Problem 58 - University Physics with Modern Physics

Question

A thin lens with a focal length of 6.00 \(\mathrm{cm}\) is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, 25.0 \(\mathrm{cm}\) from the eye, and that the lens is very close to the eye.

Step-by-Step Solution

Answer

(a) The angular magnification is 4.17. (b) The closest the object can be is approximately -7.89 cm from the lens.

Step 1: Understanding Angular Magnification

The angular magnification \( M \) obtainable with a thin lens, when the object is at the focal point, is given by the formula:\[M = \frac{D}{f}\]where \( D \) is the near point distance (25.0 cm) and \( f \) is the focal length of the lens (6.00 cm). Substituting the values, we have:\[M = \frac{25.0}{6.00} = 4.17\]This indicates that the angular magnification is 4.17 times.

Step 2: Finding the Closest Object Distance

When the object is examined through the lens, and the image is at the near point, the closest it can be brought to the lens can be calculated using the lens formula:\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]Here, \( v \) is the image distance, which is 25.0 cm (near point), and \( f \) is 6.00 cm. Solving for \( u \) (the object distance):\[\frac{1}{6} = \frac{1}{25} - \frac{1}{u}\]Rearranging gives:\[\frac{1}{u} = \frac{1}{25} - \frac{1}{6}\]\[\frac{1}{u} = \frac{1 - 25}{150}\]\[\frac{1}{u} = \frac{-19}{150}\]\[u = -7.89 \, \text{cm (approx.)}\]Thus, the object can be brought as close as approximately -7.89 cm from the lens.

Key Concepts

Angular Magnification

Angular magnification helps us understand how a lens increases the apparent size of an object. When using a thin lens as a magnifying tool, angular magnification can be calculated by comparing the near point distance of the eye to the focal length of the lens. The formula for angular magnification is:

• \( M = \frac{D}{f} \)

where:

• \( M \) is the angular magnification.
• \( D \) is the near point distance of the eye, usually 25 cm.
• \( f \) is the focal length of the lens.

For our example with a focal length of 6.00 cm, the angular magnification is calculated as \( M = \frac{25.0}{6.00} \), resulting in an angular magnification of approximately 4.17. This means that the object appears 4.17 times larger when viewed through the lens.

Lens Formula

The lens formula relates the object distance, image distance, and focal length of a lens. It's a powerful tool to determine how far from the lens an object must be to produce a clear image. The formula is expressed as:

• \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)

where:

• \( f \) is the focal length of the lens.
• \( v \) is the image distance.
• \( u \) is the object distance.

In the original problem, the image distance \( v \) is given as 25.0 cm. By rearranging the formula, we solve for \( u \), the object distance. Here, after computations, \( u \) came out to be approximately -7.89 cm, indicating that the object can be positioned very close to the lens for optimal viewing.

Focal Length

Focal length is a critical parameter of any lens. It defines how strongly the lens converges or diverges light. A shorter focal length signifies a lens that can focus light more tightly, leading to a stronger magnifying power. For our thin lens discussed here, the focal length is 6.00 cm. This relatively short focal length makes it a strong magnifier, suitable for viewing objects in great detail at a closer range. The importance of focal length in the lens formula cannot be overstated, as it directly influences the position and size of the image formed by the lens.

Near Point

The near point is the closest distance at which the eye can focus on an object comfortably. For most people, this distance is around 25 cm. When using a lens as a magnifier, positioning the image at the near point ensures maximum comfort and clarity for the viewer. In this scenario, the image created by the lens is at the near point of 25.0 cm, allowing the user to see the magnified image clearly without eye strain. The concept of the near point is vital in understanding how to use lenses effectively, especially in tools like magnifying glasses and microscopes, where viewing comfort is just as important as magnification.