Problem 53
Question
(I) A magnifier is rated at \(3.0 \times\) for a normal eye focusing on an image at the near point. (a) What is its focal length? (b) What is its focal length if the \(3.0 \times\) refers to a relaxed eye?
Step-by-Step Solution
Verified Answer
(a) 12.5 cm for near point; (b) 8.33 cm for relaxed eye.
1Step 1: Understanding Magnification Rating
A magnifier rated at 3.0× indicates that the angular magnification when used by a normal eye is 3.0. This means that the image appears 3 times larger when viewed through the magnifier as compared to the unaided eye focusing at the near point.
2Step 2: Using the Formula for Angular Magnification (Near Point)
For a near point condition, the angular magnification (M) is given by the formula \( M = 1 + \frac{D}{f} \), where \( D \) is the near point distance (typically 25 cm or 0.25 m for a normal eye) and \( f \) is the focal length. Set \( M = 3.0 \) and solve for \( f \):\[3.0 = 1 + \frac{0.25}{f}\].
3Step 3: Solving for Focal Length (Near Point)
Rearrange the equation \(3.0 = 1 + \frac{0.25}{f}\) to solve for the focal length \(f\):\[3.0 - 1 = \frac{0.25}{f}\] \[2 = \frac{0.25}{f}\] \[f = \frac{0.25}{2}\] \[f = 0.125 \text{ m or } 12.5 \text{ cm}\].
4Step 4: Using the Formula for Angular Magnification (Relaxed Eye)
For a relaxed eye, the angular magnification formula is \( M = \frac{D}{f} \) without the additional 1. Set \( M = 3.0 \) and use \( D = 0.25 \text{ m} \) to find \( f \):\[3.0 = \frac{0.25}{f}\].
5Step 5: Solving for Focal Length (Relaxed Eye)
Rearrange the equation \(3.0 = \frac{0.25}{f}\) to solve for \( f \):\[f = \frac{0.25}{3}\] \[f = 0.0833 \text{ m or } 8.33 \text{ cm}\].
Key Concepts
Focal LengthNear PointAngular Magnification
Focal Length
Focal length is a fundamental concept in optics. It refers to the distance between the lens and the point where parallel rays of light converge to a focal point. This distance determines how strongly the lens converges or diverges light. For a magnifying glass, a shorter focal length means stronger magnification.
When dealing with optical instruments, such as magnifiers, calculating the focal length is essential. In the exercise, the focal length differs based on whether the viewer's eye is relaxed or focusing at the near point.
When dealing with optical instruments, such as magnifiers, calculating the focal length is essential. In the exercise, the focal length differs based on whether the viewer's eye is relaxed or focusing at the near point.
- For a normal eye at the near point, the magnification equation is \( M = 1 + \frac{D}{f} \). Here, \( D \) is the near point distance, which is generally 25 cm for a standard human eye.
- When the eye is focused at the near point with a 3.0× magnifier, the focal length \( f \) calculates as 12.5 cm.
- For a relaxed eye, there's no added 1 in the equation, so the magnification is calculated by \( M = \frac{D}{f} \). This leads to a focal length of 8.33 cm for the same magnifier.
Near Point
The near point of the human eye is the closest distance at which an object can be comfortably and clearly focused. For an average adult, this distance is about 25 cm from the eye.
- When using a magnifying glass, the near point is important because it influences how the magnification is perceived.
- With magnifying lenses, if you focus at your near point, you can effectively perceive larger images.
- This aspect of vision is utilized in the formula for angular magnification: \( M = 1 + \frac{D}{f} \), where \( D \) represents the near point distance.
Angular Magnification
Angular magnification measures how much larger an object appears when viewed through an optical device, compared to viewing without it. This concept is central to devices like magnifiers and microscopes.
- It describes the increase in the angle of view. As the angle increases, so does the perceived size of the object.
- The equation changes depending on whether the eye is focused at the near point or relaxed.
- For the near point, the formula is \( M = 1 + \frac{D}{f} \).
- For a relaxed eye, it simplifies to \( M = \frac{D}{f} \), as relaxation involves less strain.
Other exercises in this chapter
Problem 49
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