Problem 122 - Physics

Question

An object is placed to the left of a lens, and a real image is formed to the right of the lens. The image is inverted relative to the object and is onehalf the size of the object. (a) What kind of lens, converging or diverging, is used to produce this image? (b) How is the height \(h_{\mathrm{i}}\) of the image related to the height \(h_{0}\) of the object? Don't forget to take into account the fact that the image is inverted relative to the object. (c) What is the ratio \(d_{i} / d_{0}\) of the image distance to the object distance? For the situation described in the Concept Questions, the distance between the object and the image is \(90.0 \mathrm{~cm}\). (a) How far from the lens is the object? (b) What is the focal length of the lens?

Step-by-Step Solution

Answer

(a) Converging lens. (b) \( \frac{h_i}{h_0} = -0.5 \). (c) \( \frac{d_i}{d_0} = 0.5 \). Object is 60 cm from lens, focal length is 20 cm.

Step 1: Determine the Lens Type

For a real and inverted image formed on the opposite side of the object by a lens, a converging lens is used. Diverging lenses create virtual images that are upright. In this case, given the conditions of the image, the lens must be converging.

Step 2: Relate Image Height to Object Height

The magnification formula relates the heights of the image and the object. It's given by \( m = \frac{h_i}{h_0} \). Since the image is inverted and half the size: \( \frac{h_i}{h_0} = -0.5 \). The negative sign is because the image is inverted relative to the object.

Step 3: Calculate Image Distance to Object Distance Ratio

Since magnification \( m = \frac{-d_i}{d_0} \), substituting the magnification value from the previous step gives \( \frac{-d_i}{d_0} = -0.5 \). Therefore, \( \frac{d_i}{d_0} = 0.5 \).

Step 4: Calculate Object and Image Distance from Lens

Given that the total distance between the object and the image is 90 cm, we apply the relationship from Step 3. If \( d_i = 0.5 \times d_0 \), we have: \( d_0 + d_i = 90 \, \text{cm} \) or \( d_0 + 0.5d_0 = 90 \). This simplifies to \( 1.5d_0 = 90 \), solving for \( d_0 \) gives \( d_0 = 60 \, \text{cm} \). Then \( d_i = 0.5 \times 60 = 30 \, \text{cm} \).

Step 5: Calculate the Focal Length of the Lens

Using the lens formula \( \frac{1}{f} = \frac{1}{d_0} + \frac{1}{d_i} \), substitute \( d_0 = 60 \, \text{cm} \) and \( d_i = 30 \, \text{cm} \): \[ \frac{1}{f} = \frac{1}{60} + \frac{1}{30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20} \] Thus, \( f = 20 \, \text{cm} \).

Key Concepts

Magnification Formula

The magnification formula is a crucial concept when analyzing images formed by lenses. This formula helps you understand how the size of an image relates to the object's size. If you're given the height of the object, denoted as \( h_0 \), and the height of the image as \( h_i \), then the formula can be expressed as:

• \( m = \frac{h_i}{h_0} \)

Magnification \( m \) provides us with the size relationship between the object and image. In cases where the image is inverted, as it is here, the magnification will be negative. This indicates not only a size change but also a flip. For example, if the magnification is \(-0.5\), the image is half the size of the object and inverted. Understanding this formula helps you determine both the size and orientation of the image formed.

Real and Inverted Image

A real and inverted image is precisely what it sounds like—an image not only has size, but it also flips upside down relative to the object. This is a common characteristic of images formed by converging lenses. These lenses, also known as convex lenses, are designed to converge light rays to a point. Real images result because the rays of light actually meet at a point after passing through the lens. This is different from virtual images, which appear to diverge from a point behind the lens. For real images, because the light physically converges, we can project these images onto a screen. It's important to note that such images tend to be inverted compared to the object due to the way light rays intersect after passing through the lens.

Focal Length Calculation

Calculating the focal length of a lens is essential for understanding its optical properties. The focal length, denoted \( f \), is the distance from the lens where parallel rays of light converge to a point.We use the lens formula to find \( f \) when object distance \( d_0 \) and image distance \( d_i \) are known:

• \( \frac{1}{f} = \frac{1}{d_0} + \frac{1}{d_i} \)

Let's break it down with a practical example. Suppose an object is placed 60 cm away from the lens, and the image forms 30 cm on the opposite side. Plug these values into the formula:\[\frac{1}{f} = \frac{1}{60} + \frac{1}{30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20}\]From this calculation, the focal length \( f \) is found to be 20 cm. This distance indicates the point where all parallel light rays focus, showing the lens's strength in converging light.