Size of Image on Retina

We want to find the height of the image $h_{\mathrm{i}}$, given the height of the object is $h_{\mathrm{o}}=1\text{.}\text{20}\times {\text{10}}^{-2}$ cm. We also know that the object is 60.0 cm away, so that $d_{\mathrm{o}}=\mathrm{60.0\; cm}$. For clear vision, the image distance must equal the lens-to-retina distance, and so $d_{\text{i}}=\mathrm{2.00\; cm}$ . The equation $\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}=m$ can be used to find $h_{\mathrm{i}}$ with the known information.

The only unknown variable in the equation $\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}=m$ is $h_{\text{i}}$:

Rearranging to isolate $h_{\text{i}}$ yields

Substituting the known values gives

This truly small image is not the smallest discernible—that is, the limit to visual acuity is even smaller than this. Limitations on visual acuity have to do with the wave properties of light and will be discussed in the next chapter. Some limitation is also due to the inherent anatomy of the eye and processing that occurs in our brain.