Let's solve the problem step by step: --- ## **Given Data** - Two convex lenses, each focal length \( f = 4 \) cm. - The lenses are separated by their focal length: **distance between them = 4 cm**. - Object O is placed 12 cm next to one of the lenses (let’s call it **Lens 1**). - We need to find the **position** and **magnification** of the final image after both lenses. --- ## **Step 1: Find the Image Formed by Lens 1** Let’s define the setup: - Place **Lens 1** at \( x = \). - **Lens 2** is at \( x = 4 \) cm. - The **object** is at \( x = -12 \) cm (since it is 12 cm to the left of Lens 1). **Lens Formula:** \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] Where: - \( f \) = focal length of Lens 1 = \( +4 \) cm (convex, so positive) - \( u = -12 \) cm (object is on the left of the lens, so negative) Plug in the values: \[ \frac{1}{4} = \frac{1}{v_1} - \frac{1}{-12} \] \[ \frac{1}{4} = \frac{1}{v_1} + \frac{1}{12} \] \[ \frac{1}{v_1} = \frac{1}{4} - \frac{1}{12} = \frac{3 - 1}{12} = \frac{2}{12} = \frac{1}{6} \] \[ v_1 = 6 \text{ cm} \] **Interpretation:** The image formed by Lens 1 is **6 cm to the right** of Lens 1. --- ## **Step 2: Use Image from Lens 1 as Object for Lens 2** - **Lens 2** is at \( x = 4 \) cm. - The image from Lens 1 is at \( x = + 6 = 6 \) cm. So, the **object distance for Lens 2** (relative to Lens 2): \[ u_2 = \text{(position of image from Lens 1)} - \text{(position of Lens 2)} \] \[ u_2 = 6 \text{ cm} - 4 \text{ cm} = +2 \text{ cm} \] **Positive sign:** The object is on the right side of Lens 2 (real object for Lens 2). **Lens 2 focal length:** \( f = 4 \) cm Apply the lens formula for Lens 2: \[ \frac{1}{f} = \frac{1}{v_2} - \frac{1}{u_2} \] \[ \frac{1}{4} = \frac{1}{v_2} - \frac{1}{2} \] \[ \frac{1}{v_2} = \frac{1}{4} + \frac{1}{2} = \frac{1 + 2}{4} = \frac{3}{4} \] \[ v_2 = \frac{4}{3} \text{ cm} \] **Interpretation:** The final image is **4/3 cm to the right of Lens 2**. --- ## **Step 3: Position of the Final Image Relative to Lens 1** - Lens 2 is at \( x = 4 \) cm. - Final image is \( 4/3 \) cm to the right of Lens 2: \[ x_{\text{final image}} = 4 \text{ cm} + \frac{4}{3} \text{ cm} = \frac{12}{3} + \frac{4}{3} = \frac{16}{3} \approx 5.33 \text{ cm} \] So the final image is **5.33 cm to the right of Lens 1**. --- ## **Step 4: Calculate the Total Magnification** Magnification by Lens 1: \[ m_1 = -\frac{v_1}{u_1} = -\frac{6}{-12} = .5 \] Magnification by Lens 2: \[ m_2 = -\frac{v_2}{u_2} = -\frac{4/3}{2} = -\frac{2}{3} \] **Total Magnification:** \[ m = m_1 \times m_2 = .5 \times \left(-\frac{2}{3}\right) = -\frac{1}{3} \] --- ## **Final Answers** ### **(a) Position of the final image:** **5.33 cm to the right of Lens 1.** ### **(b) Total Magnification:** **-1/3** ### **(c) Geometrical Construction (Ray Diagram):** - Object is placed far left. - First lens forms a real, inverted, reduced image between the two lenses. - Second lens forms an even smaller, inverted image to the right of the second lens. --- ### **Summary Table** | Step | Position (cm) | Magnification | |---------------|---------------|--------------| | Lens 1 Image | 6 right of L1 | .5 | | Lens 2 Image | 4/3 right of L2 | -2/3 | | **Final Image (from L1)** | 5.33 right of L1 | **-1/3** | --- ### **Boxed Final Answers** \[ \boxed{ \begin{array}{l} \text{Position of final image:} \quad 5.33\, \text{cm to the right of Lens 1} \\ \text{Total magnification:} \quad -\frac{1}{3} \end{array} } \]