A concave lens produces a virtual image of an object. The positions of the object image and lens are shown in figure one. Note that the figure is not drawn to scale. The distance between the lens and the image is the same as the distance between the
object and the image. What is the magnification of the lens?
We see in figure one an object and a concave lens, which helps to form an image of
that object. There are a couple things we can notice about this figure. First, we see that the object is bigger than the image that’s formed, and second we
see that both the object and the image are upright. Neither one of them is inverted or flipped below the optical axis. And moreover, we’re told that the distance between the lens and the image is the same
as the distance between the image and the object. Knowing all this, we want to solve for the magnification of this concave lens.
Let’s start solving for this magnification by introducing a bit of shorthand. Let’s say that the symbol 𝑑 sub 𝑜 is equal to the distance between the object and
the lens. So whenever we write 𝑑 sub 𝑜, that’s the distance we’re referring to. And let’s also say that 𝑑 sub 𝑖 is equal to the distance between the image and the
lens. So from now on, instead of writing out distance between image and lens, we’ll write
𝑑 sub 𝑖, and instead of writing distance between object and lens, we’ll write 𝑑
With that settled, we’re ready to start talking about the magnification of this
concave lens. If we represent the magnification of a lens by capital 𝑀, then we can say that, for
upright objects and images, that magnification is equal to 𝑑 sub 𝑖 over 𝑑 sub
𝑜. In other words, it’s a ratio of distances of the image and object, respectively, to
Speaking of distances, recall that the problem statement tells us the distance
between the lens and the image is the same as the distance between the image and the
object. Writing this in terms of our image and object distances, we can say that this means
𝑑 sub 𝑜 is equal to two times 𝑑 sub 𝑖.
So now when we apply our magnification equation, we can say that the magnification of
our concave lens is 𝑑 sub 𝑖 over 𝑑 sub 𝑜, which is equal to 𝑑 sub 𝑖 over two
𝑑 sub 𝑖. The factor of 𝑑 sub 𝑖 cancels from numerator and denominator, leaving us with a
fraction positive one-half. That’s the magnification of this concave lens.
Next, let’s consider some data gathered from moving the object and image around. The distance from that object to the lens and from the image to the lens are shown in
table one. The values in table one are plotted in the graph in figure two. This graph is incomplete. Complete the graph in figure two by plotting the missing data and then drawing a line
of best fit.
Looking at table one, we see it consists of five data points. Each one has a distance from the object to the lens and a distance from the image to
the lens. If we look over to figure two, we see that three of these five data points are
plotted already in the figure. We want to do two things. First, we wanna plot the data in the table which is not yet in the graph. And then we want to draw a line of best fit for those five points.
First, let’s figure out which of the points in the table are already shown in the
graph. Looking at the horizontal axis, we see that the points already plotted have values of
1.0, 2.0, and 4.0. And looking at this axis, we see that those correspond to the distance from the
object to the lens in centimeters.
Looking at our table, we see that those three points correspond to the first, the
second, and the fourth points. That means we have yet to plot the third and fifth points on our graph. We’ll do that now.
The third data point in the table has a value on the horizontal axis of 3.0
centimeters and a value of 0.375 centimeters on the vertical axis. That puts this point just about where we’ve drawn it there on our graph. Then the last point to plot from our data table has a horizontal value of 5.0
centimeters and a vertical value of 0.357 centimeters. That occurs just about at the place we’ve put the 𝑥. So we’re halfway there.
We’ve now plotted the missing data points, and next we wanna draw a line of best fit
through these five points. When we draw this line of best fit, we’ll want to have a curve smoothly through our
five points. Now we might wonder, why not draw lines which directly connect our five points
straight from one to the other?
The reason we don’t do this is these five points are the only data that can possibly
exist. We know that these five points indicate a larger trend when it comes to image
distance versus object distance. So instead of drawing straight lines that directly connect our five data points, we
draw a smooth curve that moves through all of them.
Our graph now has all five points plotted on it as well as a line of best fit through
them. Next, let’s use this graph to solve for the lens magnification. What is the magnification produced by the lens when the object is 1.5 centimeters
away from the lens? Give your answer correct to two decimal places.
To answer this question, we’ll rely on two things: first the information shown in
figure two and second, our recollection of the magnification of a lens being equal
to the distance from the image formed to the lens divided by the distance from the
object to the lens.
