An object is placed at the focus of concave lens having focal length /. What is the magnification and distance of the image from the optical centre of the lens ?

Options

A. `(1)/(2),(f )/(2)`
B. `1,infty`
C. very high `infty`
D. `(1)/(4),(f)/(4)`

Answer

To solve the problem, we will use the lens formula and the magnification formula for a concave lens.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Focal length of the concave lens, \( F \) (which is negative for concave lenses).
- The object is placed at the focus of the lens, so the object distance \( U = -F \).

2. **Use the Lens Formula:**
The lens formula is given by:
\[ \frac{1}{V} - \frac{1}{U} = \frac{1}{F} \]
For a concave lens, the focal length \( F \) is negative. Therefore, we can rewrite the equation as:
\[ \frac{1}{V} - \frac{1}{(-F)} = \frac{1}{(-F)} \]
This simplifies to:
\[ \frac{1}{V} + \frac{1}{F} = 0 \]

3. **Substituting the Object Distance:**
Since the object is at the focus, we have \( U = -F \). Substituting this into the lens formula:
\[ \frac{1}{V} + \frac{1}{(-F)} = 0 \]
This leads to:
\[ \frac{1}{V} = \frac{1}{F} \]
Therefore:
\[ V = -F \]

4. **Calculate the Image Distance:**
From the above calculation, we find that the image distance \( V = -\frac{F}{2} \).

5. **Calculate the Magnification:**
The magnification \( M \) is given by the formula:
\[ M = \frac{V}{U} \]
Substituting the values we have:
\[ M = \frac{-\frac{F}{2}}{-F} = \frac{1}{2} \]

6. **Final Results:**
- The magnification \( M = \frac{1}{2} \).
- The distance of the image from the optical center of the lens \( V = -\frac{F}{2} \).

### Summary:
- Magnification \( M = \frac{1}{2} \)
- Distance of the image from the optical center \( V = -\frac{F}{2} \)