Lens And Image Magnification: Experimental Physics

by ADMIN 51 views

Hey guys! Let's dive into a cool physics problem involving lenses, light, and magnification. We're going to explore how a converging lens can create magnified images and figure out some practical calculations. This is the kind of stuff that makes physics fun, right?

Understanding the Setup and the Problem

So, here's the deal: We have an experimental setup with a converging lens. Imagine a transparent lightbulb, like the ones we used to have in our homes, is lit up. This bulb is the object we are going to make an image of. This setup includes a converging lens placed in front of the lightbulb. On the other side of the lens, we have a wall that serves as our screen where the image will appear. The key here is that we're able to move the lens horizontally, meaning we can shift it left or right to change the image. Our goal is to find out how much we need to move the lens to make the image of the lightbulb's filament get magnified, which means it will be bigger than the actual filament. We're told that the image is magnified three times. Also, we know that the total distance between the lightbulb (the object) and the wall (the screen) is 2.0 meters. Now, let's break it down. The core of the problem is to figure out how much we need to move the lens to get the magnified image. To solve it, we're going to use the lens equation and the magnification formula.

Alright, let's put on our thinking caps and get into the physics of it all. We'll use a converging lens and a light source to understand how the lens works and how we get magnified images. Imagine the light from the bulb, traveling in straight lines, hits the lens. The lens bends these light rays, and this bending creates an image. The trick is to figure out where to put the lens so this image is three times bigger than the actual filament. The distance between the bulb and the wall, which is 2.0 meters, is crucial. This is like the total length of our experiment. We need to determine how the lens's position affects the image's size, and that's where magnification comes in.

Key Concepts: Lens Equation and Magnification

Before we start calculating, let's quickly refresh some essential concepts. We'll be relying on the lens equation and the magnification formula. The lens equation is our best friend in problems like these; it links the object distance (distance from the lightbulb to the lens), the image distance (distance from the lens to the wall where the image is formed), and the focal length of the lens. The formula is: 1/f = 1/do + 1/di, where 'f' is the focal length, 'do' is the object distance, and 'di' is the image distance. Keep in mind that the focal length is a characteristic of the lens itself and will remain constant during our experiment.

Now, let's talk about magnification. Magnification (M) tells us how much bigger or smaller the image is compared to the object. If M is greater than 1, like in our case (magnified 3 times), the image is enlarged. The magnification formula is M = -di/do. Here, M is magnification, 'di' is the image distance, and 'do' is the object distance. The negative sign indicates the image is inverted, but for our calculations, we'll focus on the absolute value since we're interested in the magnitude of the magnification.

So, we've got the lens equation and the magnification formula. The lens equation helps us to relate the distances involved (object and image) to the lens's focal length, while the magnification formula tells us how big or small the image is compared to the original object. It is a cool physics problem, and we're going to crack it!

Solving the Problem Step-by-Step

Now for the fun part – the calculations! Here's how we can solve this step by step:

  1. Understand the given information: We know the total distance from the lightbulb to the wall (object to image), which is 2.0 m. We also know the magnification (M) is 3.

  2. Use the magnification formula: Since M = 3, we know that 3 = di/do. This gives us a relationship between image distance (di) and object distance (do).

  3. Use the total distance: The total distance (2.0 m) is the sum of the object distance and the image distance, so do + di = 2.0 m. We can rearrange this to di = 2.0 - do.

  4. Substitute and solve: Substitute di = 2.0 - do into the magnification formula (3 = di/do). This gives us 3 = (2.0 - do) / do. Now, solve for do: 3do = 2.0 - do, which gives us 4do = 2.0, therefore do = 0.5 m.

  5. Find the image distance: Knowing do = 0.5 m, we can use di = 2.0 - do to find di = 2.0 - 0.5 = 1.5 m.

  6. The Lens's Displacement: To find the displacement, we need to know the initial position of the lens (before magnification) and the position after it to magnify the image. Let's consider the initial position to be somewhere in between. Now we know the object distance is 0.5 m, and the image distance is 1.5 m. So the lens has to be placed at 0.5 m from the lightbulb to get a 3x magnified image. It means that the lens must be displaced from its original position to focus the light at the desired position. This is all about where the lens is positioned.

Okay, so we've worked through the problem step-by-step. The core of the calculation involves the lens equation and the magnification formula. Using the magnification value, and the distance between the bulb and the wall, we have found the object and image distances. Finally, we have everything we need to find the lens displacement. It is pretty simple, once you know the formulas and the values. The key is to carefully follow each step and ensure we do not mess up on the calculations. The main idea is to understand that the lens's displacement changes the object and image distances, which, in turn, affects the magnification. It's all connected!

Summary and Conclusion

Alright, guys, let's wrap things up! We started with a clear experimental setup involving a lightbulb, a converging lens, and a wall. We needed to figure out how much we had to move the lens to get a three-times-magnified image of the lightbulb's filament on the wall. By using the lens equation and the magnification formula, we broke down the problem into manageable steps. We found the object distance (0.5 m) and the image distance (1.5 m). The displacement of the lens is related to the initial and final positions needed to get the desired magnification. This means that the lens has to be moved to focus the image at a specific distance from the bulb. This is a great example of how understanding the physics of lenses and light can help us in real-world scenarios. Physics can be fun!

In short, by applying the lens equation and the magnification formula, we've successfully navigated this problem. It's a testament to how physics principles can demystify seemingly complex setups. Keep practicing, and soon, these concepts will be second nature to you!