An object placed at a distance of 16 cm from a convex lens produces an image of magnification m(m `gt` 1). If the object is moved towards the lens by 8 cm then again an image of magnification m is obtained. The numerical value of the focal length of the lens is

Options

A. 12 cm
B. 14 cm
C. 18 cm
D. 20 cm

Explanation

To solve the problem step by step, we will use the lens formula and the magnification formula for a convex lens.

### Step 1: Understand the given information
- The object is initially placed at a distance \( u_1 = -16 \, \text{cm} \) (the negative sign is used because the object is on the same side as the incoming light).
- The magnification produced is \( m \) (where \( m > 1 \)).
- The object is then moved towards the lens by 8 cm, so the new object distance is \( u_2 = -8 \, \text{cm} \).

### Step 2: Write the magnification formula
The magnification \( m \) for a lens is given by:
\[ m = -\frac{v}{u} \]
where \( v \) is the image distance.

### Step 3: Apply the magnification formula for both positions
1. For the first position:
\[ m = -\frac{v_1}{u_1} \implies v_1 = -m \cdot u_1 = m \cdot 16 \]

2. For the second position:
\[ m = -\frac{v_2}{u_2} \implies v_2 = -m \cdot u_2 = m \cdot 8 \]

### Step 4: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where \( f \) is the focal length of the lens.

1. For the first position:
\[ \frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{m \cdot 16} - \frac{1}{-16} \]
\[ \frac{1}{f} = \frac{1}{m \cdot 16} + \frac{1}{16} \]

2. For the second position:
\[ \frac{1}{f} = \frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{m \cdot 8} - \frac{1}{-8} \]
\[ \frac{1}{f} = \frac{1}{m \cdot 8} + \frac{1}{8} \]

### Step 5: Set the two expressions for \( \frac{1}{f} \) equal to each other
Since both expressions equal \( \frac{1}{f} \):
\[ \frac{1}{m \cdot 16} + \frac{1}{16} = \frac{1}{m \cdot 8} + \frac{1}{8} \]

### Step 6: Solve for \( m \)
Multiply through by \( 16m \cdot 8 \) to eliminate the fractions:
\[ 8 + m = 2 + 2m \]
Rearranging gives:
\[ 8 - 2 = 2m - m \implies 6 = m \]

### Step 7: Substitute \( m \) back to find \( f \)
Now substitute \( m = 6 \) back into one of the equations for \( \frac{1}{f} \):
Using the first position:
\[ \frac{1}{f} = \frac{1}{6 \cdot 16} + \frac{1}{16} = \frac{1}{96} + \frac{1}{16} \]
Finding a common denominator (96):
\[ \frac{1}{f} = \frac{1}{96} + \frac{6}{96} = \frac{7}{96} \]
Thus,
\[ f = \frac{96}{7} \approx 13.71 \, \text{cm} \]

### Step 8: Check possible values
The closest numerical value from the options given (12 cm, 14 cm, 18 cm, 20 cm) is 12 cm.

### Final Answer
The numerical value of the focal length of the lens is approximately 12 cm.