Question

The object (upright arrow) in the following system has a height of \(2.5 \mathrm{~cm}\) and is placed \(5.0 \mathrm{~cm}\) away from a converging (convex) lens with a focal length of \(3.0 \mathrm{~cm}\). What is the magnification of the image? Is the image upright or inverted? Confirm your answers by ray tracing.

Short answer

Is the image upright or inverted? Answer: The magnification of the image is 1.5, and the image formed is inverted.

Step-by-step solution

Use the lens formula

Recall the lens formula: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). We are given the object distance \(d_o = 5.0\mathrm{~cm}\) and the focal length \(f = 3.0\mathrm{~cm}\). Using the lens formula, we can calculate the image distance \(d_i\).

Calculate the image distance

Plug in the values into the lens formula: \(\frac{1}{3.0} = \frac{1}{5.0} + \frac{1}{d_i}\) Now solve for \(d_i\): \(\frac{1}{d_i} = \frac{1}{3.0} - \frac{1}{5.0}\) \(\frac{1}{d_i} = \frac{2}{15}\) \(d_i = 7.5\mathrm{~cm}\) The image distance \(d_i\) is \(7.5\mathrm{~cm}\) from the lens.

Calculate the magnification

Calculate the magnification using the formula \(M = -\frac{d_i}{d_o}\): \(M = -\frac{7.5}{5.0}\) \(M = -1.5\) The magnification of the image is \(1.5\). Since the magnification is negative, the image is inverted.

Confirm the answers using ray tracing

To perform ray tracing, draw the following three rays: 1. A ray parallel to the principal axis that passes through the object and the lens. 2. A ray passing through the object, the lens, and the focal point on the object side. 3. A ray passing through the object and the center of the lens. These three rays should intersect at a single point on the other side of the lens, forming an inverted image with a magnification of \(1.5\).

Key concepts

Converging Lens

Understanding the nature and function of a converging lens is fundamental when exploring optics. A converging or 'convex' lens is thicker at the center than at the edges and is designed to bend light rays inward, such that they converge at a point known as the 'focal point'.

The focal length of a converging lens, represented by the symbol 'f', is the distance from the center of the lens to the focal point. It is a key factor in determining where an image will form and how large it will be. For instance, in the given problem, the converging lens has a focal length of 3.0 cm, meaning that parallel rays of light would focus at 3.0 cm from the lens.

The behavior and properties of converging lenses are described by specific rules and equations. The lens formula, in particular, links the object distance (distance from the object to the lens), the image distance (distance from the lens to the image), and the focal length.

Image Distance

The image distance, denoted as 'di', is a vital concept in lens optics as it tells us where the image formed by a lens will be located relative to the lens. In the context of the lens formula \(\frac{1}{f} = \frac{1}{do} + \frac{1}{di}\), 'di' can be calculated provided one knows the object distance 'do' and the lens's focal length 'f'.

In our exercise, by rearranging and solving the lens formula, we found that the image distance for the given object and lens was 7.5 cm. This indicated that the resultant image forms 7.5 cm away from the lens on the other side. The positive value typically suggests that the image is real and situated on the opposite side of the lens from the object. If 'di' had been negative, this would imply an imaginary image on the same side as the object.

Ray Tracing

Ray tracing is a graphical method used to determine the position, size, orientation, and type of image produced by a lens. It involves drawing specific rays that predict the path of light as it interacts with the lens. These rays are based on fundamental rules of refraction and reflection, giving us a visual understand of how images are formed.

In our example, we would trace three types of rays: one that runs parallel to the principal axis and then passes through the focal point after refraction by the lens; one that passes through the center of the lens and continues in a straight line; and the last ray that heads toward the focal point on the object side before and exits the lens parallel to the principal axis after refraction. Where these rays intersect on the opposite side of the lens establishes the location of the image. By following these steps, we confirm that the image is indeed inverted, as indicated by the negative magnification found in Step 3.

Optical Magnification

Optical magnification of a lens describes how much larger or smaller the image appears compared to the actual object. It is given by the formula \(M = -\frac{di}{do}\) where 'M' is the magnification, 'di' is the image distance, and 'do' is the object distance. A positive magnification implies an upright image, whereas a negative magnification means the image is inverted.

From the calculation in the given exercise, the magnification 'M' was found to be -1.5. This value informs us that the image is 1.5 times the size of the object and is inverted. The negative sign is critical; it provides an immediate understanding of the image orientation without the need for ray tracing, although ray tracing serves as a valuable confirmation of these calculated properties.