Convex Lens Image Formation Calculation And Explanation

Hey guys! Let's dive into a fascinating physics problem involving image formation with a convex lens. This is a classic scenario that beautifully illustrates the principles of optics. We're going to break down the problem step-by-step, ensuring you not only get the answer but also understand the underlying concepts. So, let's get started!

The Problem: Object Placement and Image Location

Our main keyword here is convex lens image formation, and we're tackling a specific question related to it. Imagine an object placed 15 cm in front of a convex lens. This lens has a focal length of 10 cm. The big question is: Where does the image form? This is a common scenario in optics, and understanding how to solve it is crucial for anyone studying physics or working with lenses. We'll use the lens formula to figure this out, but first, let's make sure we understand what each term means and why it's important. When dealing with convex lenses, the position of the object and the focal length play a crucial role in determining the characteristics of the image formed. These characteristics include whether the image is real or virtual, upright or inverted, and magnified or diminished. Understanding these relationships is fundamental to solving problems related to lenses and image formation. Specifically, the distance of the object from the lens and the focal length of the lens dictate where the image will be formed. The formula we'll use, often called the thin lens equation, provides a mathematical relationship between these variables. To solve our problem effectively, we must accurately identify and apply these key parameters in the formula, ensuring we arrive at the correct image distance. Moreover, the sign conventions used in the lens formula are essential. For instance, distances measured from the lens to the object and image are considered positive if they are on the same side as the light travels, and negative if they are on the opposite side. The focal length of a convex lens is typically considered positive, while that of a concave lens is negative. These conventions help us determine not just the location of the image, but also its nature—whether it's real or virtual. Getting these signs right is a critical step in obtaining the correct solution. Therefore, careful attention to detail and a thorough understanding of the sign conventions are vital when working with lens problems. By mastering these basics, you'll be well-equipped to tackle a variety of optical scenarios.

Understanding the Lens Formula

The core of solving this problem lies in the lens formula, often written as:

1/f = 1/d₀ + 1/dᵢ

Where:

• f is the focal length of the lens.
• d₀ is the object distance (distance of the object from the lens).
• dᵢ is the image distance (distance of the image from the lens) – this is what we want to find!

Another way to express this formula, which is often more convenient for direct calculation of the image distance, is:

dᵢ = (d₀ * f) / (d₀ - f)

This is the formula you see in the problem statement, just using slightly different notation (d₁ for dᵢ). Let's break down each component of this formula and why it's so crucial in optics. The focal length (f) is an inherent property of the lens, determined by its shape and the material it's made from. It represents the distance at which parallel rays of light converge (in the case of a convex lens) or appear to diverge from (in the case of a concave lens). The focal length is essential because it dictates the lens's ability to bend light, and thus, where images will form. For a convex lens, the focal length is always positive, while for a concave lens, it's negative. Getting this sign right is crucial for accurate calculations. Next, the object distance (d₀) is the distance between the object and the lens. This distance is critical because it influences the path the light rays take as they pass through the lens. A closer object will have light rays diverging more as they reach the lens, requiring the lens to bend them more sharply to form an image. Conversely, a farther object will have nearly parallel rays, requiring less bending. The object distance is typically considered positive, as it's measured from the object to the lens in the direction the light travels. Finally, the image distance (dᵢ) is the distance between the lens and the image formed. This is the value we are usually trying to find, as it tells us where the image will be located. The sign of the image distance is also important. A positive image distance indicates a real image, which means the light rays actually converge to form the image on the opposite side of the lens from the object. A negative image distance indicates a virtual image, where the light rays appear to diverge from a point on the same side of the lens as the object. Understanding the relationships between these distances and the focal length is vital for predicting image formation in optical systems. This formula allows us to quantitatively determine how lenses shape and position images, making it a fundamental tool in optics.

Plugging in the Values

Now comes the fun part! Let's plug the given values into our formula:

• d₀ (object distance) = 15 cm
• f (focal length) = 10 cm

Using the formula:

dᵢ = (d₀ * f) / (d₀ - f)

We get:

dᵢ = (15 cm * 10 cm) / (15 cm - 10 cm)

dᵢ = 150 cm² / 5 cm

dᵢ = 30 cm

So, the image distance is 30 cm. This means the image forms 30 cm away from the lens. Let’s walk through the calculation step by step to ensure clarity and accuracy. First, we identify the given values: the object distance d₀ is 15 cm, and the focal length f is 10 cm. It's crucial to correctly assign these values to their respective variables in the lens formula. Next, we substitute these values into the formula dᵢ = (d₀ * f) / (d₀ - f). This gives us dᵢ = (15 cm * 10 cm) / (15 cm - 10 cm). Now, we perform the multiplication in the numerator: 15 cm * 10 cm = 150 cm². Simultaneously, we subtract the values in the denominator: 15 cm - 10 cm = 5 cm. This simplifies our equation to dᵢ = 150 cm² / 5 cm. Finally, we divide the numerator by the denominator: 150 cm² / 5 cm = 30 cm. The result is an image distance dᵢ of 30 cm. This positive value indicates that the image formed is real, meaning it can be projected onto a screen. It also tells us that the image is formed on the opposite side of the lens from the object. This step-by-step approach helps break down the problem into manageable parts, making it easier to understand and less prone to errors. By carefully plugging in the values and performing the calculations sequentially, we arrive at the correct image distance. Understanding each step not only helps in solving the problem but also builds a strong foundation in the principles of lens optics. This methodical approach is valuable for tackling more complex optical problems in the future. Moreover, reviewing each step ensures that we have not overlooked any critical details or made any calculation errors, leading to a confident and accurate solution. This thoroughness is a hallmark of good problem-solving skills in physics and other scientific disciplines.

