The magnifying power of an astronomical telescope is 5, the focal power of its eye piece is 10 diopters. The focal power of its objective (in diopters) is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To find the focal power of the objective lens of an astronomical telescope, we can use the relationship between the magnifying power (M), the focal length of the objective (F_o), and the focal length of the eyepiece (F_e).

### Step-by-Step Solution:

1. **Understanding the Given Values**:
- Magnifying power (M) = 5
- Focal power of the eyepiece (P_e) = 10 diopters

2. **Convert Focal Power to Focal Length**:
- The focal power (P) in diopters is related to the focal length (F) in meters by the formula:
\[ P = \frac{1}{F} \quad \text{(in meters)} \]
- Therefore, the focal length of the eyepiece (F_e) can be calculated as:
\[ F_e = \frac{1}{P_e} = \frac{1}{10 \text{ diopters}} = 0.1 \text{ m} = 10 \text{ cm} \]

3. **Using the Magnifying Power Formula**:
- The magnifying power of an astronomical telescope is given by:
\[ M = -\frac{F_o}{F_e} \]
- Rearranging this formula to find the focal length of the objective (F_o):
\[ F_o = -M \cdot F_e \]
- Substituting the known values:
\[ F_o = -5 \cdot 10 \text{ cm} = -50 \text{ cm} \]
- Since we are interested in the absolute value for focal length, we take:
\[ F_o = 50 \text{ cm} \]

4. **Calculating the Focal Power of the Objective**:
- Now, we can find the focal power of the objective lens (P_o):
\[ P_o = \frac{1}{F_o \text{ (in meters)}} \]
- Converting 50 cm to meters:
\[ F_o = 0.5 \text{ m} \]
- Therefore, the focal power of the objective is:
\[ P_o = \frac{1}{0.5} = 2 \text{ diopters} \]

5. **Final Answer**:
- The focal power of the objective lens is **2 diopters**.