Problem 66 - Physics

Question

Two converging lenses \(\left(f_{1}=9.00 \mathrm{~cm}\right.\) and \(\left.f_{2}=6.00 \mathrm{~cm}\right)\) are separated by \(18.0 \mathrm{~cm} .\) The lens on the left has the longer focal length. An object stands \(12.0 \mathrm{~cm}\) to the left of the lefthand lens in the combination. (a) Locate the final image relative to the lens on the right. (b) Obtain the overall magnification. (c) Is the final image real or virtual? With respect to the original object, is the final image (d) upright or inverted and is it (e) larger or smaller?

Step-by-Step Solution

Answer

The final image is 11.25 cm to the right of the second lens, real, inverted, and smaller.

Step 1: Determine Image Position for First Lens

We will use the lens formula for the first lens to find the image position, where \( f_1 = 9.00 \text{ cm} \) and the object distance \( d_o = -12.00 \text{ cm} \). The lens formula is given by \( \frac{1}{f} = \frac{1}{d_i} + \frac{1}{d_o} \). Solving for \( d_i \), we have: \[ \frac{1}{d_i} = \frac{1}{f_1} - \frac{1}{d_o} = \frac{1}{9} - \frac{1}{-12} = \frac{12 + 9}{108} = \frac{21}{108} \] \( d_i = \frac{108}{21} = 5.14 \text{ cm} \) (positive, to the right of the first lens).

Step 2: Find Object Distance for Second Lens

The image from the first lens becomes the object for the second lens. The distance between the lenses is \( 18.0 \text{ cm} \), and the image is located \( 5.14 \text{ cm} \) to the right of the first lens, meaning the object distance \( d_o' \) for the second lens is \( 18.0 \text{ cm} - 5.14 \text{ cm} = 12.86 \text{ cm} \).

Step 3: Calculate Image Position for Second Lens

Using the lens formula for the second lens with \( f_2 = 6.00 \text{ cm} \) and \( d_o' = 12.86 \text{ cm} \): \[ \frac{1}{f_2} = \frac{1}{d_i'} + \frac{1}{d_o'} \] \[ \frac{1}{6} = \frac{1}{d_i'} + \frac{1}{12.86} \] Solving for \( d_i' \), we get: \[ \frac{1}{d_i'} = \frac{1}{6} - \frac{1}{12.86} = \frac{12.86 - 6}{77.16} = \frac{6.86}{77.16} \] \( d_i' = \frac{77.16}{6.86} = 11.25 \text{ cm} \).

Step 4: Determine Overall Magnification

Magnification for each lens is given by \( m_i = -\frac{d_i}{d_o} \). Calculate separately and then multiply them. For the first lens: \( m_1 = -\frac{5.14}{-12} = 0.428 \). For the second lens: \( m_2 = -\frac{11.25}{12.86} = -0.875 \). The total magnification \( M = m_1 \times m_2 = 0.428 \times -0.875 = -0.3745 \).

Step 5: Analyze Characteristics of the Final Image

For the final image: \( d_i' = 11.25 \text{ cm} \) is positive, indicating a real image located \( 11.25 \text{ cm} \) to the right of the second lens. Since the total magnification \( M = -0.3745 \) is negative, the final image is inverted relative to the object. The absolute value of \( M \) is less than 1, so the image is smaller than the object.

Key Concepts

Converging Lenses

Converging lenses, also known as convex lenses, are thicker in the middle than at the edges. These lenses have the ability to bend parallel rays of light such that they meet at a point known as the focal point. As light passes through a converging lens, it is refracted towards the center, causing the light rays to converge. This property makes them invaluable in devices like cameras, glasses, and microscopes, allowing these devices to focus light to form clear images.

Converging lenses come in various forms, such as double convex, plano-convex, and positive meniscus. Each type varies in its specific shape and focal properties, serving different applications based on their focal length and curvature.

In practical applications, the position of the focal point, determined by the lens's curvature and material, is crucial in shaping how images form and their characteristics.

Focal Length

The focal length of a lens is a measure of how strongly the lens converges or diverges light. It is the distance from the lens to its focal point where the light rays converge to form an image. For converging lenses, the focal length is positive, which means the light is directed to converge.

Focal length plays a crucial role in determining how large or small an image will appear. A lens with a shorter focal length brings light to focus more quickly, thus forming smaller images that provide a wider field of view. Conversely, a longer focal length converges light more gradually, magnifying details but offering a narrower view.

In the given exercise, the two lenses had focal lengths of 9 cm and 6 cm, respectively. These differences are crucial when calculating how and where images form, impacting the final magnification and orientation of the image.

Image Formation

Image formation occurs when light rays converge after passing through lenses, creating a visible likeness of the object. This concept relies heavily on the principles of refraction, wherein the bending of light at varying angles carries information about the object's shape and position.

The lens formula, \( \frac{1}{f} = \frac{1}{d_i} + \frac{1}{d_o} \), helps to calculate image distance \( d_i \) for converging lenses. Here, \( f \) stands for the focal length, and \( d_o \) is the object distance. Understanding this formula is crucial for determining where an image will appear in relation to the lens.

Through this exercise, you can see the strategic use of the lens formula to determine image positions, starting from the first lens to the second, illustrating the sequential process of image formation in optical systems.

Magnification

Magnification is a measure of how much larger or smaller an image is compared to the actual object. For lenses, magnification is determined by the ratio of the image distance to the object distance, expressed as \( m = -\frac{d_i}{d_o} \).

The overall magnification of a system with multiple lenses is the product of the magnifications of each lens involved. A positive magnification indicates an upright image, while a negative magnification indicates an inverted image.

In the exercise, understanding magnification was key to interpreting the image's size and orientation. By calculating the magnification for each lens and multiplying them, we discovered that the final image was inverted and smaller than the original object. This insight helps to predict how an assembly of lenses will alter an image's appearance, which is crucial in designing optical instruments.

Real and Virtual Images

Real and virtual images are two types of images formed by lenses, differentiated by the path of light rays. A real image is created when the refracted light rays actually converge at a point. Such images can be captured on a screen because the light physically sticks together at the image location.

Contrarily, a virtual image forms when the rays of light only appear to originate from a common point behind the lens; in reality, they do not actually meet. These images cannot be projected onto a screen, as seen in magnifying glasses where the image appears larger and upright.

In the context of the exercise, the final image was determined to be real because the light rays intersected on the same side as the actual image formation, signified by the positive image distance value. Understanding whether an image is real or virtual is critical for applications requiring precise image placement, such as in photography and microscopy.