An object is placed in front of two convex lenses one by one at a distance u from the lens. The focal lengths of the lenses are 30 cm and 15 cm respectively. If the size of image formed in the two cases is same, then u is

Options

A. 15 cm
B. 20 cm
C. 25 cm
D. 30 cm

Answer

To solve the problem, we need to determine the object distance \( u \) such that the size of the images formed by two convex lenses with different focal lengths is the same. The focal lengths of the lenses are given as \( f_1 = 30 \, \text{cm} \) and \( f_2 = 15 \, \text{cm} \).

### Step-by-Step Solution:

1. **Understand the Lens Formula**:
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]

2. **Express Magnification**:
The magnification \( m \) produced by a lens is given by:
\[ m = \frac{v}{u} \]
From the lens formula, we can express \( v \) in terms of \( u \) and \( f \):
\[ v = \frac{fu}{u - f} \]
Therefore, the magnification can be rewritten as:
\[ m = \frac{fu}{u - f} \cdot \frac{1}{u} = \frac{f}{u - f} \]

3. **Set Up the Equations for Both Lenses**:
For the first lens with focal length \( f_1 = 30 \, \text{cm} \):
\[ m_1 = \frac{30}{u - 30} \]
For the second lens with focal length \( f_2 = 15 \, \text{cm} \):
\[ m_2 = \frac{15}{u - 15} \]

4. **Equate the Magnifications**:
Since the sizes of the images formed by both lenses are the same, we have:
\[ m_1 = -m_2 \]
This gives:
\[ \frac{30}{u - 30} = -\frac{15}{u - 15} \]

5. **Cross-Multiply and Solve for \( u \)**:
Cross-multiplying yields:
\[ 30(u - 15) = -15(u - 30) \]
Expanding both sides:
\[ 30u - 450 = -15u + 450 \]
Bringing all terms involving \( u \) to one side:
\[ 30u + 15u = 450 + 450 \]
\[ 45u = 900 \]
Dividing both sides by 45 gives:
\[ u = 20 \, \text{cm} \]

### Final Answer:
The object distance \( u \) is \( 20 \, \text{cm} \).