*Before you read this, I suggest you read* post 22.1.

The magnifying glass is a bi-convex lens that uses the light from an object to trick the eye into thinking the object is bigger than it really is. The way is works can be explained by ray-tracing, as shown in the picture below.

In the picture above, the origin (0, 0) of a Cartesian coordinate system is defined to be at the centre of the lens. As in post 22.1, a ray of light is considered to pass into the lens in the positive *x*-axis direction. The foot of the object is at (-*u*, 0) where *u* is less than *f*, the focal length of the lens. A ray from the top of the object, at (-*u*, *h*), parallel to the principal axis of the lens, is bent and meets this axis at the point (*f*, 0), the focal point of the lens. A ray from the top of the object, that passes through the centre of the lens, is not bent – as explained in post 22.1. In the picture rays are shown in red.

The picture shows that these rays appear to come from the point P. All other points appear in an image of the object at the same distance from the lens, as shown in the picture below. So, an image appears of the object appears with its foot at (-*v*, 0), and its top at (-*v*, *h’*).

We could reach the same conclusions, about the image seen through a magnifying glass, by using equation 3 from post 22.1 which states that

1/*f* = 1/*v* – 1/*u*

where *f* is positive for a bi-convex lens, as explained in post 22.1. When we use a magnifying glass, we move the lens to satisfy this equation – another way of saying this is that we focus the image. The image appears to be at a distance given by

1/*v* = 1/*f* + 1/*u* = (*u* + *f*)/*uf* so that *v* = *uf*/(*u* + *f*)

Since *f* is positive and greater than *u*, *u* + *f* is positive. So, when we give *u* a numerical value, it is negative (because it is in the minus *x*-axis direction) with the result that *uf* is negative but *u* + *f* is positive (because *f* > *u*); with therefore, the value of *v* is also negative. This means that the image appears to be on the same side of the lens as the object. Dividing the top and bottom of the final equation by *u* gives

*v* = *f*/[1 + (*f*/*u*)] = *f*/(1 + *w*) = –*f*/(*w* – 1).

where *w* = *f*/*u*. Since *f > u*, *w* must be greater than 1. This means that *v* must be further from the lens than its focal point (and, therefore, from the object) in the negative *x*-axis direction.

In post 22.2, we saw that the magnification, *m*, of a projector was defined by *h*’/*h*, with the result that

*m* = *v*/*u*.

Appendix 1 shows that the same result is true for a magnifying glass. Now the numerical values of *u* and *v* will both be negative, so that the value of *m* is positive. This means that, unlike a projector, the image seen in a magnifying glass is not inverted (see post 22.2 for more details).

In appendix 2, I give a numerical example to make all this easier to understand.

Remember, the image we see through a magnifying glass isn’t really there. Unlike a projected image you can’t go to it and draw around its features (see post 22.2). The rays don’t form an image directly – instead the eye is tricked into believing that the image exists, as shown by the ray-tracing pictures. This type of image is called a *virtual image*.

Finally, the magnifying glass is perhaps becoming obsolete – you can get a magnifying glass app for your phone.

Related posts

22.1 Refraction at curved surfaces

22.2 The projector

Appendix 1

In the picture above, the red line is a ray that is not bent because it passes through the centre of the lens, so it makes a constant angle with the *x*-axis (the principal axis of the lens). The tangent of this angle is given by *h*’/(-*v*) and *h*/(-*u*); the negative signs arise because *u* and *v* are in the minus x-axis direction. So

*h*’/*v* = *h*/*u* or *h*’/*h* = *v*/*u*.

But *h*’/*h* is the magnification, *m*, of the lens. Therefore *m* = *v*/*u*.

Appendix 2

*Let’s suppose that you are most comfortable reading normal text at a distance of 25 cm (the *near point* of a normal adult human eye). You want to see something in small type magnified to double its normal size (×2 magnification). What is the focal length of the lens that you need?*

The magnification of the lens *m* = 2 = *v*/*u* so that *u* = *v*/2.

So 1/*f* = 1/*v* – 2/*v* = -1/*v*.

For comfortable reading, you want the image to appear at *v* = -25 cm. The minus sign arises because you want the image to appear in the minus *x*-axis direction (see the second paragraph of the main text). Then

1/*f* = -1/(-25) = 1/25 or *f* = 25 cm.