Question 1190967: A converging lens has a focal length of 4.0 cm. By drawing principal rays, find the position and size of the image if an object 1.5 cm high is placed 5.0 cm in front of the lens.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to find the image position and size using principal rays for this converging lens scenario:
**1. Key Information**
* **Object Height (h₀):** 1.5 cm
* **Object Distance (d₀):** 5.0 cm
* **Focal Length (f):** 4.0 cm (positive for converging 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/d₀ + 1/dᵢ = 1/f
Let's solve for image distance (dᵢ):
* 1/dᵢ = 1/f - 1/d₀
* 1/dᵢ = 1/4.0 cm - 1/5.0 cm
* 1/dᵢ = 0.25 cm⁻¹ - 0.2 cm⁻¹
* 1/dᵢ = 0.05 cm⁻¹
* dᵢ = 20.0 cm
**3. Magnification (for Size)**
* **Magnification (M):** -dᵢ/d₀
Let's calculate:
* M = -20.0 cm / 5.0 cm
* M = -4.0
The negative magnification means the image is inverted.
* **Image Height (hᵢ):** M * h₀
* hᵢ = -4.0 * 1.5 cm
* hᵢ = -6.0 cm
**4. Principal Rays for the Diagram**
Here's how to draw the principal rays for a converging lens:
* **Ray 1: Parallel Ray**
* A ray traveling parallel to the principal axis before the lens refracts and passes through the focal point on the *opposite* 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: Focal Ray**
* A ray passing through the focal point on the *same* side of the lens refracts and travels parallel to the principal axis.
**5. Image Formation**
* The image is formed where the refracted rays intersect. In this case, the rays converge, forming a real image on the opposite side of the lens from the object.
**Summary**
* **Image Position:** 20.0 cm from the lens (real image).
* **Image Size:** 6.0 cm tall and inverted.
**Key Points for Your Diagram**
* The image is real, so it will be on the opposite side of the lens from the object.
* The image is inverted.
* The image is larger than the object.
* Be sure to draw all three principal rays to confirm the image location.
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