Problem 68 - Physics

Question

A student is reading a lecture written on a blackboard. Contact lenses in her eyes have a refractive power of 57.50 diopters; the lens-to-retina distance is \(1.750 \mathrm{~cm}\). (a) How far (in meters) is the blackboard from her eyes? (b) If the writing on the blackboard is \(5.00 \mathrm{~cm}\) high, what is the size of the image on her retina?

Step-by-Step Solution

Answer

The blackboard is approximately 19.47 meters away, and the image on the retina is about -0.0045 cm tall.

Step 1: Understand the Lens Formula

The lens formula is given by \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f\) is the focal length, \(d_o\) is the object distance (distance from the blackboard), and \(d_i\) is the image distance (distance from the lens to the retina). We need to find \(d_o\).

Step 2: Convert Diopters to Focal Length

Since the refractive power is given as 57.50 diopters, the focal length \(f\) is the inverse of this power in meters, so \(f = \frac{1}{57.50} \approx 0.01739\) meters.

Step 3: Set Known Values from the Problem

We know \(f = 0.01739\) meters (from Step 2) and \(d_i = 1.750\) cm which is \(0.0175\) meters. We need to substitute these into the lens formula to find \(d_o\).

Step 4: Calculate the Object Distance \(d_o\)

Rearrange the lens formula to solve for \(d_o\): \(\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}\). Substitute \(f = 0.01739\) and \(d_i = 0.0175\) meters into the equation:\[ \frac{1}{d_o} = \frac{1}{0.01739} - \frac{1}{0.0175} \]Calculate to find \(d_o \approx 19.47\) meters.

Step 5: Find the Size of the Image

The magnification \(M\) can be expressed as \(M = -\frac{d_i}{d_o}\), and this is related to image height (\(h_i\)) and object height (\(h_o\)) as \(M = \frac{h_i}{h_o}\). We know \(h_o = 5.00\) cm, so:\[ h_i = M \times h_o = -\frac{0.0175}{19.47} \times 5.00 \]Calculate to find \(h_i \approx -0.0045\) cm.

Key Concepts

Lens Formula

The lens formula is fundamental in geometric optics for analyzing how lenses form images. It is represented 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.
• \( d_i \) is the distance from the lens to the image formed.

This formula helps us calculate any one of the three variables if the other two are known. It relies on the optical principle that the total optical power of a system accounts for how light converges or diverges.
This relation is particularly helpful for eyeglass lenses, camera lenses, and contact lenses in focusing an accurate image onto a surface like the retina in the eye. In the exercise, knowing the focal length and the image distance allows us to find out how far the object (the blackboard) is from the lens.

Refractive Power

Refractive power is a measure of how strongly a lens converges or diverges light. It is denoted in diopters (D) and is mathematically represented by:\[ P = \frac{1}{f} \] where \( P \) is the refractive power and \( f \) is the focal length in meters.
The higher the refractive power, the shorter the focal length, meaning the lens can bend light rays more sharply.
In the original problem, the refractive power of the student's contact lens is given as 57.50 diopters, which we converted to a focal length of approximately 0.01739 meters.
This high refractive power is typical for corrective lenses designed to adjust the user's field of vision by bringing objects into clear focus.

Image Magnification

Magnification tells us how much larger or smaller the image is compared to the actual object. It is determined using the formula:\[ M = -\frac{d_i}{d_o} \] where:

• \(M\) is the magnification.
• \(d_i\) is the image distance.
• \(d_o\) is the object distance.

The negative sign indicates that the image formed is inverted relative to the object.
The magnification is also related to the size of the image (\(h_i\)) and the object (\(h_o\)):\[ M = \frac{h_i}{h_o} \]
For example, if the object height is known and the object distance and image distance are calculated from the lens formula, the size of the image can easily be found using magnification. In the exercise, the blackboard's writing size and distances are used to calculate the retina image size.

Focal Length Calculation

Calculating the focal length is critical for determining how lenses focus light. When the refractive power is provided, the focal length can be found using \[ f = \frac{1}{P} \] where \( P \) is the refractive power in diopters.
In our scenario, substituting the given power of 57.50 diopters results in a focal length of 0.01739 meters.
Understanding and calculating focal length helps assess a lens's ability to focus light at different distances, essential for designing lenses for specific visual correction needs.
This aspect of geometric optics is crucial in fields like vision care, photography, and optical engineering, where precise control over image formation is necessary.