12ª A meniscus concave glass (n₁ = 1.5) thin lens (see Fig. 5.12) has radii of curvature of +20.0 cm and +10.0 cm. If an object is placed 20.0 cm in front of the lens, show that the image distance will be - 13.3 cm. Describe that image and draw a ray diagram.

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**5.12** A meniscus concave glass (nₗ = 1.5) thin lens (see Fig. 5.12) has radii of curvature of +20.0 cm and +10.0 cm. If an object is placed 20.0 cm in front of the lens, show that the image distance will be -13.3 cm. Describe that image and draw a ray diagram.

**Explanation:**

1. **Lens Description**: The text describes a thin lens made of concave glass. The refractive index \( n_l \) is 1.5, which is common for glass materials.

2. **Radii of Curvature**: The lens has two surfaces with radii of curvature of +20.0 cm and +10.0 cm. These values indicate the curvature of the lens surfaces, essential for determining lens behavior using the Lensmaker's Equation.

3. **Lensmaker's Equation**: This equation is typically used to find the focal length \( f \) of a lens:
   \[
   \frac{1}{f} = (n - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)
   \]
   where \( n \) is the refractive index, \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

4. **Object Placement**: The object is placed 20.0 cm in front of the lens. This distance serves as the object distance (\( o \)) to be used in the lens formula.

5. **Image Distance**: The task is to determine that the image distance is -13.3 cm using the lens formula:
   \[
   \frac{1}{f} = \frac{1}{o} + \frac{1}{i}
   \]
   where \( o \) is the object distance, and \( i \) is the image distance.

6. **Ray Diagram**: A ray diagram should be drawn to visually represent the behavior of light rays through the lens, showing how the image distance is derived.

This exercise demonstrates the application of geometric optics principles, particularly how lenses form images and how mathematics can predict physical outcomes.
Transcribed Image Text:**5.12** A meniscus concave glass (nₗ = 1.5) thin lens (see Fig. 5.12) has radii of curvature of +20.0 cm and +10.0 cm. If an object is placed 20.0 cm in front of the lens, show that the image distance will be -13.3 cm. Describe that image and draw a ray diagram. **Explanation:** 1. **Lens Description**: The text describes a thin lens made of concave glass. The refractive index \( n_l \) is 1.5, which is common for glass materials. 2. **Radii of Curvature**: The lens has two surfaces with radii of curvature of +20.0 cm and +10.0 cm. These values indicate the curvature of the lens surfaces, essential for determining lens behavior using the Lensmaker's Equation. 3. **Lensmaker's Equation**: This equation is typically used to find the focal length \( f \) of a lens: \[ \frac{1}{f} = (n - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] where \( n \) is the refractive index, \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces. 4. **Object Placement**: The object is placed 20.0 cm in front of the lens. This distance serves as the object distance (\( o \)) to be used in the lens formula. 5. **Image Distance**: The task is to determine that the image distance is -13.3 cm using the lens formula: \[ \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \] where \( o \) is the object distance, and \( i \) is the image distance. 6. **Ray Diagram**: A ray diagram should be drawn to visually represent the behavior of light rays through the lens, showing how the image distance is derived. This exercise demonstrates the application of geometric optics principles, particularly how lenses form images and how mathematics can predict physical outcomes.
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