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The question appears to involve optical principles, specifically related to the **rules of refraction** and the formation of an image through a convex lens. Let me transcribe and analyze the problem for solving it.

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**Translation of the problem:**

1. **What are the rules of refraction?**  
2. An object of 3 cm height is placed along the principal axis of a convex lens. The focal length of the lens is 30 cm, and the object is placed 10 cm away from the lens. Determine:  
   - The position of the image,  
   - Its nature (real/virtual),  
   - Its size.  

Also, can this image be projected onto a screen?

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Let me solve this step by step:  
We will use the **lens formula** and magnification to solve the problem.

### Lens Formula:
\[
\frac{1}{f} = \frac{1}{v} - \frac{1}{u}
\]
Where:  
- \(f = 30 \, \text{cm}\) (focal length of the convex lens),  
- \(u = -10 \, \text{cm}\) (object distance, taken as negative for convex lens).  

Substituting values into the formula:
\[
\frac{1}{30} = \frac{1}{v} - \frac{1}{-10}
\]
\[
\frac{1}{30} = \frac{1}{v} + \frac{1}{10}
\]
\[
\frac{1}{v} = \frac{1}{30} - \frac{1}{10}
\]
\[
\frac{1}{v} = \frac{1 - 3}{30} = -\frac{2}{30} = -\frac{1}{15}
\]

Thus:
\[
v = -15 \, \text{cm}
\]

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### Interpretation of \(v = -15\):
The image is formed **15 cm away** on the same side as the object. Since the image distance (\(v\)) is negative, it indicates that the image is **virtual** and upright.

---

### Magnification:
The magnification (\(M\)) is given by:
\[
M = \frac{v}{u} = \frac{-15}{-10} = 1.5
\]

The height of the image (\(h_i\)) can be found using:
\[
h_i = M \cdot h_o = 1.5 \cdot 3 = 4.5 \, \text{cm}
\]

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### Final Results:
1. **Position of the Image**: 15 cm on the same side as the object.  
2. **Nature of the Image**: Virtual and upright.  
3. **Size of the Image**: 4.5 cm (larger than the object).  
4. **Can it be projected?**: No, a virtual image cannot be projected onto a screen.

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Would you like me to explain any part in more detail?  

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### Related Questions:
1. What are the conditions for an image to be real or virtual in a convex lens?  
2. How does the focal length affect the image size and position?  
3. Why is object distance (\(u\)) taken as negative in the lens formula?  
4. How do you differentiate between concave and convex lenses using ray diagrams?  
5. What practical applications involve using a convex lens to create virtual images?  

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### Tip:
Always use sign conventions carefully when applying the lens formula. For convex lenses, object distances are negative, but focal length is positive.

अपवर्तन के क्या नियम हैं? 3 सेमी की वस्तु उत्तल लेंस के मुख्य अक्ष के लंबवत रखी है। लेंस का फोकस दूरी 30 सेमी और लेंस से वस्तु की दूरी 10 सेमी है। प्रतिबिंब की स्थिति, आकार एवं प्रकार ज्ञात कीजिए। क्या हम इस प्रतिबिंब को पर्दे पर देख सकते हैं?

Explore the rules of refraction and learn how to calculate the position, size, and type of image formed by a convex lens. Includes key formulas such as the lens formula and magnification. Perfect for students in Grades 9-10.

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