image distance
to object distance.
Think back to our amazingly
helpful ray diagram.
The virtual rays that
trace back from the lens
to where the image forms
begin at the top of the object,
it's height.
And they meet at the top
of the image, it's height.
It turns out that
in normal cases,
the ratio of image
to object height is equal
to the negative of image
to object distance.
We can write that in
equation form like this.
Image magnification is
equal to image height,
divided by object height,
which also equals the negative
of image distance,
divided by object distance.
So the image magnification
in our case is
negative 47 centimeters,
divided by 15 centimeters,
which is 3.1.
That means the image formed
by this object and lens
is a little over three times
bigger than the object itself.
We'll finish this closer look
with an example of
a different type of lens.
One that diverges or
spreads out rays of light.
Like converging lenses,
diverging lenses have
a particular focal length,
but diverging lenses are
different in a few ways.
In this example,
we have a diverging lens
with a focal length
of 40 centimeters.
An object distance
of 55 centimeters,
and an object height
of 18 centimeters.
We want to know,
what is the image height
and distance to the image?
First thing's first.
Let's draw a sketch
of the given information,
and then we'll do
a principal ray diagram.
Our sketch should
look like this.
Now let's trace in
three principal rays,
which will show us
where to expect the image.
We know that one ray goes
horizontally until it reaches
the diverging lens.
The lens bends the ray upward,
and we have to trace
the bent ray backwards
to intersect the first
focal point, F1.
The next ray goes straight
through the middle of the lens.
It isn't bent at all.
The last principal ray
is pointed toward
the second focal point, F2.
But rather than
passing through F2,
this ray is bent by the lens,
so it runs horizontally.
By tracing this ray backward,
we see where it intersects
the other two rays.
Like we expected
for a diverging lens,
the image is virtual,
smaller than the object
and upright.
Try it for yourself.
Press pause.
Take a couple minutes to
figure it out, then come back
and check your work.
Okay, did you find
the image distance is
negative 23 centimeters?
If so, great work.
If not, let's take
a look at it together.
We use the lens equation
so we can solve for
the image distance, d sub i.
The object distance is
given as 55 centimeters.
It's focal length magnitude
is 40 centimeters.
Since we're working
with a diverging lens,
remember that the focal length
is negative.
So when we insert numbers
into our lens equation,
the focal length, F,
is negative 40 centimeters.
This makes all the difference.
Negative 40 centimeters
times 55 centimeters,
divided by 55,
minus negative 40 centimeters,
all equals negative
23 centimeters.
This result is a great reminder
how important it is
to know sign conventions.
They're not always easy
to recall, but in long run,
learning them will
make things easier.
Now let's move to solving
for the height of the image.
We'll find this out
by using the fact that
magnification can be found
using two different ratios.
First, magnification equals
the ratio of image
to object height.
Second, it equals
the negative of image
to object distance.
We can solve for
the second ratio,
and then use that to
figure out image height.
The image magnification
is negative
negative 23 centimeters,
divided by 55 centimeters,
which equals 0.42.
That value is equal
to the image height,
divided by the object height.
So image height equals
object height, times 0.42.
We know the object height
is 18 centimeters.
18 centimeters times 0.42
is 7.6 centimeters.
That's the height of the image.
That's a closer look on lenses.
For more "Physics in Motion",
go to our homepage,
where the entire series
is available to you.
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