Answer: solution

Explanation:

(i) **Kind of Lens:**

The student is using a converging lens or a convex lens. This is because the focal length is positive, and for a converging lens, the focal length is considered positive.

(ii) **Object Position Calculation:**

The lens formula relates the object distance (u), image distance (v), and focal length (f) for a lens:

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

Given that the focal length (\( f \)) is 5 m, and the image distance (\( v \)) is 30 m (positive, indicating a real image), we can find the object distance (\( u \)):

\[ \frac{1}{5} = \frac{1}{30} - \frac{1}{u} \]

Solving for \( u \):

\[ \frac{1}{u} = \frac{1}{30} - \frac{1}{5} \]

\[ \frac{1}{u} = \frac{1}{30} - \frac{6}{30} \]

\[ \frac{1}{u} = -\frac{5}{30} \]

\[ u = -6 \, \text{m} \]

So, the object should be placed 6 meters in front of the lens.

(iii) **Diagram:**

The diagram for the image formation with the given conditions can be drawn as follows:

1. Draw a converging lens with a focal length of 5 m.

2. Place the object 6 m in front of the lens (on the left side).

3. Draw a ray from the top of the object parallel to the principal axis. After passing through the lens, it should pass through the focal point on the other side (on the right side).

4. Draw a ray from the top of the object that passes through the optical center of the lens (straight through the center without deviation).

5. The point where these two rays intersect on the right side of the lens is the location of the real, inverted, and enlarged image.

Ensure the diagram is labeled with the focal point (F), optical center (O), object (O'), and image (I').

This setup illustrates the formation of a real, inverted, and enlarged image when the object is placed 6 m from the converging lens.

Answer:

i)convex lens

ii)it must placed between 2f1 and f1 to get real and inverted enlarged image.

iii) refer up