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?

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.

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: $$h_i = M_1 \times h_o = -2.5 \times 20.0 \mathrm{~cm} = -50.0 \mathrm{~cm}$$

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: $$d'_o = 175\mathrm{~cm} - d_i = 175\mathrm{~cm} - (-75\mathrm{~cm}) = 250\mathrm{~cm}$$

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 \(d'_i\): $$\frac{1}{50} = \frac{1}{250} + \frac{1}{d'_i}$$ $$\frac{1}{d'_i} = \frac{1}{50} - \frac{1}{250}$$ $$d'_i = \frac{50 \times 250}{250 - 50} = 62.5 \mathrm{~cm}$$

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.