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.

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?

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: $v = -15 \, \text{cm}$

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: $h_i = M \cdot h_o = 1.5 \cdot 3 = 4.5 \, \text{cm}$

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.

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

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?

Tip:

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