Question

An object is located \(16.0 \mathrm{cm}\) in front of a converging lens with focal length \(12.0 \mathrm{cm} .\) To the right of the converging lens, separated by a distance of \(20.0 \mathrm{cm},\) is a diverging lens of focal length \(-10.0 \mathrm{cm} .\) Find the location of the final image by ray tracing and verify using the lens equations.

Short answer

Answer: The final image is located 40.0 cm to the right of the converging lens.

Step-by-step solution

Calculate the image location after the converging lens

We have a converging lens with focal length \(f_1 = 12.0 cm\) and an object distance \(d_{01} = 16.0 cm\). We can calculate the image distance \(d_{i1}\) using the lensmaker equation: $$\frac{1}{f_1} = \frac{1}{d_{o1}} + \frac{1}{d_{i1}}$$ Rearranging the equation and solving for \(d_{i1}\), we get: $$\frac{1}{d_{i1}} = \frac{1}{f_1} - \frac{1}{d_{o1}}$$ Plug in the given values: $$\frac{1}{d_{i1}} = \frac{1}{12.0 cm} - \frac{1}{16.0 cm}$$ Now, solve for \(d_{i1}\): $$d_{i1} = \frac{1}{\frac{1}{12.0} - \frac{1}{16.0}} = 48.0 \mathrm{cm}$$

Calculate the object distance for the diverging lens

Now that we know the image distance (\(d_{i1}\)) after the converging lens, we can calculate the object distance (\(d_{o2}\)) for the diverging lens. The given distance between the lenses is \(20.0 \mathrm{cm}\). Therefore, the object distance for the diverging lens is: $$d_{o2} = d_{i1} - D = 48.0 \mathrm{cm} - 20.0 \mathrm{cm} = 28.0 \mathrm{cm}$$

Calculate the image location after the diverging lens

We have a diverging lens with focal length \(f_2 = -10.0 cm\) and object distance \(d_{o2} = 28.0 cm\). We can calculate the image distance \(d_{i2}\) using the lensmaker equation again: $$\frac{1}{f_2} = \frac{1}{d_{o2}} + \frac{1}{d_{i2}}$$ Rearranging the equation and solving for \(d_{i2}\), we get: $$\frac{1}{d_{i2}} = \frac{1}{f_2} - \frac{1}{d_{o2}}$$ Plug in the given values: $$\frac{1}{d_{i2}} = \frac{1}{-10.0 cm} - \frac{1}{28.0 cm}$$ Now, solve for \(d_{i2}\): $$d_{i2} = \frac{1}{\frac{1}{-10.0} - \frac{1}{28.0}} = 20.0 \mathrm{cm}$$

Determine the location of the final image

Since the image after the diverging lens is \(d_{i2} = 20.0 \mathrm{cm}\) from the lens, we can calculate the total image distance from the converging lens as follows: $$d_i = D + d_{i2} = 20.0 \mathrm{cm} + 20.0 \mathrm{cm} = 40.0 \mathrm{cm}$$ This is the location of the final image. It is formed \(40.0 \mathrm{cm}\) to the right of the converging lens.