Chapter 4: Problem 15
An object \(2.50 \mathrm{~cm}\) high is located \(15.0 \mathrm{~cm}\) in front of a lens of \(+5.0 \mathrm{~cm}\) focal length. A lens with a focal length of \(-12.0 \mathrm{~cm}\) is placed \(2.50 \mathrm{~cm}\) beyond this converging lens. Find \((a)\) the position and \((b)\) the size of the final image.
Short Answer
Step by step solution
Calculate the Image Position by the First Lens
Determine the Object Distance for the Second Lens
Calculate the Final Image Position by the Second Lens
Calculate the Magnification of Each Lens
Calculate the Final Image Size
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Lens Formula
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}. \]This fundamental equation applies to both converging and diverging lenses. By rearranging the lens formula, you can solve for any of the values if the other two are known. It's important to remember:
- For converging lenses, the focal length \(f\) is positive.
- For diverging lenses, the focal length \(f\) is negative.
- When the image distance \(d_i\) is positive, the image is real and formed on the opposite side of the object.
- When \(d_i\) is negative, the image is virtual and appears on the same side as the object.
Image Position
By using the lens formula, we calculated the image distance \(d_{i1} = 7.5 \mathrm{~cm}\), indicating that the image is formed behind the lens. For the second diverging lens, the object distance is derived from this image position, acting as a virtual object, resulting in a negative value.
The image position relative to a lens helps determine characteristics like size, orientation, and type (real or virtual). Adjusting the object or the type of lens will influence where this image forms.
Focal Length
In the exercise, the first lens has a focal length of \(+5.0 \mathrm{~cm}\), indicating it is a converging lens. It brings light rays to a focus, creating a real image. On the other hand, the second lens has a focal length of \(-12.0 \mathrm{~cm}\), classifying it as a diverging lens. Such lenses spread light rays outward, usually forming virtual images unless accompanied by other lenses.
Understanding focal length is crucial in lens systems, as it helps predict how lenses will act on incoming light, and it aids in calculating image distances and sizes in optical setups.
Magnification
\[ m = -\frac{d_i}{d_o}, \]where \(m\) represents magnification, \(d_i\) is the image distance, and \(d_o\) is the object distance.
Magnification indicates not only the size of the image relative to the object but also its orientation. A negative magnification value indicates an inverted image, while a positive value signifies an upright image.
- The first lens gave a magnification of \(-0.5\), suggesting the image was half the size of the object and inverted.
- The second lens, a diverging lens, significantly enhanced the size with a magnification of \(6\), resulting in a much larger image.
Virtual Object
In the exercise, the first lens creates an image that acts as a virtual object for the second lens. Virtual objects are considered to have negative distances, as they are not physically located on the same side as the camera or eye observing the second lens.
This concept allows lenses to manipulate light to achieve complex image formations, crucial in intricate optical designs like microscopes and telescopes.