Problem 29 - University Physics with Modern Physics

Question

A converging lens forms an image of an 8.00 -mm-tall real object. The image is 12.0 \(\mathrm{cm}\) to the left of the lens, 3.40 \(\mathrm{cm}\) tall, and erect. What is the focal length of the lens? Where is the object located?

Step-by-Step Solution

Answer

The object's location is 2.82 cm left of the lens; the focal length is -1.69 cm.

Step 1: Identify the given quantities and symbols

We have a real object with an actual height \( h = 8.00 \text{ mm} = 0.80 \text{ cm} \). The image height is \( h' = 3.40 \text{ cm} \), and it is erect, which indicates a positive value. The image distance \( d_i = -12.0 \text{ cm} \) because it is on the same side as the object for a converging lens.

Step 2: Use the magnification formula

The magnification formula is \( m = \frac{h'}{h} = \frac{d_i}{d_o} \), where \( d_o \) is the object distance we need to find. Substituting the known values gives \( m = \frac{3.40}{0.80} = \frac{-12.0}{d_o} \).

Step 3: Calculate the object distance \( d_o \)

From \( m = \frac{-12.0}{d_o} = \frac{3.40}{0.80} \), solve for \( d_o \): \[ d_o = \frac{-12.0 \times 0.80}{3.40} = -2.82 \text{ cm} \]. This means the object is positioned 2.82 cm to the left of the lens.

Step 4: Use the lens formula to find focal length

The lens formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Substitute known values: \( \frac{1}{f} = \frac{1}{-2.82} + \frac{1}{-12.0} \). Calculating this gives its reciprocal: \[ \frac{1}{f} \approx -0.509 + (-0.083) = -0.592 \Rightarrow f \approx \frac{1}{-0.592} \approx -1.69 \text{ cm} \].

Key Concepts

Converging Lens

A converging lens, commonly known as a convex lens, is a lens that focuses incoming parallel light rays to a single point known as the focus. These lenses are thicker in the middle than at the edges and can be made of various materials like glass or plastic. They are used in a variety of applications, including cameras, eyeglasses, and in scientific instruments like microscopes and telescopes.
A key characteristic of the converging lens is its ability to produce real and virtual images. A real image is formed when light rays physically converge, whereas a virtual image appears when light rays seem to diverge from a point behind the lens. The real image produced by a converging lens can be projected onto a screen, and its properties are dictated by the relative distances of the object and lens.

Focal Length Calculation

Determining the focal length of a converging lens is crucial in understanding its optical nature. The focal length (\( f \)) is the distance between the lens and its focus. It reflects the "strength" of the lens in converging or diverging light.
To find the focal length, we use the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \);where \( d_o \) is the object distance and \( d_i \) is the image distance.
In the given problem, the image distance is negative because the image is on the same side of the lens as the object, indicating a virtual image. After solving the formula with the given values, we find that the focal length is approximately -1.69 cm. This negative value indicates that the lens forms a virtual image.

Magnification Formula

The magnification of a lens tells us how much larger or smaller the image is compared to the object. The magnification formula is \( m = \frac{h'}{h} = \frac{d_i}{d_o} \),where \( h' \) is the image height, \( h \) is the object height, \( d_i \) is the image distance, and \( d_o \) is the object distance.
This formula helps in understanding how the image is affected by the positioning of the object. If \( m > 1 \), the image is larger than the object; \( m < 1 \) indicates the image is smaller.
In the problem, substituting the known heights and solving for the object distance helps establish the relative positions of the object and lens, showing \( m = 4.25 \), which means the image is 4.25 times taller than the object.

Image Formation

The formation of images by a lens depends on several factors, including the type of lens, object distance, and light path. For a converging lens like the one in the exercise, an image can be real or virtual, inverted or erect, enlarged or reduced.
In this particular example, the image was erect and taller than the object, implying a virtual image, as virtual images formed by converging lenses are usually upright.

• A virtual image cannot be projected, as it appears on the same side of the lens as the object.
• The orientation and size provide essential clues about the image type and the relative distances from the lens.

The calculated values in the exercise confirm the virtual nature and the relative positions of the object and image in relation to the converging lens, offering insights into the practical behavior of optical systems.