Let’s start by looking at the graph and locating the point where the object is 1.5
centimeters away from the lens. On the horizontal axis, we find that location, and we see it’s between our first and
second data points. Knowing the object distance tells us 𝑑 sub 𝑜, so we’ll want to use this distance
and our graph to solve for 𝑑 sub 𝑖, the image distance. To do that, we can start at 1.5 centimeters on our horizontal axis and move up until
we reach our best fit line. Once there, we’ll move horizontally to the left until we run into our vertical
axis. We see that we’ve end up just between 0.44 and 0.42 centimeters. Let’s call it 0.43. So we’ll say that 𝑑𝑖 is 0.43 centimeters, and 𝑑𝑜 we’re told is 1.5
To find the magnification produced by the lens, we simply divide 𝑑𝑖 by 𝑑𝑜, or
0.43 centimeters by 1.5 centimeters, and notice that the units cancel out. Magnification is a unitless number. When we enter 0.43 divided by 1.5 on our calculator, to two decimal places, that’s
equal to 0.29. That’s the magnification produced by this lens, which means our image is about three
tenths as large as our object.
Next, let’s look at another graph which will help us solve not for magnification but
for image distance. The graph in figure three shows how the magnification of the lens changes as the
distance from the object to the lens changes. Show how figure three can be used to find the distance from the image to the
lens. Justify your answer with a calculation.
Looking at figure three, we see on the vertical axis the magnification of the lens
plotted, and on the horizontal axis we see the distance from the object to the lens,
what we’ve been calling 𝑑 sub 𝑜. We want to use this information to solve for the distance from the image to the lens
and recall that the image distance, object distance, and magnification are related
by an equation we’ve seen.
The magnification of a lens, 𝑀, is equal to the image distance divided by the object
distance. So- so if we multiply both sides of this equation by the object distance, 𝑑 sub 𝑜,
we find that 𝑑 sub 𝑖 is equal to 𝑑 sub 𝑜 times 𝑀, the magnification. Looking at the data points plotted in figure three, this means that if we find a
value for 𝑑 sub 𝑜 corresponding to magnification 𝑀, then the product of those two
values will be equal to the image distance.
We could choose any of the five points that are plotted in figure three as an example
for this calculation to solve for 𝑑 sub 𝑖. We may as well choose a point that lies along clearly marked-out vertical and
horizontal lines. We can choose the second data point which occurs at an object distance of 2.0
centimeters and a magnification of 0.20. That means these are the two values we’ll plug in for 𝑑 sub 𝑜 and 𝑀, respectively,
to solve for 𝑑 sub 𝑖.
As we do plug in, we recall that the object distance is in units of centimetres,
while the magnification has no units. 2.0 centimeters times 0.20 is equal to 0.40 centimeters. That’s the image distance corresponding to this object distance and
Finally, let’s draw a ray diagram of this situation. Complete the ray diagram in figure four to show how a concave lens produces an image
of an object. The diagram should show how the size of the image depends on the size of the object
and the distances between the object, the image, and the lens.
Looking at figure four, we see it consist of an object, a focal point marked capital
𝐹, and a concave lens along our vertical axis bisecting this figure. We’re asked to draw a ray diagram for this situation, and we know that those rays
will originate at the tip of the object.
We can see that, without any further specification, we could have an infinite number
of rays. They can move out in any direction from the object. But there’s a small subset of all these infinitely many rays called the principal
rays, which help us see where the image is formed by this object and the concave
lens. We know the general location of our concave lens, and we’ll imagine that this lens is
infinitely thin. Our goal then is to draw in the principal rays in this ray diagram so that we see
where the image is produced in figure four.
We know those rays all will begin at the tip of the object, and we can start drawing
the first one by drawing a horizontal line that reaches the concave lens. At this point, we know that the lens will bend the light. That’s called refraction. And the question is, how will the lens bend the light?
Since this principle ray that we’ve started to draw in is moving parallel to the
optical axis, that means that when it’s bent or refracted by our concave lens, that
bending will happen in such a direction that if we traced that bent ray backwards,
the trace would pass through the focal point marked capital 𝐹. So that’s one principal ray that we’ve drawn in.
But we can see one ray doesn’t specify where the image is formed. We’ll need another ray. A second principle ray that we can always draw starts from the tip of the object and
moves through the very center of the lens. And in this case, the lens doesn’t bend the light at all. It moves straight through.
At this point, looking at our diagram, we see that we have an intersection of
rays. If we take this ray that moves through the center of the lens, we see it intersects
with the line that traced back from our first principle ray. That intersection point is where the image is formed. We can draw in our image like this, and we can see that our image is smaller than the
object and it’s also upright. It’s not flipped upside down or inverted. We see, moreover, that the size of our image depends on the size of our object as
well as the distances between object, image, and lens. This diagram then shows how an image is produced by this object in this lens.