A Diverging Lens Has A Focal Length Of Magnitude 13 Cm . At What Object Distance Will The Magnification is a question that delves into the fundamental principles of optics, specifically concerning diverging lenses. Understanding the relationship between focal length, object distance, and magnification is essential for students, educators, and professionals working with optical devices. In this article, we will explore the concepts behind diverging lenses, derive the relevant formulas, and work through the problem step-by-step to determine the object distance for a given magnification.
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Understanding Diverging Lenses
What Is a Diverging Lens?
A diverging lens, also known as a concave lens, is a type of optical device that causes parallel rays of light to spread apart or diverge after passing through it. These lenses are thinner at the center than at the edges and are commonly used in applications like eyeglasses for myopia correction, optical instruments, and laser beam expanders.Properties of Diverging Lenses
- Focal length (f): Negative for diverging lenses, indicating the direction of the focus.
- Image formation: Produces virtual, erect, and diminished images for objects placed beyond the focal point.
- Refraction: The bending of light occurs inward, causing divergence.
Fundamental Concepts and Formulas
Lens Formula
The relationship between the object distance (u), the image distance (v), and the focal length (f) is given by the lens formula:\[
\frac{1}{f} = \frac{1}{v} + \frac{1}{u}
\]
Where:
- \(f\) is the focal length of the lens.
- \(v\) is the image distance from the lens.
- \(u\) is the object distance from the lens (measured from the lens to the object).
Note: Sign conventions are vital:
- For diverging lenses, \(f\) is negative.
- Object and image distances are positive if they are real and on the same side as the incoming light.
Magnification Formula
Magnification (M) describes how much larger or smaller the image appears compared to the object:
\[
M = \frac{v}{u}
\]
- Positive magnification indicates an erect image.
- Negative magnification indicates an inverted image.
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Given Data and Objective
The problem states:
- The focal length magnitude of the diverging lens is 13 cm, so \(f = -13\,cm\).
- The goal is to find the object distance \(u\) when the magnification \(M\) is known or specified.
However, in the original problem statement, the specific value of magnification is not provided, implying that the task involves understanding how to find \(u\) for a given \(M\). For the purpose of this explanation, assume we are asked to determine the object distance for a specific magnification value, say \(M = 0.5\), but the process applies to any value.
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Deriving the Relationship Between Object Distance, Focal Length, and Magnification
Step 1: Express \(v\) in terms of \(u\) and \(M\)
From the magnification formula:\[
v = M \times u
\]
Since the lens formula involves \(v\), substitute \(v = M u\):
\[
\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{M u} + \frac{1}{u}
\]
Step 2: Simplify to find \(u\)
Combine the right side over a common denominator:\[
\frac{1}{f} = \frac{1 + M}{M u}
\]
Rearranged:
\[
u = \frac{(1 + M)}{(1/f) \times M}
\]
Given that \(f = -13\,cm\):
\[
u = \frac{(1 + M)}{(1/ -13) \times M} = \frac{(1 + M) \times (-13)}{M}
\]
Simplify:
\[
u = -13 \times \frac{(1 + M)}{M}
\]
This formula allows us to compute the object distance \(u\) for any given magnification \(M\).
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Calculating Object Distance for a Specific Magnification
Suppose we want to find the object distance when the magnification \(M = 0.5\).
\[
u = -13 \times \frac{1 + 0.5}{0.5} = -13 \times \frac{1.5}{0.5} = -13 \times 3 = -39\,cm
\]
The negative sign indicates that the object is on the same side of the lens as the incoming light, which is the usual convention for real objects.
Interpretation:
- The object is located 39 cm in front of the diverging lens.
- The image formed will be virtual, upright, and diminished, consistent with the properties of diverging lenses.
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General Approach to Find Object Distance for Any Magnification
To determine the object distance for a diverging lens with a known focal length and desired magnification, follow these steps:
- Identify the focal length \(f\) (negative for diverging lenses).
- Determine the desired magnification \(M\).
- Use the formula: \[ u = -f \times \frac{1 + M}{M} \]
- Calculate \(u\) to find the object distance.
Note: The sign conventions are critical. Negative \(u\) indicates the object is in front of the lens, consistent with real object placement.
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Practical Applications and Examples
Example 1: Magnification of 1 (Equal Size Image)
Suppose the magnification is 1, meaning the image is the same size as the object:\[
u = -13 \times \frac{1 + 1}{1} = -13 \times 2 = -26\,cm
\]
The object should be placed 26 cm in front of the diverging lens to produce a virtual, upright, same-sized image.
Example 2: Magnification of 2 (Enlarged Image)
For a magnification of 2:\[
u = -13 \times \frac{1 + 2}{2} = -13 \times \frac{3}{2} = -19.5\,cm
\]
The object must be approximately 19.5 cm from the lens.
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Conclusion
Understanding the relationships between focal length, object distance, and magnification is essential in optical physics, especially when working with diverging lenses. By applying the lens formula alongside the magnification equation, one can precisely determine the object distance for any desired magnification. These calculations are crucial not only in theoretical physics but also in designing optical devices, correcting vision problems, and various scientific applications.
Remember, sign conventions and careful substitution are vital to arriving at correct solutions. Whether you're an educator guiding students or an engineer designing optical systems, mastering these fundamental concepts ensures accurate and effective use of diverging lenses in a wide range of contexts.
