An object is placed at 10 cm from a lens and real image is formed with magnification of 0.5. Then the lens is

Options

A. concave with focal length of 10/3 cm
B. convex with focal length of 10/3 cm
C. concave with focal length of 10 cm
D. convex with focal length of 10 cm

Explanation

To determine the type of lens used when an object is placed at 10 cm from the lens and a real image is formed with a magnification of 0.5, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Object distance (U) = -10 cm (the object distance is taken as negative in lens formula convention)
- Magnification (M) = 0.5

2. **Use the Magnification Formula:**
The magnification (M) is given by the formula:
\[ M = \frac{V}{U} \]
where V is the image distance. Rearranging this gives:
\[ V = M \times U \]
Substituting the known values:
\[ V = 0.5 \times (-10) = -5 \text{ cm} \]
Since the image is real, V should be positive, indicating that we need to correct our understanding of the sign convention. Thus, we take the absolute value:
\[ V = 5 \text{ cm} \]

3. **Apply the Lens Formula:**
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
Substituting the values of V and U:
\[ \frac{1}{F} = \frac{1}{5} - \frac{1}{-10} \]
This simplifies to:
\[ \frac{1}{F} = \frac{1}{5} + \frac{1}{10} \]

4. **Finding a Common Denominator:**
The common denominator for 5 and 10 is 10. Thus:
\[ \frac{1}{F} = \frac{2}{10} + \frac{1}{10} = \frac{3}{10} \]

5. **Calculate Focal Length (F):**
Taking the reciprocal gives:
\[ F = \frac{10}{3} \text{ cm} \]

6. **Determine the Type of Lens:**
Since the focal length (F) is positive, this indicates that the lens is a **convex lens**.

### Final Answer:
The lens is a **convex lens**.