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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.

Short Answer

Expert verified
(a) 0.02 cm, toward the retina.

Step by step solution

01

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.
02

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 \( f \).
03

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, \( f = \frac{100}{13.5} = 7.41 \ mm \).
04

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 \( d_i' \).
05

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, \( d_i' \approx \frac{1}{0.12833} \approx 7.79 \ mm \).
06

Calculate Lens Movement

Find the difference \( \Delta d_i = 7.79 \ mm - 8 \ mm = -0.21 \ mm \). Since 1 cm = 10 mm, \( -0.21 \ mm = -0.021 \ cm \).
07

Determine Lens Movement Direction

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Given that frogs are nearsighted in air, which statement is most likely to be true about their vision in water? (a) They are even more nearsighted; because water has a higher index of refraction than air, a frog's ability to focus light increases in water. (b) They are less nearsighted, because the cornea is less effective at refracting light in water than in air. (c) Their vision is no different, because only structures that are internal to the eye can affect the eye's ability to focus. (d) The images projected on the retina are no longer inverted, because the eye in water functions as a diverging lens rather than a converging lens.

A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as the object. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.50 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

A camera with a 90-mm-focal-length lens is focused on an object 1.30 m from the lens. To refocus on an object 6.50 m from the lens, by how much must the distance between the lens and the sensor be changed? To refocus on the more distant object, is the lens moved toward or away from the sensor?

If you run away from a plane mirror at 3.60 m\(/\)s, at what speed does your image move away from you?

The focal length of a simple magnifier is 8.00 cm. Assume the magnifier is a thin lens placed very close to the eye. (a) How far in front of the magnifier should an object be placed if the image is formed at the observer's near point, 25.0 cm in front of her eye? (b) If the object is 1.00 mm high, what is the height of its image formed by the magnifier?

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