In a laboratory five convex tenses `L_1, L_2, L_3, L_4` and `L_5` of focal lengths, 3 cm, 6 cm, 9 cm, 12 cm and 18 cm respectively are available. If two of these lenses form a telescope of length 12 cm and magnifying power 3, then objective and eye piece lenses respectively are

Options

A. `L_5, L_2`
B. `L_3, L_1`
C. ` L_4,L_2`
D. `L_2,L_1`

Answer

To solve the problem, we need to find two lenses from the given options that can form a telescope with a total length of 12 cm and a magnifying power of 3.

### Step-by-Step Solution:

1. **Understanding the Telescope Formulae**:
- For a telescope in normal adjustment, the length \( L \) is given by:
\[ L = f_o + f_e \]
where \( f_o \) is the focal length of the objective lens and \( f_e \) is the focal length of the eyepiece lens.
- The magnifying power \( M \) of the telescope is given by:
\[ M = -\frac{f_o}{f_e} \]
Since we are interested in the magnitude, we can write:
\[ M = \frac{f_o}{f_e} \]

2. **Setting Up the Equations**:
- From the problem, we know:
\[ L = 12 \text{ cm} \]
\[ M = 3 \]
- This gives us two equations:
1. \( f_o + f_e = 12 \) (Equation 1)
2. \( f_o = 3f_e \) (Equation 2)

3. **Substituting Equation 2 into Equation 1**:
- Substitute \( f_o \) from Equation 2 into Equation 1:
\[ 3f_e + f_e = 12 \]
\[ 4f_e = 12 \]
\[ f_e = 3 \text{ cm} \]

4. **Finding the Focal Length of the Objective Lens**:
- Now substitute \( f_e \) back into Equation 2 to find \( f_o \):
\[ f_o = 3f_e = 3 \times 3 = 9 \text{ cm} \]

5. **Identifying the Lenses**:
- The focal lengths we have found are:
- \( f_o = 9 \text{ cm} \) (Objective lens)
- \( f_e = 3 \text{ cm} \) (Eyepiece lens)
- From the given lenses:
- \( L_1 \) has a focal length of 3 cm.
- \( L_3 \) has a focal length of 9 cm.
- Therefore, the objective lens is \( L_3 \) and the eyepiece lens is \( L_1 \).

### Final Answer:
The objective lens is \( L_3 \) (9 cm) and the eyepiece lens is \( L_1 \) (3 cm).