Q108PP - University Physics with Modern Physics

Question

The eyes of amphibians such as frogs have a much flatter cornea but a more strongly curved (almost spherical) lens than do the eyes of air-dwelling mammals. In mammalian eyes, the shape (and therefore the focal length) of the lens changes to enable the eye to focus at different distances. In amphibian eyes, the shape of the lens doesn’t change. Amphibians focus on objects at different distances by using specialized muscles to move the lens closer to or farther from the retina, like the focusing mechanism of a camera. In air, most frogs are near-sighted; correcting the distance vision of a typical frog in air would require contact lenses with a power of about -6.0 D .A frog can see an insect clearly at a distance of 10cm. At that point the effective distance from the lens to the retina is 8 mm. If the insect moves farther from the frog, by how much and in which direction does the lens of the frog’s eye have to move to keep the insect in focus? (a) 0.02 cm toward the retina; (b)0.02 cm, away from the retina; (c)0.06 cm, toward the retina; (d)0.06 cm, away from the retina.

Step-by-Step Solution

Answer

Option (b) is correct,0.02 cm, towards the retina.

Step 1: Define the focal length.

The focal length is the distance between the convex or concave mirror and the focal point of the mirror.

The relation between the distance of object $u$, the distance of the image $u\text{'}$and the focal length $f$ is

$\frac{1}{f}=\frac{1}{u}+\frac{1}{u\text{'}}$

Step 2: Determine the focal length.

Given that,

$$\begin{gathered} u=10.0 \mathrm{cm} \\ u\text{'}=0.8 \mathrm{cm} \end{gathered}$$

Substitute the value in focal length formula:

$$\begin{gathered} \frac{1}{f}=\frac{1}{u}+\frac{1}{u\text{'}} \\ =\frac{1}{10}+\frac{1}{0.8\mathrm{cm}} \\ =1.35 {\mathrm{cm}}^{-1} \end{gathered}$$

The focal length of the lens is $1.35 {\mathrm{cm}}^{-1}.$

Step 3: Determine the distance lens moved.

The distance of image $u\text{'}$ is

$$\begin{gathered} \frac{1}{u\text{'}}=\frac{1}{f}-\frac{1}{s} \\ =1.35-\left(\frac{1}{10}\right) \\ =1.28 {\mathrm{cm}}^{-1} \end{gathered}$$

The lens should move by

$$\begin{gathered} d=0.800-0.779 \\ =0.021 \mathrm{cm} \end{gathered}$$

Hence, the lens moved by 0.02 cm.