# 2.4 Thin lenses  (Page 5/13)

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$\frac{{n}_{2}}{{d}_{\text{o}}^{\prime }}+\frac{{n}_{1}}{{d}_{\text{i}}}=\frac{{n}_{1}-{n}_{2}}{{R}_{2}}.$

The image is real and on the opposite side from the object, so ${d}_{\text{i}}>0$ and ${d}_{\text{o}}^{\prime }>0$ . The second surface is convex away from the object, so ${R}_{2}<0$ . [link] can be simplified by noting that ${d}_{\text{o}}^{\prime }=|{d}_{\text{i}}^{\prime }|+t$ , where we have taken the absolute value because ${d}_{\text{i}}^{\prime }$ is a negative number, whereas both ${d}_{\text{o}}^{\prime }$ and t are positive. We can dispense with the absolute value if we negate ${d}_{\text{i}}^{\prime }$ , which gives ${d}_{\text{o}}^{\prime }=\text{−}{d}_{\text{i}}^{\prime }+t$ . Inserting this into [link] gives

$\frac{{n}_{2}}{-{d}_{\text{i}}^{\prime }+t}+\frac{{n}_{1}}{{d}_{\text{i}}}=\frac{{n}_{1}-{n}_{2}}{{R}_{2}}.$

$\frac{{n}_{1}}{{d}_{\text{o}}}+\frac{{n}_{1}}{{d}_{\text{i}}}+\frac{{n}_{2}}{{d}_{\text{i}}^{\prime }}+\frac{{n}_{2}}{-{d}_{\text{i}}^{\prime }+t}=\left({n}_{2}-{n}_{1}\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right).$

In the thin-lens approximation    , we assume that the lens is very thin compared to the first image distance, or $t\ll {d}_{\text{i}}^{\prime }$ (or, equivalently, $t\ll {R}_{1}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{R}_{2}$ ). In this case, the third and fourth terms on the left-hand side of [link] cancel, leaving us with

$\frac{{n}_{1}}{{d}_{\text{o}}}+\frac{{n}_{1}}{{d}_{\text{i}}}=\left({n}_{2}-{n}_{1}\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right).$

Dividing by ${n}_{1}$ gives us finally

$\frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}=\left(\frac{{n}_{2}}{{n}_{1}}-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right).$

The left-hand side looks suspiciously like the mirror equation that we derived above for spherical mirrors. As done for spherical mirrors, we can use ray tracing and geometry to show that, for a thin lens,

$\frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}=\frac{1}{f}$

where f is the focal length of the thin lens (this derivation is left as an exercise). This is the thin-lens equation. The focal length of a thin lens is the same to the left and to the right of the lens. Combining [link] and [link] gives

$\frac{1}{f}=\left(\frac{{n}_{2}}{{n}_{1}}-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)$

which is called the lens maker’s equation . It shows that the focal length of a thin lens depends only of the radii of curvature and the index of refraction of the lens and that of the surrounding medium. For a lens in air, ${n}_{1}=1.0$ and ${n}_{2}\equiv n$ , so the lens maker’s equation reduces to

$\frac{1}{f}=\left(n-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right).$

## Sign conventions for lenses

To properly use the thin-lens equation, the following sign conventions must be obeyed:

1. ${d}_{\text{i}}$ is positive if the image is on the side opposite the object (i.e., real image); otherwise, ${d}_{\text{i}}$ is negative (i.e., virtual image).
2. f is positive for a converging lens and negative for a diverging lens.
3. R is positive for a surface convex toward the object, and negative for a surface concave toward object.

## Magnification

By using a finite-size object on the optical axis and ray tracing, you can show that the magnification m of an image is

$m\equiv \frac{{h}_{\text{i}}}{{h}_{\text{o}}}=\text{−}\frac{{d}_{\text{i}}}{{d}_{\text{o}}}$

(where the three lines mean “is defined as”). This is exactly the same equation as we obtained for mirrors (see [link] ). If $m>0$ , then the image has the same vertical orientation as the object (called an “upright” image). If $m<0$ , then the image has the opposite vertical orientation as the object (called an “inverted” image).

## Using the thin-lens equation

The thin-lens equation and the lens maker’s equation are broadly applicable to situations involving thin lenses. We explore many features of image formation in the following examples.

Consider a thin converging lens. Where does the image form and what type of image is formed as the object approaches the lens from infinity? This may be seen by using the thin-lens equation for a given focal length to plot the image distance as a function of object distance. In other words, we plot

${d}_{\text{i}}={\left(\frac{1}{f}-\frac{1}{{d}_{\text{o}}}\right)}^{-1}$

for a given value of f . For $f=1\phantom{\rule{0.2em}{0ex}}\text{cm}$ , the result is shown in part (a) of [link] .

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