# Two-Lens System: Image Location and Nature ## Given Information - **Lens 1 focal length:** \( f_1 \) - **Lens 2 focal length:** \( f_2 \) - **Relationship:** \( f_1 = 3f_2 \) - **One focal point is shared:** The right focal point of Lens 1 coincides with the left focal point of Lens 2. - **Object position:** \( 2.5f \) to the left of Lens 1 (with \( f = f_2 \), so \( 2.5f_2 \)) - **Find:** Final image location (relative to Lens 2) and nature (real or virtual). --- ## Step 1: Assign Focal Lengths Let \( f_2 = f \). Then \( f_1 = 3f \). --- ## Step 2: Lens Positions and Focal Points Let’s choose coordinates: - Lens 1 at \( x = \) - Lens 2 at \( x = d \) ### Shared Focal Point The right focal point of Lens 1 is at \( x = f_1 = 3f \). The left focal point of Lens 2 is at \( x = d - f_2 = d - f \). Setting these equal: \[ 3f = d - f \implies d = 4f \] So, **Lens 2 is 4f to the right of Lens 1.** --- ## Step 3: Object Distance for Lens 1 Object is at \( x = -2.5f \) (left of Lens 1). Object distance: \[ s_1 = 2.5f \] (Lens 1 is at \( x = \), object at \( x = -2.5f \).) --- ## Step 4: Image by Lens 1 Lens formula: \[ \frac{1}{f_1} = \frac{1}{s_1'} + \frac{1}{s_1} \] where \( s_1' \) is the image distance from Lens 1. Plug in values: \[ \frac{1}{3f} = \frac{1}{s_1'} + \frac{1}{2.5f} \] \[ \frac{1}{s_1'} = \frac{1}{3f} - \frac{1}{2.5f} \] \[ \frac{1}{s_1'} = \frac{2.5 - 3}{7.5f} = \frac{-.5}{7.5f} = -\frac{1}{15f} \] \[ s_1' = -15f \] **Interpretation:** The image is 15f to the *left* of Lens 1 (negative sign = same side as object; virtual image). --- ## Step 5: Image Position Relative to Lens 2 - Lens 2 is at \( x = 4f \). - Image from Lens 1 is at \( x = -15f \). Distance from image to Lens 2: \[ s_2 = x_{\text{image}} - x_{\text{Lens 2}} = (-15f) - (4f) = -19f \] So, for Lens 2, the object (which is the image from Lens 1) is *19f to the left* (virtual object). --- ## Step 6: Image by Lens 2 Lens 2 formula: \[ \frac{1}{f_2} = \frac{1}{s_2'} + \frac{1}{s_2} \] Recall \( f_2 = f \), \( s_2 = -19f \): \[ \frac{1}{f} = \frac{1}{s_2'} + \frac{1}{-19f} \] \[ \frac{1}{s_2'} = \frac{1}{f} + \frac{1}{19f} = \frac{19 + 1}{19f} = \frac{20}{19f} \] \[ s_2' = \frac{19f}{20} \] **Interpretation:** The final image is **\( .95f \) to the right of Lens 2** (since \( s_2' > \), real image). --- ## Step 7: Summary ### Final Image - **Location:** \( .95f \) to the right of Lens 2 - **Nature:** Real image --- ## Answer > **The final image is located \( \frac{19}{20}f_2 \) (or \( .95f_2 \)) to the right of Lens 2, and the image is real.** --- ### Diagram Description - **Object:** \( 2.5f_2 \) to the left of Lens 1 - **Lens 1:** At \( x = \), focal length \( 3f_2 \) - **Lens 2:** At \( x = 4f_2 \), focal length \( f_2 \) - **Image:** \( .95f_2 \) to the right of Lens 2 --- **[No images can be rendered here, but you may sketch two lenses separated by \( 4f_2 \), with the object at \( -2.5f_2 \), and the final image just right of Lens 2.]**