Responder
**a.** The image is formed 30 cm from the mirror.
**b.** The magnification is –1.
**c.** The image is real, inverted, and the same size as the object.
**d.** Two principal rays from the object converge at the image point 30 cm from the mirror, confirming the image is real, inverted, and the same size as the object.
Solución
Let's solve each part of the problem step-by-step.
### **Given:**
- **Object height (\( h_o \))**: 4 cm
- **Object distance (\( u \))**: 30 cm
- **Focal length (\( f \))**: 15 cm
- **Radius of curvature (\( R \))**: \( R = 2f = 30 \) cm
Since the object is placed at the center of curvature (\( C \)) of the concave mirror, we can use the mirror equations and magnification formulas to find the desired quantities.
---
### **a. Location of the Image**
**Mirror Equation:**
\[
\frac{1}{f} = \frac{1}{u} + \frac{1}{v}
\]
**Given:**
- \( f = 15 \) cm (concave mirror, so \( f \) is positive)
- \( u = 30 \) cm (object distance, positive as per standard sign conventions for mirrors)
**Plugging in the values:**
\[
\frac{1}{15} = \frac{1}{30} + \frac{1}{v}
\]
**Solving for \( v \):**
\[
\frac{1}{v} = \frac{1}{15} - \frac{1}{30} = \frac{2 - 1}{30} = \frac{1}{30}
\]
\[
v = 30 \text{ cm}
\]
**Conclusion:**
- The image is formed **30 cm** from the mirror, which means it is located at the **center of curvature**.
---
### **b. Magnification of the Image**
**Magnification Formula:**
\[
m = \frac{h_i}{h_o} = -\frac{v}{u}
\]
**Given:**
- \( v = 30 \) cm
- \( u = 30 \) cm
**Calculating Magnification:**
\[
m = -\frac{30}{30} = -1
\]
**Conclusion:**
- The magnification is **-1**, meaning the image is **the same size** as the object but inverted.
---
### **c. Nature of the Image**
From the magnification and image distance:
- **Real vs. Virtual**:
- Since \( v \) is positive, the image is **real**.
- **Orientation**:
- The magnification is negative (\( m = -1 \)), indicating the image is **inverted**.
**Conclusion:**
- The image is **real**, **inverted**, and the **same size** as the object.
---
### **d. Ray Tracing to Confirm Results**
To confirm the image formation, consider tracing two principal rays from the top of the object:
1. **Parallel Ray:**
- **Path**: Travels parallel to the principal axis, reflects through the focal point (\( F \)).
2. **Focal Ray:**
- **Path**: Passes through the focal point before reflection and then travels parallel to the principal axis.
**Process:**
1. **Draw the Principal Axis:**
- Mark the **mirror** at one end.
- Locate the **focal point (\( F \))** at 15 cm from the mirror.
- Locate the **center of curvature (\( C \))** at 30 cm from the mirror.
2. **Position the Object:**
- Place the object at \( C \) (30 cm from the mirror).
3. **Trace the Rays:**
- **Parallel Ray**: From the top of the object, draw a ray parallel to the principal axis, reflecting through \( F \) and converging at the image point at \( C \).
- **Focal Ray**: From the top of the object, draw a ray passing through \( F \) before hitting the mirror, reflecting parallel to the principal axis and converging at the image point at \( C \).
**Conclusion:**
- Both rays converge at the same point at 30 cm on the same side as the object, confirming the image is **real**, **inverted**, and **same size** as the object.
---
### **Summary of Answers**
**a.** The image is formed 30 cm from the mirror (at the center of curvature).
**b.** The magnification is –1.
**c.** The image is real, inverted, and the same size as the object.
**d.** *(As described above)* Two principal rays from the object converge at the image point 30 cm from the mirror, confirming the image’s characteristics.
Revisado y aprobado por el equipo de tutoría de UpStudy
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