Choose the correct alternative corresponding to the object distance 'u', image distance 'v' and the focal length 'F' of a converging lens from the following .
(i) The average speed of the image as the object moves with the uniform speed from distance `(3F)/(4)` to `(F)/(2)` is greater than the average speed of the image as the object moves with same speed from distance `(F)/(2) "to" (F)/(4)`
(ii) The minimum distance between a real object and its real image in case of a converging lens is `4F` where F is its focal length.

Options

A. both are correct
B. both are incorrect
C. (i) is correct , (ii) is incorrect
D. (i) is incorrect , (ii) is correct

Answer

To solve the problem, we need to analyze the two statements provided regarding the behavior of a converging lens.

### Step 1: Analyze Statement (i)

**Statement (i)**: The average speed of the image as the object moves with uniform speed from distance \( \frac{3F}{4} \) to \( \frac{F}{2} \) is greater than the average speed of the image as the object moves with the same speed from distance \( \frac{F}{2} \) to \( \frac{F}{4} \).

1. **Using the Lens Formula**:
The lens formula is given by:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Where \( v \) is the image distance, \( u \) is the object distance, and \( f \) is the focal length.

2. **Calculate Image Distance for the First Case**:
- For \( u = -\frac{3F}{4} \):
\[ \frac{1}{v_1} = \frac{1}{f} + \frac{4}{3F} \]
- Solving gives:
\[ v_1 = \frac{3F}{1} = 3F \]

- For \( u = -\frac{F}{2} \):
\[ \frac{1}{v_2} = \frac{1}{f} + \frac{2}{F} \]
- Solving gives:
\[ v_2 = 2F \]

3. **Calculate Average Speed**:
- The average speed of the image when the object moves from \( \frac{3F}{4} \) to \( \frac{F}{2} \):
\[ \text{Distance covered} = v_1 - v_2 = 3F - 2F = F \]
- Time taken:
\[ t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{F}{\alpha_0} \]
- Average speed:
\[ \text{Average speed}_1 = \frac{F}{t_1} = \frac{F}{\frac{F}{\alpha_0}} = \alpha_0 \]

- For the second case from \( \frac{F}{2} \) to \( \frac{F}{4} \):
- Calculate \( v \) for both positions similarly and find the average speed.

4. **Comparison**:
- After calculating both average speeds, we find that the average speed from \( \frac{3F}{4} \) to \( \frac{F}{2} \) is indeed greater than from \( \frac{F}{2} \) to \( \frac{F}{4} \).

### Conclusion for Statement (i):
The statement is **true**.

---

### Step 2: Analyze Statement (ii)

**Statement (ii)**: The minimum distance between a real object and its real image in case of a converging lens is \( 4F \).

1. **Using the Lens Formula Again**:
- The total distance \( x \) between the object and the image is given by:
\[ x = u + v \]

2. **Find Minimum Distance**:
- From the lens formula, we know:
\[ v = \frac{u \cdot f}{u + f} \]
- Substitute \( v \) into \( x \):
\[ x = u + \frac{u \cdot f}{u + f} \]
- Differentiate \( x \) with respect to \( u \) and set it to zero to find the minimum.

3. **Calculate Minimum**:
- After differentiation and solving, we find \( u = 2f \).
- Substitute \( u \) back to find \( v \):
\[ v = 2f \]
- Therefore, the minimum distance \( x = u + v = 2f + 2f = 4f \).

### Conclusion for Statement (ii):
The statement is **true**.

---

### Final Answer:
Both statements are true.