(for A Lens) Given F=+12cm, Calculate Di, L, And M When Do Isa. 36cm,b. 24cm,c. 18cm
Understanding the fundamental principles of lenses is crucial in optics, especially when calculating image distances, magnification, and object distances for various lens configurations. In this article, we will explore how to determine the image distance (Di), the object distance (L), and the magnification (M) for a converging lens with a focal length (F) of +12 cm, given different object distances: 36 cm, 24 cm, and 18 cm. These calculations are essential for applications ranging from photography to optical instrument design and are based on the lens formula and magnification concepts.
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Understanding Lens Formula and Magnification
Before diving into specific calculations, it is important to review the primary equations used in lens optics.
Lens Formula
The lens formula relates the focal length (F), the object distance (L), and the image distance (Di):\[ \frac{1}{F} = \frac{1}{L} + \frac{1}{D_i} \]
- F is the focal length of the lens (positive for converging lenses).
- L is the object distance from the lens.
- Di is the image distance from the lens.
Magnification (M)
Magnification describes how large the image appears relative to the object:
\[ M = \frac{hi}{ho} = \frac{D_i}{L} \]
- \( hi \) and \( ho \) are the heights of the image and object, respectively.
- Alternatively, it can be expressed as:
\[ M = \frac{D_i}{L} \]
- Positive magnification indicates an upright image, while negative indicates an inverted image.
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Given Data and Approach
In our scenario:
- Focal length, \( F = +12\,cm \)
- Object distances, \( L = 36\,cm, 24\,cm, 18\,cm \)
Our goal is to find:
- Image distance, \( D_i \)
- Magnification, \( M \)
- Confirm the nature of the image (real or virtual, upright or inverted)
The calculation process involves applying the lens formula to each object distance, then calculating the magnification.
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Calculations for Each Object Distance
Case A: Object Distance \( L = 36\,cm \)
Using the lens formula:
\[ \frac{1}{F} = \frac{1}{L} + \frac{1}{D_i} \]
Substituting the known values:
\[ \frac{1}{12} = \frac{1}{36} + \frac{1}{D_i} \]
Rearranged to solve for \( D_i \):
\[ \frac{1}{D_i} = \frac{1}{12} - \frac{1}{36} \]
Calculating:
\[ \frac{1}{D_i} = \frac{3}{36} - \frac{1}{36} = \frac{2}{36} = \frac{1}{18} \]
Thus,
\[ D_i = 18\,cm \]
Next, calculate the magnification:
\[ M = \frac{D_i}{L} = \frac{18}{36} = 0.5 \]
Since \( M \) is positive, the image is upright, and because \( D_i \) is positive, the image is real and formed on the same side as the outgoing light.
Summary:
- \( D_i = 18\,cm \)
- \( M = 0.5 \)
- Image is real, inverted or upright? Because the magnification is positive and the image distance is positive, the image is upright and real.
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Case B: Object Distance \( L = 24\,cm \)
Applying the lens formula again:
\[ \frac{1}{12} = \frac{1}{24} + \frac{1}{D_i} \]
Solving for \( D_i \):
\[ \frac{1}{D_i} = \frac{1}{12} - \frac{1}{24} = \frac{2}{24} - \frac{1}{24} = \frac{1}{24} \]
Therefore,
\[ D_i = 24\,cm \]
Calculating magnification:
\[ M = \frac{D_i}{L} = \frac{24}{24} = 1 \]
Here, the image is of equal size to the object, upright, and real.
Summary:
- \( D_i = 24\,cm \)
- \( M = 1 \)
- Image is real, upright, same size.
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Case C: Object Distance \( L = 18\,cm \)
Applying the lens formula:
\[ \frac{1}{12} = \frac{1}{18} + \frac{1}{D_i} \]
Rearranged:
\[ \frac{1}{D_i} = \frac{1}{12} - \frac{1}{18} \]
Calculating:
\[ \frac{1}{D_i} = \frac{3}{36} - \frac{2}{36} = \frac{1}{36} \]
Thus,
\[ D_i = 36\,cm \]
Magnification:
\[ M = \frac{D_i}{L} = \frac{36}{18} = 2 \]
Positive magnification indicates an upright image, and the image is real and magnified.
Summary:
- \( D_i = 36\,cm \)
- \( M = 2 \)
- Image is real, upright, magnified.
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Summary of Results
| Object Distance \( L \) | Image Distance \( D_i \) | Magnification \( M \) | Nature of Image |
|---------------------------|--------------------------|-----------------------|------------------------------|
| 36 cm | 18 cm | 0.5 | Real, upright, diminished |
| 24 cm | 24 cm | 1 | Real, upright, same size |
| 18 cm | 36 cm | 2 | Real, upright, magnified |
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Interpreting the Results
These calculations reveal how the object distance influences the image properties:
- When the object is far from the lens (36 cm), the image is real, upright, and smaller.
- At a closer distance (24 cm), the image maintains the same size.
- When the object is very close (18 cm), the image is magnified.
It is noteworthy that all images are real and upright in these cases because the object distances are greater than the focal length, and the calculated image distances are positive, consistent with real image formation in converging lenses.
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Practical Implications and Applications
Understanding these calculations is essential in designing optical devices:
- Camera lenses: Adjusting object distances affects image size and clarity.
- Magnifying glasses: Positioning the object within the focal length produces virtual, magnified, upright images.
- Projectors and microscopes: Precise calculations ensure accurate image formation.
Additionally, knowing how to compute these parameters helps in troubleshooting and optimizing optical systems.
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Conclusion
Calculating image and object distances and magnifications for a converging lens involves straightforward application of the lens formula and magnification equations. For a lens with a focal length of +12 cm, different object distances yield various image characteristics, from real, reduced images to magnified ones. Mastery of these calculations is fundamental for students and professionals working in optics, photography, and related fields, enabling them to predict and manipulate image formation effectively.
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Remember: Always verify the sign conventions—positive for real images and converging lenses, negative for virtual images and diverging lenses—to interpret results correctly.
