The magnifytion power of the astronomical telescope for normal adjustment is 50cm. The focal length of the eyepiece is 2cm. The required length of the telescope for normal adjustment is

Options

A. 102cm
B. 100cm
C. 98cm
D. 25cm

Answer

To solve the problem, we need to determine the required length of the astronomical telescope for normal adjustment. The magnifying power (M) of an astronomical telescope in normal adjustment is given by the formula:

\[ M = \frac{f_o}{f_e} \]

where:
- \( f_o \) is the focal length of the objective lens,
- \( f_e \) is the focal length of the eyepiece lens.

Given:
- Magnifying power \( M = 50 \)
- Focal length of the eyepiece \( f_e = 2 \, \text{cm} \)

### Step 1: Calculate the Focal Length of the Objective Lens

Using the magnifying power formula:

\[ M = \frac{f_o}{f_e} \]

Substituting the known values:

\[ 50 = \frac{f_o}{2} \]

To find \( f_o \), we can rearrange the equation:

\[ f_o = 50 \times 2 \]

\[ f_o = 100 \, \text{cm} \]

### Step 2: Calculate the Length of the Telescope

The length of the telescope (L) in normal adjustment is the sum of the focal lengths of the objective and the eyepiece:

\[ L = f_o + f_e \]

Substituting the values we found:

\[ L = 100 \, \text{cm} + 2 \, \text{cm} \]

\[ L = 102 \, \text{cm} \]

### Conclusion

The required length of the telescope for normal adjustment is:

\[ \boxed{102 \, \text{cm}} \]

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