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 solve the problem, we will follow these steps:

### Step 1: Identify the given values
- The object distance (u) is given as 10 cm. Since the object is placed in front of the lens, we take it as negative in lens formula:
\[ u = -10 \, \text{cm} \]
- The magnification (m) is given as 0.5.

### Step 2: Use the magnification formula
The magnification (m) for a lens is given by the formula:
\[ m = \frac{h'}{h} = -\frac{v}{u} \]
Where:
- \( h' \) is the height of the image,
- \( h \) is the height of the object,
- \( v \) is the image distance,
- \( u \) is the object distance.

Given that \( m = 0.5 \), we can rearrange the magnification formula to find \( v \):
\[ 0.5 = -\frac{v}{-10} \]
This simplifies to:
\[ v = 0.5 \times 10 = 5 \, \text{cm} \]

### Step 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 can be rewritten as:
\[ \frac{1}{f} = \frac{1}{5} + \frac{1}{10} \]

### Step 4: Find a common denominator and solve for \( f \)
The least common multiple of 5 and 10 is 10. Thus:
\[ \frac{1}{f} = \frac{2}{10} + \frac{1}{10} = \frac{3}{10} \]
Now, taking the reciprocal to find \( f \):
\[ f = \frac{10}{3} \, \text{cm} \]

### Step 5: Determine the type of lens
Since the focal length \( f \) is positive, we conclude that the lens is a **convex lens**.

### Final Answer
The lens is a **convex lens** with a focal length of \( \frac{10}{3} \, \text{cm} \).

---