An astronomical telesope has objective and eyepiece of focal lengths `40cm` and `4 cm` respectively. To view an object `200cm` away from the objective, the lenses must be separated by a distance `:`

Options

A. 46.0cm
B. 50.0cm
C. 54.0cm
D. 37.3cm

Explanation

To solve the problem of finding the distance between the objective and the eyepiece of an astronomical telescope, we will follow these steps:

### Step 1: Identify the given values
- Focal length of the objective lens, \( f_o = 40 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 4 \, \text{cm} \)
- Object distance from the objective lens, \( u = -200 \, \text{cm} \) (the negative sign is used because the object is on the same side as the incoming light)

### Step 2: Use the lens formula to find the image distance from the objective lens
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging this formula gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
Substituting the values for the objective lens:
\[ \frac{1}{v_o} = \frac{1}{40} + \frac{1}{-200} \]

### Step 3: Calculate the right-hand side
Calculating the right-hand side:
\[ \frac{1}{v_o} = \frac{1}{40} - \frac{1}{200} \]
Finding a common denominator (which is 200):
\[ \frac{1}{v_o} = \frac{5}{200} - \frac{1}{200} = \frac{4}{200} = \frac{1}{50} \]

### Step 4: Find the image distance \( v_o \)
Taking the reciprocal gives:
\[ v_o = 50 \, \text{cm} \]
This means the image is formed 50 cm from the objective lens.

### Step 5: Determine the distance between the lenses for normal adjustment
In normal adjustment of the telescope, the image formed by the objective lens should be at the focus of the eyepiece lens. The distance from the objective lens to the eyepiece lens in normal adjustment is given by:
\[ d = v_o + f_e \]
Substituting the values:
\[ d = 50 \, \text{cm} + 4 \, \text{cm} = 54 \, \text{cm} \]

### Final Answer
The lenses must be separated by a distance of \( 54 \, \text{cm} \).

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