Problem 43 - College Physics

Question

A converging lens with a focal length of \(20 \mathrm{~cm}\) is used to produce an image on a screen that is \(2.0 \mathrm{~m}\) from the lens. What are the object distance and the lateral magnification of the image?

Step-by-Step Solution

Answer

The object distance is approximately 22.22 cm, and the magnification is 9.

Step 1: Convert Units

First, convert all measurements to the same units. The 2.0 meters for the image distance is converted to centimeters: \[ 2.0 \, \text{m} = 200 \, \text{cm} \] So, the image distance \(q\) is \(200\, \text{cm}\) and the focal length \(f\) is \(20 \text{ cm}\).

Step 2: Use the Lens Formula

We use the lens formula to find the object distance \(p\): \[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \] Plugging in the known values, we get \[ \frac{1}{20} = \frac{1}{p} + \frac{1}{200} \]

Step 3: Solve for Object Distance

Rearrange the equation from the previous step to solve for \(\frac{1}{p}\):\[ \frac{1}{p} = \frac{1}{20} - \frac{1}{200} \]Calculate:\[ \frac{1}{p} = \frac{10}{200} - \frac{1}{200} = \frac{9}{200} \]Thus, the object distance is:\[ p = \frac{200}{9} \approx 22.22 \text{ cm} \]

Step 4: Calculate Lateral Magnification

The lateral magnification \(m\) is given by the ratio of the image distance to the object distance:\[ m = \frac{q}{p} = \frac{200}{22.22} \approx 9 \]This indicates that the image is 9 times the size of the object.

Key Concepts

Lens Formula

When dealing with lenses, understanding the lens formula is crucial. It helps you determine how an object and its image relate through a lens. The formula states that the reciprocal of the focal length (\(f\)) of the lens is equal to the sum of the reciprocals of the object distance (\(p\)) and the image distance (\(q\)):\[\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\]This equation is derived from geometrical optics principles, specifically the behavior of light through converging lenses.

• The focal length (\(f\)) is a constant for any given lens and is the point where light rays converge.
• The object distance (\(p\)) is how far the object is placed from the lens.
• The image distance (\(q\)) is where the resulting image forms on the opposite side of the lens.

By rearranging this formula, you can solve for any missing variable if two others are known. In our exercise, we used it to find the object distance, demonstrating how it connects real-world lens setups by providing a mathematical relationship between the key distances.

Object Distance

The object distance (\(p\)) is a core aspect when working with lenses. It measures how far the object causing an image is positioned from the lens itself.In the context of our problem, we needed to find \(p\) by using the already calculated image distance (\(q\)) of 200 cm and the focal length (\(f\)) of 20 cm:1. Start with the lens formula:\[\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\]2. Rearrange to solve for \(\frac{1}{p}\):\[\frac{1}{p} = \frac{1}{f} - \frac{1}{q}\]3. Substitute the known values:\[\frac{1}{p} = \frac{1}{20} - \frac{1}{200}\]4. Simplify to find \(p\):\[\frac{1}{p} = \frac{9}{200}\]\[p = \frac{200}{9} \approx 22.22 \text{ cm}\]This calculation tells us the position of the object required to create the specific image observed. Understanding this position is necessary for precise optical setups.

Lateral Magnification

Lateral magnification (\(m\)) is an important concept that gives insight into the size relationship between the object and its image as formed by a lens. It is defined as the ratio of the image distance (\(q\)) to the object distance (\(p\)):\[m = \frac{q}{p}\]This simple yet powerful calculation provides the scale at which the image is magnified compared to the physical object. A magnification greater than 1 indicates the image is larger, while less than 1 means it's smaller.After calculating the object distance as 22.22 cm, substituting the known values for \(q\) and \(p\) gives:\[m = \frac{200}{22.22} \approx 9\]This result shows the image is 9 times larger than the object itself. Understanding this allows grasp of how much the lens system enhances or reduces what is observed, crucial for practical applications like cameras and projectors.