The magnifications for two positions of image of the same lens are 3 and 2, of a fix point object on a fix screen when the distance between the two positions of lens is 30 cm .
The focal lengths of the lens is

Options

A. 180 cm
B. 90 cm
C. 30 cm
D. 15 cm

Answer

To solve the problem, we need to find the focal length of the lens given the magnifications for two positions of the lens and the distance between these positions.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Magnification for the first position of the lens, \( m_1 = 3 \)
- Magnification for the second position of the lens, \( m_2 = 2 \)
- Distance between the two positions of the lens, \( d = 30 \, \text{cm} \)

2. **Use the Formula for Focal Length:**
The formula to find the focal length \( f \) of the lens when the magnifications and distance between the lens positions are known is:
\[ f = \frac{d \cdot m_1 \cdot m_2}{m_1 - m_2} \]

3. **Substitute the Values into the Formula:**
Substitute \( d = 30 \, \text{cm} \), \( m_1 = 3 \), and \( m_2 = 2 \) into the formula:
\[ f = \frac{30 \cdot 3 \cdot 2}{3 - 2} \]

4. **Calculate the Denominator:**
Calculate \( m_1 - m_2 \):
\[ 3 - 2 = 1 \]

5. **Calculate the Numerator:**
Calculate \( 30 \cdot 3 \cdot 2 \):
\[ 30 \cdot 3 = 90 \]
\[ 90 \cdot 2 = 180 \]

6. **Final Calculation of Focal Length:**
Now, substitute back into the formula:
\[ f = \frac{180}{1} = 180 \, \text{cm} \]

7. **Conclusion:**
The focal length of the lens is \( 180 \, \text{cm} \).

### Summary of the Solution:
The focal length of the lens is \( 180 \, \text{cm} \).