Interpreting the Result

A positive image distance (30 cm in our case) tells us that the image is real. This means the light rays actually converge to form the image on the opposite side of the lens from the object. If the image distance were negative, it would indicate a virtual image, where the light rays appear to diverge from a point on the same side of the lens as the object. Now, let’s delve deeper into interpreting the positive image distance and what it signifies in the context of lens optics. A positive image distance, as we've found in our calculation, is a telltale sign of a real image. Real images are formed when light rays emanating from the object converge at a specific point after passing through the lens. This convergence creates an image that can be projected onto a screen, making it visible and tangible. In contrast, a negative image distance would imply a virtual image. Virtual images are formed when light rays do not actually converge but appear to diverge from a point. These images cannot be projected onto a screen; instead, they are perceived by the eye when looking through the lens. The location of the real image, 30 cm from the lens in our case, provides further insights. Since the image distance is positive, it means the image is formed on the opposite side of the lens from the object. This is a characteristic trait of real images formed by convex lenses when the object is placed outside the focal length. Furthermore, we can infer other properties of the image from the object and image distances. For instance, by comparing the image distance to the object distance, we can determine the magnification of the image. Magnification is the ratio of the image height to the object height and can be calculated as M = -dᵢ / d₀. In our scenario, the magnification would be M = -30 cm / 15 cm = -2. The negative sign indicates that the image is inverted, and the magnitude of 2 signifies that the image is twice the size of the object. Understanding the implications of the sign and magnitude of the image distance and magnification is crucial for fully characterizing the image formed by a lens. It allows us to predict not only the location but also the size and orientation of the image. This comprehensive understanding is fundamental to many applications of lenses, such as in cameras, telescopes, and microscopes. In summary, a positive image distance of 30 cm tells us we have a real, inverted, and magnified image formed on the opposite side of the lens from the object.

The Answer

Therefore, the correct answer is:

A. 30 cm

Key Takeaways

So, what did we learn today about convex lenses and image formation? Understanding the lens formula and how to apply it is crucial for solving these types of problems. Remember to pay close attention to the signs of the distances and focal length, as they tell you a lot about the image characteristics. Let's recap the key takeaways from this problem to ensure we've solidified our understanding of convex lenses and image formation. Firstly, we've reinforced the importance of the lens formula, 1/f = 1/d₀ + 1/dᵢ or its equivalent dᵢ = (d₀ * f) / (d₀ - f). This formula is the cornerstone for solving problems involving lenses, connecting the focal length, object distance, and image distance. Mastering this formula is essential for predicting where an image will form given the characteristics of the lens and the object's position. Secondly, we've highlighted the significance of sign conventions in lens calculations. The sign of the image distance provides crucial information about the nature of the image—positive for real images and negative for virtual images. The focal length is positive for convex lenses and negative for concave lenses. These conventions are not arbitrary; they are rooted in the physics of how light travels and bends through lenses. Ignoring these signs can lead to incorrect solutions and misinterpretations of the image characteristics. Thirdly, we've emphasized the distinction between real and virtual images. Real images are formed by the actual convergence of light rays and can be projected onto a screen. Virtual images, on the other hand, are formed by the apparent divergence of light rays and cannot be projected. Understanding this difference is vital for applications such as designing optical instruments like cameras and telescopes. Fourthly, we've touched upon the concept of magnification and how it relates to image size and orientation. The magnification, calculated as M = -dᵢ / d₀, tells us how much larger or smaller the image is compared to the object and whether the image is inverted or upright. A negative magnification indicates an inverted image, while a positive magnification indicates an upright image. Finally, we've underscored the importance of a step-by-step approach to problem-solving in physics. Breaking down a complex problem into smaller, manageable steps not only makes it easier to solve but also reduces the likelihood of errors. This methodical approach involves identifying given values, selecting the appropriate formula, substituting values correctly, performing calculations accurately, and interpreting the results in the context of the problem. By consistently applying these takeaways, you'll be well-prepared to tackle a wide range of problems related to lenses and optics. The principles discussed here are fundamental not only to academic physics but also to numerous real-world applications, from photography to vision correction.

I hope this explanation was helpful, guys! Keep practicing, and you'll become pros at lens problems in no time!