An astronimical telescope is to be designed to have a magnifying power of 50 in normal adjustment. If the length of the tube is 102 cm, find the powers of the objective and the eyepiece.

Answer

To solve the problem of designing an astronomical telescope with a magnifying power of 50 and a tube length of 102 cm, we can follow these steps:

### Step 1: Understand the relationship between magnifying power and focal lengths
The magnifying power (M) of an astronomical telescope in normal adjustment is given by the formula:
\[ M = \frac{f_o}{f_e} \]
where:
- \( f_o \) = focal length of the objective lens
- \( f_e \) = focal length of the eyepiece lens

Given that the magnifying power \( M = 50 \), we can write:
\[ \frac{f_o}{f_e} = 50 \]
This implies:
\[ f_o = 50 f_e \] (Equation 1)

### Step 2: Use the length of the telescope
The length of the telescope (L) is the sum of the focal lengths of the objective and the eyepiece:
\[ L = f_o + f_e \]
Given that the length of the telescope is 102 cm, we can write:
\[ f_o + f_e = 102 \] (Equation 2)

### Step 3: Substitute Equation 1 into Equation 2
Now, we can substitute Equation 1 into Equation 2:
\[ 50 f_e + f_e = 102 \]
This simplifies to:
\[ 51 f_e = 102 \]

### Step 4: Solve for the focal length of the eyepiece
Now, we can solve for \( f_e \):
\[ f_e = \frac{102}{51} = 2 \text{ cm} \]

### Step 5: Find the focal length of the objective
Using Equation 1, we can now find \( f_o \):
\[ f_o = 50 f_e = 50 \times 2 = 100 \text{ cm} \]

### Step 6: Calculate the powers of the lenses
The power (P) of a lens is given by the formula:
\[ P = \frac{1}{f} \]
where \( f \) is in meters.

1. **Power of the eyepiece (P_e)**:
\[ P_e = \frac{1}{f_e} = \frac{1}{2 \text{ cm}} = \frac{1}{0.02 \text{ m}} = 50 \text{ diopters} \]

2. **Power of the objective (P_o)**:
\[ P_o = \frac{1}{f_o} = \frac{1}{100 \text{ cm}} = \frac{1}{1 \text{ m}} = 1 \text{ diopter} \]

### Final Answer
- Power of the eyepiece = 50 diopters
- Power of the objective = 1 diopter

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