91Ó°ÊÓ

To start, we'll use the thin lens equation for the first lens (focal length = 2 cm): \((1/\text{focal length}) = (1/\text{object distance}) - (1/\text{image distance})\) Rearranging the equation and using signs for a converging lens: \(image~~distance = \frac{object~~distance \times focal~~length}{object~~distance - focal~~length}\) Plug in values (object distance = 2.5 cm): \(image~~distance = \frac{2.5\times2}{(2.5-2)} = 10 ~~cm\) We also need to calculate the magnification for Lens 1: \(magnification = \frac{-image~~distance}{object~~distance} = \frac{-10}{2.5} = -4\)

We will do the same calculation for Lens 3, whose object distance would be the distance of Lens 3 from Lens 1's image. Object distance for Lens 3 = \((19.8\,{\rm cm} - 10\,{\rm cm}) = 9.8\,{\rm cm}\) Applying the thin lens equation for Lens 3, rearrange the equation: \(image~~distance = \frac{object~~distance \times focal~~length}{object~~distance - focal~~length}\) Plug in values (focal length = 4 cm): \(image~~distance = \frac{9.8\times4}{(9.8-4)} = 8.29 ~~cm\) Calculate the magnification for Lens 3: \(magnification = \frac{-image~~distance}{object~~distance} = \frac{-8.29}{9.8} = -0.846\) Now, multiply magnifications of Lens 1 and Lens 3: \(overall~~magnification = magnification_{1} \times magnification_{3} = -4 \times (-0.846) = 3.38\)

The overall magnification of the system is given as \(6.84\). The magnification produced by the first and third lenses is \(3.38\). As the product of the magnifications of all 3 lenses must equal the overall magnification, we can determine the magnification produced by the unknown lens: \(magnification_{unknown} = \frac{overall~~magnification}{magnification_{1+3}} = \frac{6.84}{3.38} = 2.023\) Since the image produced by the system is upright, the image produced by the unknown lens must be upright as well since all 3 lenses' magnification products give the overall magnification. This indicates that the unknown lens is a diverging lens, as diverging lenses produce upright images.

The object distance for the unknown lens is equal to the image distance from Lens 1: Object distance for unknown lens = \(10\,{\rm cm}\) We know the magnification for the unknown lens is \(2.023\). Since the lens is diverging, the magnification should be positive: \(magnification = \frac{image~~distance}{object~~distance} = 2.023\) Rearrange to solve for image distance: \(image~~distance = magnification \times object~~distance = 2.023 \times 10 = 20.23\,{\rm cm}\) Now we can plug in values to find the focal length of the unknown lens. As it is a diverging lens, we will use a negative sign for the focal length. \((1/\text{focal length}) = (1/\text{object distance}) - (1/\text{image distance})\) Plug in values and rearrange: \(focal~~length = \frac{object~~distance \times image~~distance}{object~~distance + image~~distance}= -\frac{10\times20.23}{(10+20.23)} = -6.57\,{\rm cm}\)

To create the ray diagram, we'd follow these steps: 1. Place the object, first lens, unknown lens, third lens, and final image on a straight horizontal line in respective positions. 2. Draw rays from the object to the first lens and then refracted rays converging to form the image at 10 cm. 3. Treat the formed image as the new object for the unknown lens. 4. As it's a diverging lens, draw refracted rays diverging, producing an upright image at 20.23 cm (10+20.23). 5. Finally, treat this image as the new object for the third converging lens. Draw refracted rays converging to form an upright final image at 39.8 cm. Following these steps, the ray diagram should confirm our answers: the unknown lens is diverging, and its focal length is -6.57 cm.