A Galilean telescope has `f_(o)= 200cm` and `f_(epsilon)= 1 cm`. The length the tube L and magnifying power if it is used to see object at a large distance in normal adjustment is

Options

A. L = 45, m = 10
B. L = 35, m = 20
C. L = 15, m = 15
D. L = 25, m = 30

Explanation

To solve the problem regarding the Galilean telescope, we need to find two things: the length of the tube (L) and the magnifying power (M) when the telescope is in normal adjustment.

### Step 1: Understanding the Focal Lengths
We are given:
- Focal length of the objective lens, \( f_o = 200 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 1 \, \text{cm} \)

### Step 2: Calculating the Length of the Tube (L)
In a Galilean telescope, the length of the tube (L) can be calculated using the formula:
\[ L = f_o - |f_e| \]
Here, \( |f_e| \) is the absolute value of the focal length of the eyepiece lens. Since the eyepiece lens is concave, its focal length is considered negative, but we take the absolute value for this calculation.

Substituting the values:
\[ L = 200 \, \text{cm} - |1 \, \text{cm}| = 200 \, \text{cm} - 1 \, \text{cm} = 199 \, \text{cm} \]

### Step 3: Calculating the Magnifying Power (M)
The magnifying power (M) of a Galilean telescope is given by the formula:
\[ M = -\frac{f_o}{f_e} \]
Substituting the values:
\[ M = -\frac{200 \, \text{cm}}{1 \, \text{cm}} = -200 \]
Since magnifying power is typically expressed as a positive value, we take the absolute value:
\[ M = 200 \]

### Final Answers
- Length of the tube \( L = 199 \, \text{cm} \)
- Magnifying power \( M = 200 \)

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