You are given a thin converging lens with a focal length of \(10 \mathrm{~cm}\). a) An object is placed at \(6 \mathrm{~cm}\) from the lens. Draw a ray diagram for the situation. b) Calculate the image distance. Does the calculated distance agree with the distance given by the ray diagram? Is the image real or virtual? c) The object is moved out to \(8 \mathrm{~cm}\). Draw a ray diagram for the situation. d) Calculate the image distance. Does the calculated distance agree with the distance given by the ray diagram? Is the image real or virtual. Is it larger or smaller than the image from part (a)?

To draw a ray diagram, follow these three principal rays: 1. The incident ray initially parallel to the axis of the lens passes through the focal point after refraction. 2. The incident ray passing through the center of the lens comes through without being deviated. 3. The incident ray passing through the focal point before reaching the lens will emerge parallel to the axis after refraction. Draw the lens and the object at a distance of \(6 \mathrm{~cm}\) from the lens, including the three principal rays. The intersection point of the principal rays determines the location and orientation of the image.

Use the lens formula to calculate the image distance: The lens formula is given by \begin{equation} \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \end{equation} where \(f\) is the focal length, \(d_o\) is the distance of the object to the lens, and \(d_i\) is the distance of the image to the lens. Given \(f = 10 \mathrm{~cm}\) and \(d_o = 6 \mathrm{~cm}\), we can find the image distance \(d_i\): \begin{equation} \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \end{equation} \begin{equation} d_i = \frac{1}{\frac{1}{10} - \frac{1}{6}} = 15 \mathrm{~cm} \end{equation} Verify if the calculated distance agrees with the distance given by the ray diagram. Since the distance between the intersection of the principal rays and the lens in the ray diagram is approximately \(15 \mathrm{~cm}\), the calculated distance does indeed agree with the ray diagram. Determine the image type by comparing the object distance to the focal length. The object distance is smaller than the focal length, which means the image will be virtual.

Repeat the steps for drawing a ray diagram in part (a), this time placing the object at a distance of \(8 \mathrm{~cm}\) from the lens.

Use the lens formula to calculate the image distance: Given \(f = 10 \mathrm{~cm}\) and \(d_o = 8 \mathrm{~cm}\), we can find the image distance \(d_i\): \begin{equation} \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \end{equation} \begin{equation} d_i = \frac{1}{\frac{1}{10} - \frac{1}{8}} = 40 \mathrm{~cm} \end{equation} Verify if the calculated distance agrees with the distance given by the ray diagram. As with the previous situation, the distance between the intersection of the principal rays and the lens is approximately \(40 \mathrm{~cm}\), the calculated distance does agree with the ray diagram. Determine the image type by comparing the object distance to the focal length. The object distance is greater than the focal length, which means the image will be real. To compare the size of the image in part (d) to the image in part (a), look at the magnification \(M\). To calculate the magnification, use the following formula: \begin{equation} M = \frac{d_i}{d_o} \end{equation} For the situation when the object is placed at \(8 \mathrm{~cm}\), the magnification is: \begin{equation} M = \frac{40 \ \text{cm}}{8 \ \text{cm}} = 5 \end{equation} The image from part (d) is 5 times larger than the object, while in part (a) the image was smaller than the object. Therefore, the image in part (d) is larger than the image in part (a).