Problem 59 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A converging lens of focal length \(f=12 \mathrm{~cm}\) is being used by a writer as a magnifying glass to read some fine print on his book contract. Initially, the writer holds the lens above the fine print so that its image is at infinity. To get a better look, he then moves the lens so that the image is at his \(25-\mathrm{cm}\) near point. How far, and in what direction (toward or away from the fine print) did the writer move the lens? Assume the writer's eye is adjusted to remain always very near the magnifying glass.

Step-by-Step Solution

Answer

The writer moved the lens approximately 3.89 cm towards the fine print.

Step 1: Determine the Initial Lens Position

When the image is at infinity, the distance from the lens to the object is equal to the focal length of the lens. Therefore, the initial object-lens distance is \(12\, \text{cm}\).

Step 2: Use Lens Formula for Final Position

The lens formula is given as: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]Here, \(f = 12 \text{ cm}\), and \(v = -25 \text{ cm}\) since the image position \(v\) is on the same side as the object. Substitute these into the equation to find \(u\).

Step 3: Calculate the New Object Distance

Rearrange the lens formula to solve for \(u\):\[ \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \]Substitute the known values: \[ \frac{1}{u} = \frac{1}{12} - \left(-\frac{1}{25}\right) \]Calculate the result:\[ \frac{1}{u} = \frac{1}{12} + \frac{1}{25} = \frac{25 + 12}{300} = \frac{37}{300} \]Then, find \(u\):\[ u = \frac{300}{37} \approx 8.11 \text{ cm} \]

Step 4: Determine the Lens Movement

Initially, the lens was \(12\, \text{cm}\) away from the object. It was moved to approximately \(8.11\, \text{cm}\) from the object. Hence, the lens was moved towards the fine print by:\[ 12 - 8.11 = 3.89\, \text{cm} \]

Key Concepts

Focal Length

The focal length of a lens is a fundamental property that determines how strongly the lens can converge or diverge light. For converging lenses, the focal length is positive and represents the distance from the lens to the focal point along the lens's principal axis.

In the context of our example, the converging lens has a focal length of 12 cm. This means that, with the lens, parallel rays of light will converge to a point 12 cm away from the lens. When using lenses as a magnifying glass, like our writer, knowing the focal length helps to position the lens correctly for optimal viewing clarity and magnification.

Lens Formula

The lens formula is a crucial equation in optics that connects the focal length ( ext{} ext{ ext{} ext{f}}{f}) of a lens to the distances of the image ( ext{} ext{ ext{} ext{v}}{v}) and the object ( ext{} ext{ ext{} ext{u}}{u}) relative to the lens. It is expressed as:

• \(rac{1}{f} = rac{1}{v} + rac{1}{u}\)

This formula helps us understand how images form through lenses, predicting where an image will appear for a given object position. When applying this formula in practice, it's important to take note of sign conventions for distances:

• Focal length ( ext{}{f}) is positive for converging lenses.
• Image distance ( ext{}{v}) is positive if the image forms on the opposite side of the light source and negative if it's on the same side.
• Object distance ( ext{}{u}) is always positive.

Image Distance

Image distance refers to how far the image is from the lens. It's an important measurement that indicates where the converged light rays will meet to form a visible image.

In our scenario, the writer initially held the lens in such a way that the image formed at infinity, meaning parallel rays were emerging from the lens. He then adjusted it so that the image is at his 25-cm near point. This particular position changed the image distance to -25 cm.

The negative sign here indicates that the image is virtual, a common occurrence when lenses are used for magnification, and it's on the same side as the object. Knowing how to predict and calculate image distance is key to effective lens positioning in optics.

Object Distance

Object distance is the distance from the object being viewed to the lens. This measurement is foundational in determining image characteristics through the lens formula.

Initially, when the image was at infinity, the object distance equaled the focal length, which was 12 cm. But, for a more comfortable view at his near point, the writer adjusted the lens, resulting in a new object distance of roughly 8.11 cm from the print.

• This change was calculated using the lens formula, a valuable tool for precisely determining the object distance needed for optimal image viewing.

By understanding the relationship between these distances and the mechanics of lenses, one can control and manipulate vision through lenses effectively.