Problem 101 - Physics for Scientists and Engineers with Modern Physics

Question

A 50-year-old man uses \(+2.5-\mathrm{D}\) lenses to read a newspaper \(25 \mathrm{~cm}\) away. Ten years later, he must hold the paper \(32 \mathrm{~cm}\) away to see clearly with the same lenses. What power lenses does he need now in order to hold the paper \(25 \mathrm{~cm}\) away? (Distances are measured from the lens.)

Step-by-Step Solution

Answer

The man needs lenses with power approximately \(-0.875 \text{ D}\) to read at 25 cm.

Step 1: Understanding Lens Formula

The lens formula is given by \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. Here, \( v \) is the distance where he wants to hold the paper (25 cm), and we need to find the new focal length with this distance.

Step 2: Initial Lens Power Calculation

Initially, the man uses lenses with power \(+2.5\, \mathrm{D}\). The focal length \( f \) of these lenses in meters is calculated as: \[ f = \frac{1}{P} = \frac{1}{2.5} = 0.4\, \text{m} = 40\, \text{cm} \] So, the focal length of the initial lenses is 40 cm.

Step 3: Determine the Change in Reading Distance

Initially, the man read at 25 cm. Ten years later, he reads at 32 cm. The object distance \( u \) when he holds the paper at 32 cm is \( -32 \) cm (since object distance is taken negative in lens formula convention).

Step 4: Calculate New Required Focal Length

To find the new required focal length for holding paper at 25 cm, use the lens formula where the image distance (\( v \)) is now -25 cm: \[ \frac{1}{f_{new}} = \frac{1}{-25} - \frac{1}{-32} \] Simplifying, \[ \frac{1}{f_{new}} = -\frac{1}{25} + \frac{1}{32} \] \[ = \frac{-32 + 25}{800} \] \[ = \frac{-7}{800} \] Hence, the new focal length \( f_{new} = -\frac{800}{7} \approx -114.29 \text{ cm} \).

Step 5: Calculate New Lens Power

Convert the new focal length from centimeters to meters: \( -1.1429 \text{ m} \). The new lens power is given by \( P_{new} = \frac{1}{f_{new}} \). Thus, \[ P_{new} = \frac{1}{-1.1429} \approx -0.875 \text{ D} \]

Key Concepts

Lens Formula

The lens formula is a fundamental equation in optics that helps us understand how lenses form images. It is expressed as \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \). This formula relates three key quantities: the focal length \( f \), the image distance \( v \), and the object distance \( u \). It applies to both convex and concave lenses and aids in calculating magnification and understanding image properties.
When using the lens formula, keep in mind:

• Image distance \( v \): The distance from the lens to the point where the image forms. It is positive for real images (on the opposite side of the lens as the object) and negative for virtual images (on the same side as the object).
• Object distance \( u \): The distance from the lens to the object. Conventionally taken as negative in this formula.
• Focal length \( f \): The lens's ability to converge or diverge light, influenced by the lens shape and refractive index.

By accurately substituting the distances in the formula, one can solve for any of these three variables, offering insights into optical system design and function.

Focal Length Calculation

Focal length calculation is crucial for determining how a lens will focus light. To find the focal length \( f \) of a lens, one can utilize the lens's optical power \( P \). This relationship is described by the equation \( f = \frac{1}{P} \), where the focal length is in meters and the power in diopters (D).
In our exercise, the original lens had a power of \(+2.5\,\mathrm{D}\), which means it has a focal length:

• \( f = \frac{1}{2.5} = 0.4 \text{ m} = 40 \text{ cm} \)
• Positive power indicates a convex lens capable of converging light.

When the man needs to hold the paper closer later in life, we recalculate the focal length using the lens formula with new distances. These changes allow us to determine his new optical requirements.
Remember, a negative focal length, as calculated for the new lens, signifies a diverging lens, which is required when the previous lens is no longer efficient due to changes like presbyopia.

Lens Power Calculation

Understanding lens power calculation is essential for finding the right lenses to correct vision changes. Lens power \( P \) is calculated using the formula \( P = \frac{1}{f} \), where \( f \) is the focal length in meters. It measures a lens's ability to bend light to converge or diverge.
For the exercise problem, initially, the lens power was \(+2.5\mathrm{D}\), indicating the converging power of the lens.

• After aging and needing a different focus, the calculation for the new focal length \( f_{new} = -1.1429 \text{ m} \) yielded a new power of:
• \( P_{new} = \frac{1}{-1.1429} \approx -0.875 \text{ D} \).
• Here, a negative value shows the increase in eye accommodation needs.

This lens power translates into using different lenses or adapting existing ones to maintain clear vision at the desired reading distance. The effectiveness of these calculations depends on understanding the optometric parameters and having accurate measurements of the distances involved.