Problem 49 - College Physics

Question

A convex lens has a focal length of \(0.12 \mathrm{~m}\). Where on the lens axis should an object be placed in order to get (a) a real, magnified image with a magnification of 2.0 and (b) a virtual, magnified image with a magnification of \(2.0 ?\)

Step-by-Step Solution

Answer

(a) Place the object 0.24 m from the lens for a real image. (b) Place the object 0.18 m from the lens for a virtual image.

Step 1: Understanding the Lens Formula

The lens formula relates the object distance (\(u\)), the image distance (\(v\)), and the focal length (\(f\)) of a lens. It's given by \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\). In this problem, \(f = 0.12\, \text{m}\). We will use this formula to find out where the object should be placed.

Step 2: Using Magnification for the Real Image

For part (a), the magnification (\(M\)) is defined as \( M = \frac{v}{u} = 2.0 \). This implies \( v = 2u \). Using the lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\), substitute \(v = 2u\) and \(f = 0.12\, \text{m}\) to find \(u\).

Step 3: Calculating Object Distance for Real Image

Substitute \(v = 2u\) in the lens formula to get \(\frac{1}{0.12} = \frac{1}{2u} - \frac{1}{u}\). Simplify to \(\frac{1}{0.12} = \frac{1}{2u} - \frac{2}{2u}\) or \(\frac{1}{0.12} = -\frac{1}{2u}\). Solve for \(u\) to find \(u = -0.12 \times 2\), resulting in \(u = -0.24 \, \text{m}\).

Step 4: Using Magnification for the Virtual Image

For part (b), we again have \( M = \frac{v}{u} = 2.0 \), so \( v = 2u \), but for a virtual image, the object distance is typically negative, and the image distance \(v\) is also negative to indicate a virtual image. We use \(v = -2u\) to reflect the nature of a virtual image and apply the same procedure using the lens formula.

Step 5: Calculating Object Distance for Virtual Image

Substitute \(v = -2u\) in the lens formula to get \(\frac{1}{0.12} = \frac{-1}{2u} - \frac{1}{u}\). Simplify to \(\frac{1}{0.12} = -\frac{3}{2u}\). Solve for \(u\) to find \(u = -3 \times 0.12 / 2\), giving \(u = -0.18 \, \text{m}\).

Key Concepts

Lens Formula

The lens formula provides a simple mathematical relationship connecting three essential quantities in lens optics: the object distance (\(u\)), the image distance (\(v\)), and the focal length (\(f\)). This relationship is expressed as:\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]This formula is crucial in determining where an image forms when an object is placed at a certain distance from a lens. In our exercise, a convex lens with a focal length of 0.12 meters is considered. The focal length (\(f\)) is a fixed value, and by knowing this, we can calculate the needed object distance to achieve the desired image, whether it be real or virtual.

• For a **real image**, the image distance is positive.
• For a **virtual image**, the image distance is negative.

Understanding this formula is vital because it allows us to predict practical outcomes, like where and how an image will appear. Thus, it lays the groundwork for effectively applying other concepts, like magnification.

Image Magnification

Magnification in lenses refers to how much larger or smaller an image appears compared to the object. It's defined by the ratio of the image height to the object height and is mathematically represented as:\[M = \frac{v}{u}\]In our exercise, the magnification factor is 2.0. This indicates that the image is twice the size of the object. Magnification can be positive or negative, indicating image direction:

• **Positive magnification** indicates an upright image.
• **Negative magnification** indicates an inverted image.

For part (a), where a real image is formed, the magnification being 2 means the image is twice the size and inverted. Since we have \(M = 2\), substituting this into the relation gives \(v = 2u\) for a real image. Part (b) deals with creating a virtual image, also with a magnification of 2. Here, \(v = -2u\), illustrating the inverse nature typical of virtual images produced by a convex lens.

Physics Optics

Physics optics covers the study of how light behaves and interacts with different materials, specifically through lenses and mirrors. Convex lenses are a common focus in this field, with properties like refraction and focal points taking center stage.
When light passes through a convex lens, it bends towards the principal axis, converging at a focal point. This bending is due to the lens's shape and the optical material it's made from. Convex lenses are used to focus light and project images more substantial than the actual object when placed at specific distances.
In the problem, understanding optics facilitates the application of the lens formula and magnification to place the object correctly along the lens axis. This understanding allows us to:

• Predict **image formation** - real or virtual.
• Calculate precise **object distances** for varied results.

By mastering physics optics, one can leverage these principles to solve diverse real-world challenges ranging from basic magnification in education to complex imaging systems.

Focal Length

Focal length is a fundamental concept in lens optics that describes the distance from the lens to the focal point, where parallel rays of light converge. It is denoted by \(f\) and is a key factor when assessing how a lens converges or diverges light rays.For a convex lens, a positive focal length signifies that the lens is converging, gathering parallel light rays to a point beyond the lens. In exercises like ours, knowing the focal length is crucial as it directly influences where the object must be positioned to form an image.
To find the object position for different types of images in this problem, we used a focal length of 0.12 meters:

• Determine the **object distance** for magnified real images using positive focal properties.
• Apply signs accordingly to calculate for **virtual images**, accounting for the negative implications on image distance.

Ultimately, the focal length not only helps determine the type of image (real or virtual) but also the nature of magnification achievable by the lens.