Of magnifying glasses and magnificent ants

Remember the rules of ray tracing.

You don’t need the focal length of the lens, just the sign of it. You can find this out from the type of lens.

One of the results should look strange for a magnifying glass, the other should make sense.

The job of a magnifying glass is to enlarge objects.

On which side of the lens do you see the image?

Can you project the image you get onto a screen? What does that tell you about the type of image?

The image rays should converge 25 cm away from your eye/the lens.

Ray tracing might not give you an accurate answer, but it does help you understand the situation.

Remember that the magnification factor we are looking for is 2 (2x).

You can try to see what happens with a real lens (or an online video), and then explain it using ray diagrams.

Your eye can adapt to focus on images closer than your “close point,” but it would cause strain.

You can assume the image stays between the lens and the focal point

In the case of a convex lens, you can follow the steps below for both situations. The only difference will be the position of the object:

1. Draw the optical axis (a horizontal line) and the lens symbol (perpendicular to the optical axis).
2. Draw the focal points and position the object according to the situation.
3. Draw a light ray (straight line) from the top of the object to the lens, parallel to the optical axis. After the lens, continue the line, making it pass through the focal point on the image space.
4. Draw a line from the top of the object going through the point where the lens symbol and the optical axis intersect. This line does not get deflected as it goes through the lens.
5. (Optional) Draw a line that passes through the focal point, the top of the object, and the lens. After the lens, the line continues horizontally, parallel to the optical axis.
6. If the rays don’t cross in the image space, extend them backward (dotted line) until they do. The resulting image will be virtual.
7. Draw the image. The point where the rays cross is the top of the object.

Using the fact that both focal distances are positive for a convergent lens, we can create our ray diagram. Here are 2 examples of what the resulting diagrams should look like.

In the examples, the focal distances of the image and the object are similar, which is general property of magnifying glasses. If your drawings have different focal lengths, don’t worry, the focal lengths don’t have to be exactly the same. You can check your answers using the calculator/equation using any values, as long as the objects are located in the correct positions.

In both examples, the image is larger than the object. However, only one of them shows the result we expect from a magnifying glass. When you look through a magnifying glass, the image is formed behind the lens (in the object side of the diagram). Only the diagram with the object between the focal point and the lens meets this requirement. Moreover, this set up produces a larger magnification the closer the object is to the focal point (at the cost of image quality).

To answer this question, we can either use the Thin Lens Equation Calculator, or the relevant equations. Either way, the key step is the realization that because the image is virtual (i.e., it is on the object side), the distance from the image to the lens will be negative, so we need to use y = -25.0 cm. The resulting value for x should be positive. Looking at the equation for magnification of a lens, we will get the relationship between the distance to the object and the distance to the image.
M = |y|/x
By substituting in the values we know (Magnification, M, and distance to the image, |y|) we can calculate the distance from the object to the lens, x.
2 = 25.0 cm / x => x = 25.0 cm/2 = 12.5 cm

The object is precisely in the middle of the lens and the image.
Now that we know the position of the image and the object, we can calculate the focal distance of the magnifying glass using the thin lens equation:
1/x + 1/y = 1/f
It is crucial to remember that because the image is virtual (exists on the object side), the value of y is negative (y = -25 cm). This was not important before, because we had an absolute value.
Now we substitute in the formula and solve for f (the focal distance):
1/12.5 cm - 1/25.0 cm = 1/f => 1/25.0 cm = 1/f => f = 25.0 cm
The lens has a focal distance of 25.0 cm.

You can use the calculator, ray diagrams, or both to understand what happens as the object’s location differs.

Moving away from the object:
At first, moving away from the object with the lens close to your eye slowly reduces the magnification, and make everything blurry. This happens because the image is now formed in the image space, behind your eye.
You can check this by holding the lens still and moving your head farther and farther back – you should see the image getting larger and larger. Then it becomes a blurry mess. After that moment, if you keep moving back, the image should become more and more crisp, but it’ll also be upside down. You are now seeing the image we predicted in Question 1, where the object is farther from the lens and the focal point.

Moving closer to the object:
As we move closer to the object, the magnification will reduce. Since the image will form before it reaches the closest point of your eye, you will experience eye-strain at the beginning. After a while, it will just become blurry, as your eye cannot focus on the object/image anymore. At the very end (where the lens is touching the object), the lens will have almost no effect on the image of the object, and it will act like a flat piece of glass, like a window.