Chapter 33: Problem 85

Determine the position and size of the final image formed by a system of elements consisting of an object \(2.0 \mathrm{~cm}\) high located at \(x=0 \mathrm{~m},\) a converging lens with focal length \(5.0 \cdot 10^{1} \mathrm{~cm}\) located at \(x=3.0 \cdot 10^{1} \mathrm{~cm}\) and a plane mirror located at \(x=7.0 \cdot 10^{1} \mathrm{~cm}\)

Short Answer

Expert verified

Answer: The final image is located at a distance of 75 cm from the lens and has a height of -5.0 cm.

Step by step solution

Find the image formed by the converging lens

To find the image formed by the lens, we will use the lens equation: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. In this case, the focal length \(f = 5.0 \cdot 10^1 \mathrm{~cm}\), and the object distance \(d_o = 3.0 \cdot 10^1 \mathrm{~cm}\). Plug the values into the lens equation: \(\frac{1}{5.0 \cdot 10^1} = \frac{1}{3.0 \cdot 10^1} + \frac{1}{d_i}\) Now, solve for \(d_i\): \(d_i = \frac{1}{\frac{1}{5.0 \cdot 10^1} - \frac{1}{3.0 \cdot 10^1}} \approx 7.50 \cdot 10^1 \mathrm{~cm}\)

Determine the magnification of the lens

To find the magnification of the lens, use the magnification formula: \(m = \frac{h_i}{h_o} = \frac{-d_i}{d_o}\), where \(h_o\) is the object height, and \(h_i\) is the image height. \(h_o = 2.0 \mathrm{~cm}\) Plug the values into the magnification formula: \(m = \frac{-7.50 \cdot 10^1}{3.0 \cdot 10^1}\) Solve for \(m\): \(m = -2.5\) Now, find the height of the image formed by the lens: \(h_i = mh_o = -2.5(2.0) = -5.0 \mathrm{~cm}\)

Interact with the plane mirror

Since the converging lens forms a virtual image with a height of \(-5 \mathrm{~cm}\) at a distance of \(7.50 \cdot 10^1 \mathrm{~cm}\) from the lens, and the plane mirror is located at \(x = 7.0 \cdot 10^{1} \mathrm{~cm}\), the distance between the image and the mirror is \(d_{mirror} = 7.50 \cdot 10^1 - 7.0 \cdot 10^{1} = 5.0 \mathrm{~cm}\). As the image is virtual, the mirror will form a new image on the same side of the mirror as the original image. The new image will be at the same distance from the mirror, so the final image distance from the mirror is \(5.0 \mathrm{~cm}\). The final image distance from the lens is \(7.00 \cdot 10^1 + 5.0 = 7.50 \cdot 10^1 \mathrm{~cm}\).