Lens equations


The lens equation

In a similar fashion as we did with the converging mirror, we can derive equations relating the position of the object ($o$), the position of the image ($i$), the focal length ($f$), the height of the object ($h$), and the height of the image ($h'$). These variables are identified in the figure below. As in the case of the converging mirror, we take the signs of all of the variables for the real image to be positive. (Even though in this case, the image is on the opposite side of the lens, whereas in the case of the mirror it was on the same side — we just take everything in the real image case to be positive.) Only the three easy-to-draw rays are taken through the lens in this case.

If we look at the two similar triangles outlined with heavy lines below,

we see that $h/o = h'/i$ or

$$\frac{h'}{h} = \frac{i}{o}$$

just as we had in the case of the spherical mirrors. Looking at two other similar triangles

we see that

$\frac{h'}{i - f} = \frac{h}{f}$    or   $\frac{h'}{h} = \frac{i-f}{f}$

Setting this equal to $i/o$ and cross multiplying gives us

$$if + of = oi.$$

Dividing by $iof$ yields the same result as for the spherical mirrors:

$$\frac{1}{f} = \frac{1}{i} + \frac{1}{o}$$ 

So we have the same pair of equations we had for the spherical mirrors, but with different interpretations of when the variables $h'$ and $i$ are positive (and $f$ is not simply related to the curvature of the lens).

Unified equations and sign conventions

Just as with the curved mirror equations, there are two more sets; for a virtual image and a converging lens and a virtual image with a diverging lens. (A diverging lens  never makes a real image.) As with the mirrors, if we choose a  standard orientation everything reduces to the same equation — and the same equations as for mirrors.

Notice that in our treatment of the real image at the top of the page we described all the distances as positive — even though the object was above the center line and the image was below it. We took $h$ and $h'$ as both being positive. If we had chosen y coordinates one should have been positive and the other negative. Some books  do this. We prefer to consider this the standard situation and make everything positive. Here are the orientations for the real image situation for the lens:

  • object is to the left of the lens → $o$ (object distance) is +
  • object is above the center line  → $h$ (object height) is +
  • image is to the right of the lens → $i$ (image distance) is +
  • image is below the center line → $h'$(image height) is +
  • lens is converging   → $f$ (focal length) is +

With this, we have the pair of equations with everything positive and all the signs in the equations positive.

$$\frac{1}{f} = \frac{1}{i} + \frac{1}{o} \quad\quad \frac{h'}{h} = \frac{i}{o}$$

Now for all the other situations (and even for lenses), the exact same equations also hold. The only trick is, when anything flips from the standard (real image) situation, it changes sign.

  • object is below the center line  → $h$ (object height) is -
  • image is to the left of the lens → $i$ (image distance) is -
  • image is above the center line → $h'$ (image height) is -
  • lens is diverging   → $f$ (focal length) is -

This method has a number of great advantages. There is only one pair of equations to learn and one standard situation to remember. And if we make a wrong assumption about whether an image is real or virtual, up or down, the equation will handle things for us. If the sign comes out differently from what we expected it means we just have to flip something!

But remember the "dangerous bend" we saw in the curved mirror equations! The quantities in the equations are symbols and the signs can be "hidden" in what value the symbol contains. Any symbol may be positive or negative, whatever sign is in front of it. 

Another difficult point is that with these conventions the object distance must always be positive! If the object is on the right, the whole diagram should be flipped left to right.

The conventions are displayed in the figure below. They make sense when you keep the standard situation — the real image from a converging lens with the object in the top half.

The only real difference from the mirror case is that the lens diagram is "unfolded" since the light can go through instead of having to bounce off.

Joe Redish 4/11/12

Article 712
Last Modified: April 8, 2022