Let's solve the problem step by step. **Given:** - Object height, \( h = 1.23 \) cm - Focal length of left lens, \( f_1 = 3.82 \) cm (converging, so positive) - Focal length of right lens, \( f_2 = 2.28 \) cm (converging, so positive) - Object distance from left lens, \( d_{o1} = 5.95 \) cm (to the left, so positive) - Distance between lenses, \( D = 10.9 \) cm --- ## (a) **Where is the secondary image formed by the lens on the right located relative to that lens?** ### **Step 1: Find image formed by the left lens (Lens 1)** Use the lens formula: \[ \frac{1}{f_1} = \frac{1}{d_{o1}} + \frac{1}{d_{i1}} \] where: - \( f_1 = 3.82 \) cm, - \( d_{o1} = 5.95 \) cm, - \( d_{i1} \) = image distance from lens 1 (to be found). \[ \frac{1}{d_{i1}} = \frac{1}{f_1} - \frac{1}{d_{o1}} \] \[ \frac{1}{d_{i1}} = \frac{1}{3.82} - \frac{1}{5.95} \] \[ \frac{1}{3.82} \approx 0.2620 \] \[ \frac{1}{5.95} \approx 0.1681 \] \[ \frac{1}{d_{i1}} = 0.2620 - 0.1681 = 0.0939 \] \[ d_{i1} = \frac{1}{0.0939} \approx 10.65 \text{ cm} \] So, the image formed by lens 1 is **10.65 cm to the right** of lens 1. --- ### **Step 2: Find the object distance for the right lens (Lens 2)** - The distance between the two lenses is \( D = 10.9 \) cm. - The first image is \( d_{i1} = 10.65 \) cm to the right of lens 1, so it is **to the left** of lens 2 by \( D - d_{i1} = 10.9 - 10.65 = 0.25 \) cm. Thus, **the image formed by the first lens acts as the object for the second lens**, and is located **0.25 cm to the left of lens 2**. So, the object distance for lens 2 is \( d_{o2} = +0.25 \) cm (positive, since it is to the left of lens 2). --- ### **Step 3: Find the image formed by the right lens (Lens 2)** Use the lens formula again: \[ \frac{1}{f_2} = \frac{1}{d_{o2}} + \frac{1}{d_{i2}} \] where: - \( f_2 = 2.28 \) cm, - \( d_{o2} = 0.25 \) cm, - \( d_{i2} \) = image distance from lens 2 (to be found). \[ \frac{1}{d_{i2}} = \frac{1}{f_2} - \frac{1}{d_{o2}} \] \[ \frac{1}{d_{i2}} = \frac{1}{2.28} - \frac{1}{0.25} \] \[ \frac{1}{2.28} \approx 0.4386 \] \[ \frac{1}{0.25} = 4.00 \] \[ \frac{1}{d_{i2}} = 0.4386 - 4.00 = -3.5614 \] \[ d_{i2} = \frac{1}{-3.5614} \approx -0.281 \text{ cm} \] **Interpretation:** - Negative sign means the image is **to the left of lens 2** (same side as the object for this lens). - Magnitude: \( 0.281 \) cm. **Final Answer for (a):** > **The secondary image is located 0.281 cm to the left of the right lens (lens 2).** --- ## (b) **What is the type and height of this secondary image?** ### **Type** Since the image is on the same side as the object for lens 2 (left side), it's a **virtual image** (for a converging lens, a real image forms on the opposite side). ### **Height: Find Total Magnification** Total magnification is the product of the magnifications of each lens. #### **Magnification by Lens 1:** \[ m_1 = -\frac{d_{i1}}{d_{o1}} \] \[ m_1 = -\frac{10.65}{5.95} \approx -1.79 \] #### **Magnification by Lens 2:** \[ m_2 = -\frac{d_{i2}}{d_{o2}} \] \[ m_2 = -\frac{-0.281}{0.25} = +1.124 \] #### **Total Magnification:** \[ m_{\text{tot}} = m_1 \times m_2 = (-1.79) \times (+1.124) \approx -2.01 \] #### **Height of the final image:** \[ h' = m_{\text{tot}} \times h = (-2.01) \times 1.23 \approx -2.47 \text{ cm} \] **The negative sign** indicates that the image is **inverted** relative to the original object. --- **Final Answer for (b):** > **The secondary image is virtual, inverted, and has a height of approximately 2.47 cm (inverted).** --- ## (c) **How far and in what direction should the right lens be moved to make the secondary image coincide with the original object?** We want the final image (after both lenses) to be formed **at the original object position** (i.e., left of lens 1 by 5.95 cm). Let's define: - Let the new distance between the two lenses be \( D' \). - The object is a distance \( d_{o1} = 5.95 \) cm to the left of lens 1. - We want the final image (after both lenses) to be located at the same position as the original object. ### **Step 1: Image by Lens 1** As before, image formed by lens 1 is at \( d_{i1} = +10.65 \) cm (to the right of lens 1). ### **Step 2: New Position of Lens 2** Let the new distance between the two lenses be \( D' \). The object distance for lens 2 (relative to lens 2): \[ d_{o2}' = D' - d_{i1} \] (The image is to the left of lens 2, so the sign is positive if \( D' > d_{i1} \).) ### **Step 3: Required Image Location for Lens 2** We want the **final image after lens 2** to be at the original object position, which is \( D' + d_{o1} \) **to the left of lens 2** (since lens 2 is now at a new location, shifted compared to the original setup). So, the image distance for lens 2: \[ d_{i2}' = -(D' + d_{o1}) \] (Negative because it is to the left of lens 2.) ### **Step 4: Lens Equation for Lens 2** \[ \frac{1}{f_2} = \frac{1}{d_{o2}'} + \frac{1}{d_{i2}'} \] Substitute values: - \( f_2 = 2.28 \) cm, - \( d_{o2}' = D' - 10.65 \), - \( d_{i2}' = -(D' + 5.95) \). \[ \frac{1}{2.28} = \frac{1}{D' - 10.65} + \frac{1}{-(D' + 5.95)} \] \[ \frac{1}{2.28} = \frac{1}{D' - 10.65} - \frac{1}{D' + 5.95} \] Combine right side: \[ \frac{1}{D' - 10.65} - \frac{1}{D' + 5.95} = \frac{(D' + 5.95) - (D' - 10.65)}{(D' - 10