Question

A converging lens of focal length \(f=50.0 \mathrm{~cm}\) is placed \(175 \mathrm{~cm}\) to the left of a metallic sphere of radius \(R=100 . \mathrm{cm} .\) An object of height \(h=20.0 \mathrm{~cm}\) is placed \(30.0 \mathrm{~cm}\) to the left of the lens. What is the height of the image formed by the metallic sphere?

Short answer

Answer: The height of the final image formed by the metallic sphere is -12.5 cm. The negative sign indicates that the image is inverted.

Step-by-step solution

Find the image distance formed by the lens

We use the lens formula to find the image distance formed by the lens: $$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$ where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. We are given \(f=50.0 \mathrm{~cm}\) and \(d_o=30.0 \mathrm{~cm}\). Plugging these values in, and solving for \(d_i\): $$\frac{1}{50} = \frac{1}{30} + \frac{1}{d_i}$$ $$\frac{1}{d_i} = \frac{1}{50} - \frac{1}{30}$$ $$d_i = \frac{50 \times 30}{30 - 50} = -75 \mathrm{~cm}$$ Note that the image distance is negative, which means that the image is virtual.

Find the image height formed by the lens

Now, let's find the magnification, \(M_1\), created by the lens using the following formula: $$M_1 = \frac{h_i}{h_o} = \frac{d_i}{d_o}$$ where \(h_o\) is the object height, and \(h_i\) is the image height. We are given \(h_o = 20.0 \mathrm{~cm}\) and we calculated \(d_i=-75\mathrm{~cm}\) and \(d_o=30.0\mathrm{~cm}\) previously: $$M_1 = \frac{-75}{30} = -2.5$$ The negative sign here indicates that the image is inverted. Now we can find the image height formed by the lens: $$h_i = M_1 \times h_o = -2.5 \times 20.0 \mathrm{~cm} = -50.0 \mathrm{~cm}$$

Find the object distance for the metallic sphere

The image formed by the lens is now acting as an object for the metallic sphere. Thus, the object distance, \(d'_o\), for the metallic sphere is the distance between the lens and the sphere minus the image distance formed by the lens: $$d'_o = 175\mathrm{~cm} - d_i = 175\mathrm{~cm} - (-75\mathrm{~cm}) = 250\mathrm{~cm}$$

Find the image distance for the metallic sphere

Since the metallic sphere is essentially acting as a concave mirror with the radius of curvature equal to the radius of the sphere, we can use the mirror formula to find the image distance, \(d'_i\), created by the metallic sphere: $$\frac{1}{R/2} = \frac{1}{d'_o} + \frac{1}{d'_i}$$ We are given \(R=100\mathrm{~cm}\), and we have found \(d'_o=250\mathrm{~cm}\). Plugging these values in and solving for \(d'_i\): $$\frac{1}{50} = \frac{1}{250} + \frac{1}{d'_i}$$ $$\frac{1}{d'_i} = \frac{1}{50} - \frac{1}{250}$$ $$d'_i = \frac{50 \times 250}{250 - 50} = 62.5 \mathrm{~cm}$$

Find the image height created by the metallic sphere

Now, let's find the magnification, \(M_2\), created by the metallic sphere using the following formula: $$M_2 = \frac{h'_i}{h'_o} = \frac{d'_i}{d'_o}$$ where \(h'_o\) is the object height for the metallic sphere (which is the image height formed by the lens), and \(h'_i \) is the image height created by the metallic sphere. We calculated \(d'_i=62.5\mathrm{~cm}\) and \(d'_o=250\mathrm{~cm}\) previously, and we found \(h'_o=-50.0 \mathrm{~cm}\) from step 2: $$M_2 = \frac{62.5}{250} = 0.25$$ Now we can find the image height created by the metallic sphere: $$h'_i = M_2 \times h'_o = 0.25 \times (-50.0 \mathrm{~cm}) = -12.5 \mathrm{~cm}$$ The height of the final image formed by the metallic sphere is \(-12.5\mathrm{~cm}\). The negative sign indicates that the image is inverted.