Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
(a) An object is placed 29.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Conceptualize Because the lens is diverging, the focal length is negative ✓ Categorize Because the lens is diverging, we expect it to form an upright, reduced Analyze Find the image distance (in cm) by using the following equation: 9 q= M = - Find the magnification of the image from the following equation: -음. f P (b) An object is placed 10.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION The ray diagram for this situation is shown in figure (b). Analyze Find the image distance (in cm) by using the following equation: M = cm 1 97 q= Find the magnification of the image from the following equation: 1 9 q= (c) An object is placed 6.00 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Analyze Find the image distance (in cm) by using the following equation: M = cm . The ray diagram for this situation is shown in figure (a). , virtual image for any object position. cm Find the magnification of the image from the following equation:

(a) An object is placed 29.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Conceptualize Because the lens is diverging, the focal length is negative ✓ Categorize Because the lens is diverging, we expect it to form an upright, reduced Analyze Find the image distance (in cm) by using the following equation: 9 q= M = - Find the magnification of the image from the following equation: -음. f P (b) An object is placed 10.0 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION The ray diagram for this situation is shown in figure (b). Analyze Find the image distance (in cm) by using the following equation: M = cm 1 97 q= Find the magnification of the image from the following equation: 1 9 q= (c) An object is placed 6.00 cm from the lens. Construct a ray diagram, find the image distance, and describe the image. SOLUTION Analyze Find the image distance (in cm) by using the following equation: M = cm . The ray diagram for this situation is shown in figure (a). , virtual image for any object position. cm Find the magnification of the image from the following equation:

icon
Related questions
Question
100%

I need help with this question I can ask more questions to finish the other parts.

**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).
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
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
Expert Solution
steps

Step by step

Solved in 4 steps with 18 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer