Lens Problems With Solutions Magnification And Focal Length Explained
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In the realm of optics, understanding how lenses manipulate light to form images is crucial. This article delves into two intriguing problems involving lenses, magnification, and focal length, providing a step-by-step approach to solving them. Whether you're a student grappling with physics concepts or simply curious about the science behind lenses, this guide will illuminate the principles at play. Prepare to embark on a journey through the fascinating world of optics, where we unravel the mysteries of image formation and lens behavior. By the end of this exploration, you'll have a solid grasp of how to tackle similar problems and a deeper appreciation for the elegant laws governing light and lenses.
Problem 1 Determining Lens Distance and Focal Length
Understanding the Problem Statement
The first problem presents a scenario involving a lamp filament, a screen, and a converging lens. The filament is positioned 80 cm away from the screen, and the converging lens forms a magnified image of the filament on the screen, with a magnification factor of three. The challenge lies in determining two key parameters the distance of the lens from the filament and the focal length of the lens. This problem encapsulates fundamental concepts in geometric optics, requiring a thoughtful application of lens equations and magnification principles.
Applying the Lens Formula and Magnification Equation
To solve this problem, we'll employ two fundamental equations in geometric optics the lens formula and the magnification equation. The lens formula, expressed as 1/f = 1/v - 1/u, relates the focal length (f) of the lens to the object distance (u) and the image distance (v). The magnification equation, given by m = v/u, connects the magnification (m) to the image and object distances. By skillfully manipulating these equations and incorporating the given information, we can unravel the unknowns.
Step-by-Step Solution
Let's embark on a step-by-step solution to determine the lens distance and focal length. We know that the distance between the filament (object) and the screen (image) is 80 cm. This can be expressed as u + v = 80 cm, where u is the object distance and v is the image distance. We are also given that the magnification (m) is 3, which means v/u = 3. Now, we have a system of two equations with two unknowns, allowing us to solve for u and v.
From the magnification equation, we can express v as 3u. Substituting this into the equation u + v = 80 cm, we get u + 3u = 80 cm, which simplifies to 4u = 80 cm. Solving for u, we find that the object distance u is 20 cm. Now, we can find the image distance v by substituting u back into the equation v = 3u, giving us v = 3 * 20 cm = 60 cm. Thus, the lens is 20 cm away from the filament and 60 cm away from the screen.
Now that we have the object and image distances, we can determine the focal length (f) using the lens formula 1/f = 1/v - 1/u. Plugging in the values for u and v, we get 1/f = 1/60 cm - 1/20 cm. To simplify this, we find a common denominator, resulting in 1/f = (1 - 3) / 60 cm = -2 / 60 cm. Taking the reciprocal of both sides, we find f = -30 cm. The negative sign indicates that the lens is a diverging lens. However, since the problem states that the lens is converging, there seems to be an error in the initial problem statement or the magnification value. Assuming the magnification is positive (m = 3), the image is inverted, and the correct lens formula should be 1/f = 1/v + 1/u. In this case, 1/f = 1/60 cm + 1/20 cm = (1 + 3) / 60 cm = 4 / 60 cm. Taking the reciprocal, we get f = 15 cm. Therefore, the focal length of the lens is 15 cm.
Summarizing the Solution
In summary, we determined that the lens is positioned 20 cm away from the filament and 60 cm away from the screen. The focal length of the converging lens is calculated to be 15 cm. This solution demonstrates the power of applying lens equations and magnification principles to solve problems in geometric optics. By carefully analyzing the given information and employing the appropriate formulas, we can successfully determine key parameters of lens systems.
Problem 2 Image Height and Lens Placement
Problem Overview
The second problem shifts our focus to an object of a specific height placed at a certain distance from a lens. An object 2 cm tall is positioned on the principal axis of a converging lens with a focal length of 10 cm. The problem requires us to explore different scenarios by placing the object at varying distances from the lens 15 cm, 20 cm, and 30 cm and determine the image position, image nature (real or virtual), and image height in each case. This problem provides a practical application of the lens formula and magnification concepts, allowing us to visualize how image characteristics change with object distance.
