Problem 108 - University Physics with Modern Physics

Question

A frog can see an insect clearly at a distance of 10 cm. At that point the effective distance from the lens to the retina is 8 mm. If the insect moves 5 cm 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

(a) 0.02 cm, toward the retina.

Step 1: Understand the Problem

The frog needs to keep the insect in focus as it moves from 10 cm to 15 cm distance. The lens has to adjust to maintain a clear image on the retina, considering the new object distance.

Step 2: Apply Lens Formula

Use the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. Initially, \( d_o = 10 \ cm \) or \( 100 \ mm \) and \( d_i = 8 \ mm \). Substitute to find \( f \).

Step 3: Calculate Initial Focal Length

From \( \frac{1}{f} = \frac{1}{100} + \frac{1}{8} \), calculate \( f \). Solve: \[ \frac{1}{f} = \frac{1}{100} + \frac{1}{8} = \frac{1 + 12.5}{100} = \frac{13.5}{100} \] Therefore, \( f = \frac{100}{13.5} = 7.41 \ mm \).

Step 4: Determine Image Distance for New Object Distance

With the new object distance \( d_o = 15 \ cm \) or \( 150 \ mm \), use the lens formula again: \( \frac{1}{7.41} = \frac{1}{150} + \frac{1}{d_i'} \). Replace \( f \) and solve for \( d_i' \).

Step 5: Calculate New Image Distance

Solve for \( d_i' \): \[ \frac{1}{d_i'} = \frac{1}{7.41} - \frac{1}{150} \] \[ \frac{1}{d_i'} \approx 0.135 - 0.00667 = 0.12833 \] Hence, \( d_i' \approx \frac{1}{0.12833} \approx 7.79 \ mm \).

Step 6: Calculate Lens Movement

Find the difference \( \Delta d_i = 7.79 \ mm - 8 \ mm = -0.21 \ mm \). Since 1 cm = 10 mm, \( -0.21 \ mm = -0.021 \ cm \).

Step 7: Determine Lens Movement Direction

The result \( -0.021 \ cm \) indicates that the lens needs to move towards the retina to maintain focus.

Key Concepts

Lens Formula

In optics, the lens formula is a crucial equation used to describe the relationship between the object distance, the image distance, and the focal length of a lens. The formula is given as:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where:

• \( f \) is the focal length of the lens.
• \( d_o \) is the distance from the object to the lens (object distance).
• \( d_i \) is the distance from the lens to the image formed (image distance).

This formula is essential for determining how a lens will form an image of an object. It is commonly used in various applications, such as eyeglasses, cameras, and even in the eyes of living organisms like frogs. By rearranging the formula, we can solve for any one of the three variables if the other two are known. This equation helps in adjusting the focus to keep an image sharp, like when the frog needs to keep an insect in focus as it moves farther away.

Object Distance

The object distance, denoted as \( d_o \), is the distance from the object to the lens through which it is viewed. In our frog's eye problem, this distance initially is 10 cm or 100 mm, as the frog sees the insect clearly at this point. When the insect moves 5 cm further away, the object distance becomes 15 cm or 150 mm.Understanding and adjusting for changes in object distance is crucial in optics, especially for maintaining the focus of an image on the retina. When an object moves, as with the insect moving further from the frog, the lens needs to adjust accordingly to ensure that the light converges properly on the retina. This adjustment process involves the lens moving slightly to compensate for the increased distance, essential for clear vision.

Focal Length

The focal length, symbolized by \( f \), is a measure of how strongly a lens converges or diverges light. It is a fixed property of the lens but is essential in determining how the lens focuses images. For the frog's eye, the focal length is calculated by using the initial object and image distances:\[ \frac{1}{f} = \frac{1}{100} + \frac{1}{8} \]This results in the focal length being approximately 7.41 mm.The focal length is a key component of the lens formula, as it helps define where the image will be formed. A shorter focal length implies stronger convergence of light and thus a nearer focus. While a fixed property, the effect of focal length can be adjusted by changing object and image distances to maintain a sharp image.

Image Distance

Image distance, denoted as \( d_i \), is the distance from the lens to the point where the image is formed, typically on the retina in biological systems. Originally, the frog sees the insect with an image distance of 8 mm. This changes as the object distance changes.When the insect moves to a new position at 15 cm, we need to find the new image distance using the lens formula again:\[ \frac{1}{7.41} = \frac{1}{150} + \frac{1}{d_i'} \]Solving this gives a new image distance \( d_i' \approx 7.79 \ mm \). The shortening of the image distance indicates that the lens must move towards the retina by about 0.21 mm to keep the image in focus. Understanding this movement helps in interpreting how lenses, like the eye's, adjust for clear vision despite changes in object distance.