A Galilean telescope of `10`-fold magnification has the length of `45 cm` when adjusted to infinity. Determine :
(a) the focal lengths of the telescope's objective and ocular,
(b) by what distance the ocualr should be displaced to adjust the telescope to the distance of `50 m.`

Options

A. 50 cm, 5cm
B. `50cm, -5cm`
C. `(450)/(11)cm, (45)/(11)cm`
D. `(450)/(11)cm, (-45)/(11)cm`

Explanation

To solve the problem step by step, let's break it down into two parts as per the question.

### Part (a): Determine the focal lengths of the telescope's objective and ocular.

1. **Understanding the Magnification Formula**:
The angular magnification (M) of a Galilean telescope is given by the formula:
\[ M = \frac{f_o}{f_e} \]
where \( f_o \) is the focal length of the objective lens and \( f_e \) is the focal length of the ocular lens.

2. **Given Values**:
We know that the magnification \( M = 10 \) (10-fold magnification) and the length of the telescope \( L = 45 \, \text{cm} \).

3. **Setting Up the Equation**:
From the magnification formula, we can express the focal length of the objective in terms of the focal length of the ocular:
\[ f_o = M \cdot f_e = 10 f_e \quad \text{(Equation 1)} \]

4. **Using the Length of the Telescope**:
The length of the telescope when adjusted to infinity is given by:
\[ L = f_o - f_e \]
Substituting Equation 1 into this equation gives:
\[ 45 = 10 f_e - f_e \]
Simplifying this:
\[ 45 = 9 f_e \]
Therefore, we can find \( f_e \):
\[ f_e = \frac{45}{9} = 5 \, \text{cm} \]

5. **Finding \( f_o \)**:
Now substituting \( f_e \) back into Equation 1 to find \( f_o \):
\[ f_o = 10 f_e = 10 \times 5 = 50 \, \text{cm} \]

### Summary of Part (a):
- Focal length of the ocular lens \( f_e = 5 \, \text{cm} \)
- Focal length of the objective lens \( f_o = 50 \, \text{cm} \)

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### Part (b): Determine the distance the ocular should be displaced to adjust the telescope to a distance of 50 m.

1. **Using the Lens Formula**:
We will use the lens formula for the objective lens:
\[ \frac{1}{f_o} = \frac{1}{s_o} + \frac{1}{s_d} \]
where \( s_o \) is the object distance and \( s_d \) is the image distance.

2. **Given Object Distance**:
The object distance \( s_o \) is given as \( 50 \, \text{m} = 5000 \, \text{cm} \).

3. **Calculating the Image Distance**:
Rearranging the lens formula gives:
\[ s_d = \frac{s_o \cdot f_o}{s_o - f_o} \]
Substituting the known values:
\[ s_d = \frac{5000 \times 50}{5000 - 50} = \frac{250000}{4950} \approx 50.505 \, \text{cm} \]

4. **Finding the New Length of the Telescope**:
The new length \( L' \) when adjusted to the distance of 50 m can be expressed as:
\[ L' = s_d + f_e = 50.505 + 5 = 55.505 \, \text{cm} \]

5. **Calculating the Displacement**:
The displacement \( \Delta L \) of the ocular is given by:
\[ \Delta L = L' - L = 55.505 - 45 = 10.505 \, \text{cm} \]

### Summary of Part (b):
- The ocular should be displaced by approximately \( 10.505 \, \text{cm} \) to adjust the telescope to the distance of \( 50 \, \text{m} \).

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