Solving for Image Position, Nature, and Height
To tackle this problem, we'll again rely on the lens formula (1/f = 1/v - 1/u) and the magnification equation (m = v/u = h'/h), where h is the object height and h' is the image height. By applying these equations to each scenario with different object distances, we can systematically determine the image position, nature, and height. Let's dive into each case individually.
Case 1 Object Distance of 15 cm
In the first case, the object is placed 15 cm away from the lens (u = 15 cm). The focal length of the lens is given as 10 cm. Using the lens formula, we have 1/10 cm = 1/v - 1/15 cm. Rearranging the equation to solve for v, we get 1/v = 1/10 cm + 1/15 cm. Finding a common denominator, we have 1/v = (3 + 2) / 30 cm = 5 / 30 cm. Taking the reciprocal, we find v = 6 cm. The positive value of v indicates that the image is real and formed on the opposite side of the lens.
Now, we can calculate the magnification using m = v/u = 6 cm / 15 cm = 0.4. The magnification is positive, indicating that the image is upright. To find the image height (h'), we use the magnification equation m = h'/h. Given the object height (h) is 2 cm, we have 0.4 = h' / 2 cm. Solving for h', we get h' = 0.4 * 2 cm = 0.8 cm. Therefore, in this case, the image is real, upright, and 0.8 cm tall.
Case 2 Object Distance of 20 cm
Next, let's consider the case where the object is placed 20 cm away from the lens (u = 20 cm). Again, the focal length is 10 cm. Applying the lens formula, we have 1/10 cm = 1/v - 1/20 cm. Rearranging to solve for v, we get 1/v = 1/10 cm + 1/20 cm = (2 + 1) / 20 cm = 3 / 20 cm. Taking the reciprocal, we find v = 20 / 3 cm ≈ 6.67 cm. The positive value of v indicates a real image.
Calculating the magnification, we have m = v/u = (20 / 3 cm) / 20 cm = 1 / 3 ≈ 0.33. The magnification is positive, so the image is upright. Using the magnification equation to find the image height, we have 0.33 = h' / 2 cm. Solving for h', we get h' = 0.33 * 2 cm ≈ 0.66 cm. Thus, when the object is 20 cm from the lens, the image is real, upright, and approximately 0.66 cm tall.
Case 3 Object Distance of 30 cm
Finally, let's analyze the scenario where the object is placed 30 cm from the lens (u = 30 cm). The focal length remains 10 cm. Using the lens formula, we have 1/10 cm = 1/v - 1/30 cm. Rearranging to solve for v, we get 1/v = 1/10 cm + 1/30 cm = (3 + 1) / 30 cm = 4 / 30 cm. Taking the reciprocal, we find v = 30 / 4 cm = 7.5 cm. The positive value of v indicates a real image.
Calculating the magnification, we have m = v/u = 7.5 cm / 30 cm = 0.25. The magnification is positive, so the image is upright. Using the magnification equation to find the image height, we have 0.25 = h' / 2 cm. Solving for h', we get h' = 0.25 * 2 cm = 0.5 cm. Therefore, when the object is 30 cm from the lens, the image is real, upright, and 0.5 cm tall.
Summarizing the Results
In summary, we have analyzed three cases with different object distances. When the object is 15 cm from the lens, the image is real, upright, and 0.8 cm tall. When the object is 20 cm from the lens, the image is real, upright, and approximately 0.66 cm tall. And when the object is 30 cm from the lens, the image is real, upright, and 0.5 cm tall. This analysis demonstrates how the image characteristics change as the object distance varies, highlighting the versatility of lenses in image formation.
Conclusion Mastering Lens Equations for Optical Problem Solving
Through these two comprehensive problems, we've delved into the fascinating world of lenses, magnification, and focal length. By applying the lens formula and magnification equation, we've successfully determined key parameters of lens systems, such as lens distance, focal length, image position, image nature, and image height. These problem-solving techniques are invaluable for anyone seeking to understand the principles of optics and lens behavior. As you continue your exploration of physics, remember that a solid grasp of fundamental equations and a systematic approach are your greatest allies in unraveling the mysteries of the universe.
