Question

Two identical thin convex lenses, each of focal length $f$, are separated by a distance $d=2.5f$. An object is placed in front of the first lens at a distance $d_{\mathrm{a},1}=2f.$ a) Calculate the position of the final image of an object through the system of lenses. b) Calculate the total transverse magnification of the system. c) Draw the ray diagram for this system and show the final image. d) Describe the final image (real or virtual, erect or inverted, larger or smaller) in relation to the initial object.

Short answer

Based on the calculations, the final image formed by the system of lenses is virtual, inverted, and 2 times larger than the initial object. To draw the ray diagram, follow these steps: 1. Draw the two lenses, labeled Lens 1 and Lens 2, vertically with the distance of 2.5f between them. 2. Place the object at a distance of 2f to the left of Lens 1. 3. Draw the image formed by Lens 1 at a distance of 2f to the right of Lens 1. This image will act as the object for Lens 2. 4. Draw the final image at a distance of f to the left of Lens 2. This final image is virtual, inverted, and 2 times larger than the initial object. 5. Draw at least three principal rays from the object to the final image: a. Ray 1: Parallel to the principal axis from the object, passing through the focal point after refraction through Lens 1, then parallel to the principal axis after refraction through Lens 2. b. Ray 2: Passing through the center of Lens 1 without deviation, then passing through the center of Lens 2 without deviation. c. Ray 3: Passing through the focal point of Lens 1, then parallel to the principal axis after refraction through Lens 1, and finally passing through the focal point of Lens 2. 6. Verify that all three rays intersect at the final image location. The completed ray diagram will show the system of lenses, object, and image locations, as well as the principle rays forming the final image.

Step-by-step solution

Calculate position of image formed by Lens 1

From the given information, we know that the object distance for Lens 1 is $d_{a,1}=2f$. We can use the lens equation $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$, where $d_o$ represents the object distance, $d_i$ is the image distance, and $f$ is the focal length of the lens. For Lens 1: $\frac{1}{f}=\frac{1}{2f}+\frac{1}{d_{i,1}}$ Solving for $d_{i,1}$ $d_{i,1}=2f$

Calculate object distance for Lens 2

The image formed by Lens 1 acts as the object for Lens 2. The object distance for Lens 2 ($d_{a,2}$) is the distance between the image formed by Lens 1 and Lens 2 itself: $d_{a,2}=d-d_{i,1}$ $d_{a,2}=2.5f-2f=0.5f$

Calculate the position of the final image formed by the system

Now we can find the image formed by Lens 2, using the lens equation again: For Lens 2: $\frac{1}{f}=\frac{1}{d_{a,2}}+\frac{1}{d_{i,2}}$ Substitute $d_{a,2}=0.5f$: $\frac{1}{f}=\frac{2}{f}+\frac{1}{d_{i,2}}$ Solving for $d_{i,2}$: $d_{i,2}=-f$

Calculate the total magnification of the system

To find the total magnification, we will multiply the individual magnifications of the lenses. The magnification for the lens is given by $M=-\frac{d_i}{d_o}$. For Lens 1: $M_1=-\frac{d_{i,1}}{d_{a,1}}=-\frac{2f}{2f}=-1$ For Lens 2: $M_2=-\frac{d_{i,2}}{d_{a,2}}=-\frac{-f}{0.5f}=2$ The total magnification is $M_{total}=M_1\times M_2=(-1)\times (2)=-2$

Describe the final image in relation to the object

Since the total magnification is negative, the final image is inverted compared to the object. The magnitude of the magnification is 2, so the image is 2 times larger than the initial object. The image distance $d_{i,2}=-f$ for the second lens is negative, which means the final image is virtual. So, the final image is virtual, inverted, and 2 times larger than the initial object. For the ray diagram, draw the system of lenses, object, and image locations as determined by our calculations, and show at least three principal rays passing through the lens system to form the final image.

Key concepts

Lens Equation

The lens equation is a fundamental tool in the study of optics and is crucial for understanding how thin convex lenses form images. It is given by the formula: $$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$where $f$ is the focal length of the lens, $d_o$ is the distance of the object from the lens, and $d_i$ is the distance of the image from the lens. By manipulating this equation, we can solve for any of the three variables if the other two are known, allowing us to predict the position of the image formed by a lens. For example, if an object is placed at twice the focal length from a lens ($d_o=2f$), the image will form at twice the focal length on the other side ($d_i=2f$), as shown in the step-by-step solution.

Image Magnification

Image magnification tells us how much larger or smaller the image is compared to the original object. It is defined as the ratio of the image distance to the object distance, with a negative sign for a real image formed by a convex lens:$$M=-\frac{d_i}{d_o}$$In our example, the first lens produces an image that is the same size as the object but inverted, as indicated by the magnification of -1. The second lens then magnifies the image by a factor of 2, resulting in a total magnification of -2, meaning the image is twice the size of the object and inverted.

Ray Diagrams

Ray diagrams illustrate the path of light as it passes through optical systems like lens and mirror assemblies. These diagrams are invaluable for visualizing image formation. To construct a ray diagram, you typically draw at least three principal rays emanating from a point on an object: one parallel to the lens axis, which refracts through the focal point on the other side; the second passing straight through the center of the lens without deviation; and the third aimed towards the focal point on the object's side, emerging parallel to the axis after refraction. By extending these rays, they intersect at the image point, providing a graphical representation of both the location and size of the image formed by the lens system.

Focal Length

The focal length of a lens is the distance from the lens to its focus, where rays of light that are parallel to the principal axis converge after passing through the lens. A longer focal length results in a less powerful lens that forms images closer to the lens itself for a given object distance. In the problem we're examining, each lens has a focal length $f$, and by knowing the focal lengths and the separation between the lenses, we can calculate the characteristics of the image produced by the combination of the lenses, including its position and size.

Virtual and Real Images

Images formed by lenses are categorized as real or virtual based on how the light rays converge. A real image is formed when rays of light actually converge at an image point, and it can be projected onto a screen. In contrast, a virtual image appears to be at a location where the light rays do not really converge; instead, they appear to diverge from this point. In our example, the final image formed by the lens system is virtual, as evidenced by the negative image distance for the second lens ($d_{i,2}=-f$). This means that the image cannot be projected onto a screen and will appear on the same side of the lens as the object, larger and inverted relative to the original object.