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Transcribed Image Text:**Images Formed by a Diverging Lens**
A diverging lens has a focal length of 10.0 cm. An image is formed by a diverging lens under different scenarios. The diagrams illustrate how the position of an object relative to the focal point affects image formation.
### Diagram Descriptions:
1. **Diagram (a): The object is farther from the lens than the focal point.**
- The object (represented as an arrow) is placed beyond the focal point \( F_1 \).
- Light rays diverge after passing through the lens.
- The blue ray, parallel to the principal axis, refracts and appears to come from the focal point \( F_2 \).
- The red ray, passing through the center of the lens, continues in a straight line.
- The green ray, directed towards the lens, diverges after refraction.
- The image formed is virtual, upright, and reduced in size, appearing on the same side as the object.
2. **Diagram (b): The object is at the focal point.**
- The object is positioned at the focal point \( F_1 \).
- Rays parallel to the principal axis diverge, appearing to originate from a point on the opposite side of the lens (focal point \( F_2 \)).
- The image appears at infinity. In practical terms, no image is formed on the screen because the light rays do not converge.
3. **Diagram (c): The object is closer to the lens than the focal point.**
- The object is positioned closer to the lens than \( F_1 \).
- Similar to diagram (a), rays diverge with the appearance of coming from the focal point \( F_2 \) after passing through the lens.
- The virtual image formed is upright and larger than the object.
### Key Points:
- **Focal Length:** The fixed distance from the lens (10.0 cm) where parallel rays converge or appear to diverge from.
- **Lens Behavior:** Diverging lenses cause parallel rays to spread out, resulting in virtual images.
- **Image Characteristics:** Depending on the object's position relative to the focal point, the image may vary in size and orientation but remains virtual (cannot be projected on a screen).
![The text on the image outlines exercises involving the use of a diverging lens to find image distances and magnifications using ray diagrams and calculations.
**(a)** An object is placed **29.0 cm** from the lens.
**SOLUTION**
*Conceptualize:* Because the lens is diverging, the focal length is **negative**. The ray diagram for this situation is shown in figure (a).
*Categorize:* Because the lens is diverging, we expect it to form an upright, **reduced**, virtual image for any object position.
*Analyze:* Find the image distance (in cm) by using the following equation:
\[
\frac{1}{q} = \frac{1}{f} - \frac{1}{p}
\]
Find the magnification of the image from the following equation:
\[
M = -\frac{q}{p} =
\]
---
**(b)** An object is placed **10.0 cm** from the lens.
**SOLUTION**
The ray diagram for this situation is shown in figure (b).
*Analyze:* Find the image distance (in cm) by using the following equation:
\[
\frac{1}{q} = \frac{1}{f} - \frac{1}{p}
\]
Find the magnification of the image from the following equation:
\[
M = -\frac{q}{p} =
\]
---
**(c)** An object is placed **6.00 cm** from the lens.
**SOLUTION**
*Analyze:* Find the image distance (in cm) by using the following equation:
\[
\frac{1}{q} = \frac{1}{f} - \frac{1}{p}
\]
Find the magnification of the image from the following equation:
\[
M =
\]
---
**Finalize:** For all three object positions, the image position is negative and the magnification is a positive number smaller than 1, which confirms that the image is virtual, smaller than the object, and upright.
---
**EXERCISE**
An object is placed in front of a diverging lens. If the image is **–6.00 cm** from the lens, and the magnified image is **25%** of the object, what are the object distance (in cm) and focal length (in cm) of the lens?
\[
\text{object distance](https://content.bartleby.com/qna-images/question/c6ac2863-162e-4262-bc0d-8a607c6a4bb3/b0dca87c-6bb6-4330-9002-0bdc5933f422/u2xcqo_processed.png)
Transcribed Image Text:The text on the image outlines exercises involving the use of a diverging lens to find image distances and magnifications using ray diagrams and calculations.
**(a)** An object is placed **29.0 cm** from the lens.
**SOLUTION**
*Conceptualize:* Because the lens is diverging, the focal length is **negative**. The ray diagram for this situation is shown in figure (a).
*Categorize:* Because the lens is diverging, we expect it to form an upright, **reduced**, virtual image for any object position.
*Analyze:* Find the image distance (in cm) by using the following equation:
\[
\frac{1}{q} = \frac{1}{f} - \frac{1}{p}
\]
Find the magnification of the image from the following equation:
\[
M = -\frac{q}{p} =
\]
---
**(b)** An object is placed **10.0 cm** from the lens.
**SOLUTION**
The ray diagram for this situation is shown in figure (b).
*Analyze:* Find the image distance (in cm) by using the following equation:
\[
\frac{1}{q} = \frac{1}{f} - \frac{1}{p}
\]
Find the magnification of the image from the following equation:
\[
M = -\frac{q}{p} =
\]
---
**(c)** An object is placed **6.00 cm** from the lens.
**SOLUTION**
*Analyze:* Find the image distance (in cm) by using the following equation:
\[
\frac{1}{q} = \frac{1}{f} - \frac{1}{p}
\]
Find the magnification of the image from the following equation:
\[
M =
\]
---
**Finalize:** For all three object positions, the image position is negative and the magnification is a positive number smaller than 1, which confirms that the image is virtual, smaller than the object, and upright.
---
**EXERCISE**
An object is placed in front of a diverging lens. If the image is **–6.00 cm** from the lens, and the magnified image is **25%** of the object, what are the object distance (in cm) and focal length (in cm) of the lens?
\[
\text{object distance
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