Problem 74 - Physics

Question

The contacts worn by a farsighted person allow her to see objects clearly that are as close as \(25.0 \mathrm{~cm}\), even though her uncorrected near point is \(79.0 \mathrm{~cm}\) from her eyes. When she is looking at a poster, the contacts form an image of the poster at a distance of \(217 \mathrm{~cm}\) from her eyes. (a) How far away is the poster actually located? (b) If the poster is \(0.350 \mathrm{~m}\) tall, how tall is the image formed by the contacts?

Step-by-Step Solution

Answer

(a) Poster is 128 cm away. (b) Image is 0.594 m tall.

Step 1: Understand the Problem

We are given the distances related to a farsighted person's vision with contacts and need to determine the actual and image height of the poster. The formula to use is the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \(d_o\) is the object distance and \(d_i\) is the image distance.

Step 2: Calculate the Focal Length

First, use the corrected near point to find the focal length. Set the new near point \(d_i = -25.0\) cm because this is the image formed by the corrective lens when the object is placed at the uncorrected near point \(d_o = -79.0\) cm.Use the lens formula:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{-79} + \frac{1}{-25} \]Calculate for \(f\).

Step 3: Find the Object Distance

Use the focal length obtained in Step 2 to determine the object distance \(d_o\) when the image is at \(d_i=217\) cm. The formula is:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{217} \]Solve for \(d_o\) to find the actual distance of the poster.

Step 4: Determine the Image Height

The magnification \(m\) of the lens is given by \(m = \frac{h_i}{h_o} = \frac{-d_i}{d_o}\), where \(h_o = 0.350\) m is the height of the poster and \(h_i\) is the height of the image. Calculate \(h_i\) using this formula with \(d_o\) and \(d_i\) from previous steps.

Step 5: Perform the Calculations

Complete the numerical calculations according to the formulas derived in the previous steps, with:- \(f\) from Step 2.- \(d_o\) and \(d_i\) values from Step 3.- Calculate \(h_i\) from Step 4.

Key Concepts

Focal Length Calculation

The focal length of a lens is a crucial factor in determining how light is focused, affecting the visibility of objects at various distances. In this exercise, we are given a farsighted person's near point with and without contacts. We need to determine the focal length needed for the corrective lenses.
First, the corrected image is formed at a distance of 25.0 cm from the eyes. This means the image distance, denoted as \(d_i\), is \(-25.0\, \text{cm}\) (negative because it's a virtual image formed on the same side as the object). The uncorrected near point, which is the farthest point at which the person can see clearly, is \(-79.0\, \text{cm}\).
Using the lens formula:

• The formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)
• Substituting values: \( \frac{1}{f} = \frac{1}{-79} + \frac{1}{-25} \)

By calculating, you can find the focal length \(f\), which is necessary to correct the person's vision and bring the near point to 25.0 cm.

Lens Formula

The lens formula is a key equation in geometric optics and is stated as \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. This formula allows us to determine one of these variables if the other two are known.
In practical terms, it represents the interplay between how lenses focus light and the distances involved in forming clear images.
For this exercise, after finding the focal length in our previous calculations, we use this formula again to determine how far the poster is from the person. Given \(d_i = 217\, \text{cm}\), we rearrange the formula:

• \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{217} \)
• Rearrange to solve for \(d_o\): \( \frac{1}{d_o} = \frac{1}{f} - \frac{1}{217} \)

Substituting the focal length from the previous step and solving this gives us the actual distance to the poster \(d_o\).

Image Magnification

Magnification is a measure of how much larger or smaller the image of an object is compared to the object itself. It is calculated using the formula: \( m = \frac{h_i}{h_o} = \frac{-d_i}{d_o} \). Here, \(h_o\) is the object height (in this case, the poster's height), \(h_i\) is the height of the image formed, \(d_o\) is the object distance, and \(d_i\) is the image distance.
For this exercise, given the poster's height \(h_o = 0.350\, \text{m}\) and using the distances calculated previously, the exercise shows that:

• Substitute \(d_o\) and \(d_i\) into \(m = \frac{-d_i}{d_o}\)
• Find \(h_i\) by rearranging to \(h_i = m \times h_o\)

This calculation results in the actual image height formed by the contact lenses, which can either enlarge or reduce the original object's dimensions.

Object and Image Distance

Understanding object and image distances is fundamental in designed systems such as eyeglasses or optical instruments. These distances determine how an image is projected when light passes through a lens.
Object distance \(d_o\) refers to how far the actual object is from the lens, while the image distance \(d_i\) is the distance from the lens to where the image is formed.
For the farsighted person in this problem, contacts help alter these distances to form a clearer image at the person's corrected near point. The poster's image distance, as seen by the person, is set at \(217\, \text{cm}\), but we need to use that to calculate \(d_o\) so the actual poster placement is accurate.

• Understanding this concept allows designers to optimize lens systems for clear vision in corrective eyewear like glasses.

Accurate calculations ensure effective visual corrections, showing just how essential precise measurement is in optical applications.