Question 1190968: An object 2.0 cm high is located 10.0 cm in front of a diverging lens whose focal length is -6.0 cm. Find the position and size of the image by drawing principal rays.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Absolutely! Let's break down how to find the image position and size using principal rays for this diverging lens scenario.
**1. Key Information**
* **Object Height (ho):** 2.0 cm
* **Object Distance (do):** 10.0 cm
* **Focal Length (f):** -6.0 cm (negative for diverging lens)
**2. Lens Equation (for Calculation)**
While we're focusing on principal rays for the diagram, the lens equation helps us calculate precise values to check our ray diagram against.
* **Lens Equation:** 1/do + 1/di = 1/f
Let's solve for image distance (di):
* 1/di = 1/f - 1/do
* 1/di = 1/(-6.0 cm) - 1/(10.0 cm)
* 1/di = -0.167 cm⁻¹ - 0.1 cm⁻¹
* 1/di = -0.267 cm⁻¹
* di ≈ -3.75 cm
The negative sign indicates a virtual image (on the same side of the lens as the object).
**3. Magnification (for Size)**
* **Magnification (M):** -di/do
Let's calculate:
* M = -(-3.75 cm) / 10.0 cm
* M = 0.375
The positive magnification means the image is upright.
* **Image Height (hi):** M * ho
* hi = 0.375 * 2.0 cm
* hi = 0.75 cm
**4. Principal Rays for the Diagram**
Here's how to draw the principal rays for a diverging lens:
* **Ray 1: Parallel Ray**
* A ray traveling parallel to the principal axis before the lens refracts as if it came from the focal point on the *same* side of the lens.
* **Ray 2: Central Ray**
* A ray passing through the center of the lens continues in a straight line, undeflected.
* **Ray 3: Aimed Ray**
* A ray aimed toward the focal point on the *opposite* side of the lens refracts parallel to the principal axis.
**5. Image Formation**
* The image is formed where the refracted rays (or their extended lines, in the case of virtual images) intersect. In this case, the rays diverge, so we extend the refracted rays *backwards* behind the lens. The intersection of these extended lines marks the location of the virtual image.
**Summary**
* **Image Position:** Approximately 3.75 cm in front of the lens (virtual image).
* **Image Size:** 0.75 cm tall and upright.
**Key Points for Your Diagram**
* The image is virtual, so it will be on the same side of the lens as the object.
* The image is upright.
* The image is smaller than the object.
* Be sure to extend the refracted rays backward to locate the virtual image.
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