Problem 47 - Algebra for College Students

Question

The thin lens equation relates the object distance \(S_{o},\) the image distance \(S_{i,}\) and the focal length \(F\) for a thin lens. If the object distance is \(500 \mathrm{mm}\) and the focal length is \(100 \mathrm{mm},\) then what is the image distance? $$ \frac{1}{S_{o}}+\frac{1}{S_{i}}=\frac{1}{F} $$

Step-by-Step Solution

Answer

The image distance is 125 mm.

Step 1 - Substitute Given Values

Plug in the given values into the thin lens equation. The given values are the object distance \(S_{o} = 500 \mathrm{mm} \), and the focal length \(F = 100 \mathrm{mm}\). The thin lens equation is: \[ \frac{1}{S_{o}}+\frac{1}{S_{i}}=\frac{1}{F} \] Therefore, substituting the known values, we get: \[ \frac{1}{500}+\frac{1}{S_{i}}=\frac{1}{100} \]

Step 2 - Isolate \( \frac{1}{S_{i}} \)

Subtract \( \frac{1}{500} \) from both sides to isolate \( \frac{1}{S_{i}} \). This gives: \[ \frac{1}{S_{i}}=\frac{1}{100}-\frac{1}{500} \]

Step 3 - Combine the Fractions

To subtract the fractions, find a common denominator. The common denominator for 100 and 500 is 500. Rewrite the fractions: \[ \frac{1}{100} = \frac{5}{500} \] So the equation becomes: \[ \frac{1}{S_{i}}=\frac{5}{500}-\frac{1}{500} \] \[ \frac{1}{S_{i}} = \frac{4}{500} \]

Step 4 - Simplify the Fraction

Simplify \( \frac{4}{500} \) to make it easier to work with: \[ \frac{4}{500} = \frac{2}{250} = \frac{1}{125} \] So, \[ \frac{1}{S_{i}} = \frac{1}{125} \]

Step 5 - Solve for \( S_{i} \)

Take the reciprocal of both sides to solve for \( S_{i} \): \[ S_{i} = 125 \mathrm{mm} \]

Key Concepts

Object Distance

In the context of the thin lens equation, the term 'object distance' denotes the distance from the object to the center of the lens. This is usually represented as \(S_o\). Consider an object placed 500 millimeters from the lens, as given in our problem. It's essential to use the correct units when substituting into the equation to ensure accuracy. Object distance is crucial in determining where the image will form, impacting the image distance and characteristics such as size and orientation.

Focal Length

The focal length, represented as \(F\), is the distance from the lens to the focal point, where parallel rays of light converge. Think of it as a measure of the lens's focusing power; a shorter focal length means stronger focusing ability. In our exercise, the focal length is 100 millimeters. This value directly influences the image distance. Knowing the focal length allows us to apply the thin lens equation effectively, helping in calculating the position of the image formed by the lens.

Image Distance

The image distance, denoted as \(S_i\), is the distance between the image and the lens. Using the thin lens equation, we substitute the known values (object distance and focal length) to find this distance. For instance, with \(S_o = 500 \text{ mm}\) and \(F = 100 \text{ mm}\), we isolate \(\frac{1}{S_i}\) and solve for \(S_i\). After performing the calculations, we find that \(S_i\) is 125 millimeters. Image distance helps determine where the image will appear and its nature (real or virtual, inverted or upright).

Fraction Simplification

Fraction simplification is an essential part of solving the thin lens equation. After substituting and rearranging the terms, simplifying the resulting fractions makes the calculations manageable. In our example, subtracting fractions and finding a common denominator yields: \(\frac{1}{100} = \frac{5}{500}\). This allows us to perform the subtraction: \(\frac{5}{500} - \frac{1}{500} = \frac{4}{500}\). Simplifying \(\frac{4}{500}\) to \(\frac{1}{125}\) helps us solve for the image distance efficiently. Understanding fraction simplification ensures accurate and straightforward solutions in optical calculations.