Question

Determine the position and size of the final image formed by a system of elements consisting of an object \(2.0 \mathrm{~cm}\) high located at \(x=0 \mathrm{~m},\) a converging lens with focal length \(5.0 \cdot 10^{1} \mathrm{~cm}\) located at \(x=3.0 \cdot 10^{1} \mathrm{~cm}\) and a plane mirror located at \(x=7.0 \cdot 10^{1} \mathrm{~cm}\)

Short answer

Answer: The final image is located at a distance of 75 cm from the lens and has a height of -5.0 cm.

Step-by-step solution

Find the image formed by the converging lens

To find the image formed by the lens, we will use the lens equation: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. In this case, the focal length \(f = 5.0 \cdot 10^1 \mathrm{~cm}\), and the object distance \(d_o = 3.0 \cdot 10^1 \mathrm{~cm}\). Plug the values into the lens equation: \(\frac{1}{5.0 \cdot 10^1} = \frac{1}{3.0 \cdot 10^1} + \frac{1}{d_i}\) Now, solve for \(d_i\): \(d_i = \frac{1}{\frac{1}{5.0 \cdot 10^1} - \frac{1}{3.0 \cdot 10^1}} \approx 7.50 \cdot 10^1 \mathrm{~cm}\)

Determine the magnification of the lens

To find the magnification of the lens, use the magnification formula: \(m = \frac{h_i}{h_o} = \frac{-d_i}{d_o}\), where \(h_o\) is the object height, and \(h_i\) is the image height. \(h_o = 2.0 \mathrm{~cm}\) Plug the values into the magnification formula: \(m = \frac{-7.50 \cdot 10^1}{3.0 \cdot 10^1}\) Solve for \(m\): \(m = -2.5\) Now, find the height of the image formed by the lens: \(h_i = mh_o = -2.5(2.0) = -5.0 \mathrm{~cm}\)

Interact with the plane mirror

Since the converging lens forms a virtual image with a height of \(-5 \mathrm{~cm}\) at a distance of \(7.50 \cdot 10^1 \mathrm{~cm}\) from the lens, and the plane mirror is located at \(x = 7.0 \cdot 10^{1} \mathrm{~cm}\), the distance between the image and the mirror is \(d_{mirror} = 7.50 \cdot 10^1 - 7.0 \cdot 10^{1} = 5.0 \mathrm{~cm}\). As the image is virtual, the mirror will form a new image on the same side of the mirror as the original image. The new image will be at the same distance from the mirror, so the final image distance from the mirror is \(5.0 \mathrm{~cm}\). The final image distance from the lens is \(7.00 \cdot 10^1 + 5.0 = 7.50 \cdot 10^1 \mathrm{~cm}\).

Determine the position and size of the final image

The final image will have the same height as the image formed by the lens, as plane mirrors do not change the magnification. So, the final image is located at \(x = 7.50 \cdot 10^1 \mathrm{~cm}\) from the lens and has a height of \(h_{final} = -5.0 \mathrm{~cm}\).

Key concepts

Converging Lens

A converging lens, often referred to as a convex lens, is designed to bring parallel rays of light to a single focal point. This is why it is called a "converging" lens. The characteristic shape of a converging lens, being thicker in the middle than at the edges, focuses incoming light beams. To determine how this lens affects an image, we need to know its focal length, which is the distance from the lens to the focal point. A shorter focal length implies a stronger converging power. In solving image-related problems, remember that a converging lens can create both real and virtual images depending on the object's position relative to the focal point.

Image Formation

Image formation through lenses involves directing light rays to converge or diverge, forming an image based on where these rays intersect or seem to originate. The nature of the image—whether it is real or virtual—depends on the object's location and the type of lens being used. For a converging lens:

• When the object is placed beyond the focal point, the lens forms a real and inverted image on the opposite side of the lens.
• If the object is positioned at the double focal length, the lens creates an image of equal size to the object.
• Should the object be within the focal length, the lens can't produce real images; instead, it results in a virtual and upright image on the same side of the lens as the object.

Understanding these scenarios helps predict the behavior of lenses in various settings.

Lens Equation

The lens equation is a critical tool for determining the characteristics of an image formed by a lens. It is expressed as:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]Here, \(f\) represents the focal length of the lens, \(d_o\) is the object distance (from the lens), and \(d_i\) is the image distance (from the lens). Applying the lens equation allows us to solve for unknowns when two of the three variables are known.For instance, if you know the focal length and object distance, you can compute the image distance. Such calculations help in determining whether the resulting image is real or virtual, and its corresponding placement relative to the lens.

Magnification

Magnification is the process by which a lens changes the size of an image compared to the object. The magnification formula is given by:\[m = \frac{h_i}{h_o} = \frac{-d_i}{d_o}\]Where \(m\) is the magnification, \(h_i\) and \(h_o\) are the image and object heights, and \(d_i\) and \(d_o\) are the image and object distances. Negative magnification indicates that the image is inverted.Magnification can tell you how much bigger or smaller an image is compared to the object. A magnitude greater than 1 indicates that the image is larger, and a magnitude less than 1 indicates that it is smaller. This helps in determining how an optical system alters perspectives and sizes in viewing.

Virtual Image

A virtual image is a type of image that is formed when light rays appear to diverge from a point behind the optical system, rather than converge in front of it as they do for a real image. In the context of a converging lens:

• A virtual image cannot be projected onto a screen as it doesn't exist in real space but instead forms on the same side of the lens or mirror as the object.
• Such images are upright and appear behind the lens when the object is placed within the focal length of a converging lens.
• Though virtual images cannot be captured directly, they can be viewed by looking through the lens or into the mirror that creates them.

Virtual images are crucial in understanding how devices like magnifying glasses and some optical instruments work.