Figure 34-47a shows the basic structure of the human eye. Light refracts into the eye through the cornea and is then further redirected by a lens whose shape (and thus ability to focus the light) is controlled by muscles. We can treat the cornea and eye lens as a single effective thin lens (Fig. 34-47b). A “normal” eye can focus parallel light rays from a distant object O to a point on the retina at the back of the eye, where the processing of the visual information begins. As an object is brought close to the eye, however, the muscles must change the shape of the lens so that rays form an inverted real image on the retina (Fig. 34-47c). (a) Suppose that for the parallel rays of Figs. 34-47a and b, the focal length f^’of the effective thin lens of the eye is 2.50 cm. For an object at distance p = 40 cm, what focal length f^’ of the effective lens is required for the object to be seen clearly? (b) Must the eye muscles increase or decrease the radii of curvature of the eye lens to give focal length f^’?

• $f=2.5\text{cm}$.
• $p=40\text{cm}$.

Using the given lens formula for the eye, find its new focal length. After that, using the lens maker's equation, find the decrease in the radii of the curvature of the eye lens to give focal length f^’.

The formulas are as follows:

$$\begin{gathered} \frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\left(\frac{1}{p}+\frac{1}{i}\right) \\ \frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\left(n-1\right)\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \end{gathered}$$

Here p is the pole, f is the focal length, and i is the image distance.

(a)

This problem is regarding the human eye. Model the cornea and eye lens as a single effective thin lens, with the image formed at the retina.

The lens formula is given by,

$\frac{1}{i}=\frac{1}{f}-\frac{1}{p}$

For a relaxed eye, its lens focuses far-away objects on the retina so that the image distance i is behind the lens, and the object distance is$p=\infty$, then,

$\begin{array}{rcl}\frac{1}{i}& =& \frac{1}{f}-\frac{1}{\infty }\\ \frac{1}{i}& =& \frac{1}{f}\\ i& =& f\\ & & \end{array}$

Here f is the focal length of the relaxed effective lens.

So that$i=f=2.50\text{cm}$.

Now p is the new object distance and f^’ is the new focal length, then,

According to the lens formula,

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f\text{'}}$

Butlocalid="1663029348279" $i=f$.

Hence,

$\begin{array}{rcl}\frac{1}{p}+\frac{1}{f}& =& \frac{1}{f\text{'}}\\ f\text{'}& =& {\left(\frac{1}{p}+\frac{1}{f}\right)}^{-1}\\ f\text{'}& =& \frac{pf}{p+f}\\ & & \end{array}$

Substituting the values,

localid="1663029034365" $\begin{array}{rcl}f\text{'}& =& \frac{40.0\times 2.50}{40.0+2.50}\\ f\text{'}& =& \frac{100.0}{42.5}\\ f\text{'}& =& 2.35\text{cm}\\ & & \end{array}$

Therefore, the focal lengthf^’ of the effective lens required for the object to be seen clearly is 2.35 cm.

The lens maker's equation is,

$\frac{1}{f}=\left(n-1\right)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

Where, $r_1$and $r_2$are the radii of the curvature of the two surfaces of the lens, and n is the refractive index of the material from which the lens is formed.

But $r_1$and $r_2$have the same magnitude. $r_1$ is positive and $r_2$is negative.

role="math" localid="1663029226148" $\begin{array}{rcl}\frac{1}{f}& =& \left(n-1\right)\left(\frac{1}{r_1}-\frac{1}{r_1}\right)\\ f& =& {\left(n-1\right)}^{-1}{\left(\frac{1}{r_1}-\frac{1}{r_1}\right)}^{-1}\\ & & \end{array}$

Hence, the focal length decreases the radii of the curvature also decrease.

Using the lens formula, the focal length of an effective lens of the eye required to see the object clearly can be found.