Problem 112 - Physics

Question

A stamp collector is viewing a stamp with a magnifying glass held next to her eye. Her near point is \(25 \mathrm{~cm}\) from her eye. (a) What is the refractive power of a magnifying glass that has an angular magnification of \(6.0\) when the image of the stamp is located at the near point? (b) What is the angular magnification when the image of the stamp is \(45 \mathrm{~cm}\) from the eye?

Step-by-Step Solution

Answer

(a) 20 diopters, (b) 9 times.

Step 1: Understanding the Problem

We aim to find the refractive power of a magnifying glass with specific angular magnification when the image is at different distances from the eye. The refractive power is related to the focal length, and we need to relate this to angular magnification under different conditions.

Step 2: Using the Angular Magnification Formula

The angular magnification (\(M\)) of a magnifying glass can be given by the formula:\[ M = 1 + \frac{25}{f} \]where \(f\) is the focal length of the magnifying glass in cm, and the near point is \(25\) cm.

Step 3: Solving for Refractive Power in Part (a)

For part (a), the magnification is \(6.0\). We use the formula:\[ 6.0 = 1 + \frac{25}{f} \]Solving for \(f\), we rearrange the equation:\[ 5 = \frac{25}{f} \] \[ f = \frac{25}{5} = 5 \text{ cm} \] The refractive power \(P\) is the reciprocal of the focal length in meters:\[ P = \frac{1}{0.05} = 20 \text{ diopters} \]

Step 4: Calculating Angular Magnification for Part (b)

For part (b), the image is \(45\) cm from the eye. The angular magnification formula for an image not at the near point is:\[ M = \frac{d_i}{f} \]where \(d_i\) is the image distance.Given \(f = 5\) cm from part (a), we use:\[ M = \frac{45}{5} = 9 \text{ times} \]

Key Concepts

Refractive Power

Refractive power is a measure of how strongly a lens can bend light. It is expressed in diopters (D), which is the reciprocal of the focal length of the lens in meters. This means a lens with a high refractive power bends light sharply and brings it to focus at a shorter distance. To calculate the refractive power (P), you use the formula:\[ P = \frac{1}{f} \]where \(f\) is the focal length in meters. In our example, after determining the focal length of the magnifying glass to be 5 cm (0.05 meters), the refractive power is:\[ P = \frac{1}{0.05} = 20 \text{ diopters} \]This tells us that the lens is quite powerful, making it effective for tasks like magnifying small objects such as stamps.Understanding refractive power helps in choosing the right lens for various optical applications, ensuring the desired clarity and magnification.

Angular Magnification

Angular magnification describes how much larger an object appears when viewed through an optical device compared to the naked eye. It's crucial for tools like magnifying glasses, telescopes, and microscopes. The angular magnification (M) formula varies depending on the position of the image or the object:

• If the image is at the near point:\[ M = 1 + \frac{25}{f} \]
• If the image is not at the near point:\[ M = \frac{d_i}{f} \]

In our exercise, we found two scenarios:- With the image at the near point (25 cm), the magnification is 6.- When the image is 45 cm from the eye, the magnification increases to 9.This rise in magnification at 45 cm reflects how the distance from the eye to the image affects how large the object appears.

Focal Length

Focal length is the distance between the lens and the point where it brings parallel rays of light into focus. It's a critical parameter in defining the lens's properties and its ability to enlarge objects. For our magnifying glass, the focal length (\(f\)) was derived using angular magnification:- Using the formula \[ M = 1 + \frac{25}{f} \]for a magnification of 6, rearranging gives \(f = 5\) cm.Knowing the focal length helps us determine how far from the lens an object needs to be held to achieve a certain image size. It also allows us to understand and calculate other properties such as refractive power, essential for tasks demanding precision viewing.

Image Distance

Image distance refers to the length between the lens and the image it forms. In optical systems, this is a crucial parameter as it influences magnification and clarity. For the magnifying glass scenario, we examined two image distances:

• The near point at 25 cm, where we aimed for an angular magnification of 6.
• An extended distance at 45 cm, resulting in a higher magnification of 9.

These distances affect both the size and clarity of the viewed image. A closer image distance typically means the object is more apparent but might require a stronger lens or higher refractive power. By understanding image distance, users can optimize the setup of their optical tools to best suit their vision needs.