An astronomical telscope has an eyepiece of focal-length 5 cm. If the angular magnification in normal adjustment is 10, the distance between the objective and eyepiece in cm is

Options

A. 110
B. 55
C. 50
D. 45

Answer

To solve the problem step by step, we will use the formula for angular magnification and the lens formula.

### Step 1: Understand the given data
- Focal length of the eyepiece (f) = 5 cm
- Angular magnification (M) = 10

### Step 2: Write the formula for angular magnification
In normal adjustment, the angular magnification (M) of a telescope is given by the formula:
\[ M = \frac{v}{u} \]
where:
- \( v \) = image distance from the eyepiece
- \( u \) = object distance from the objective lens

### Step 3: Relate the image distance to the object distance
For a telescope in normal adjustment, the image formed by the objective is at the focus of the eyepiece. Therefore, we can express \( v \) in terms of \( u \):
\[ v = M \cdot u \]
Substituting the value of \( M \):
\[ v = 10u \]

### Step 4: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting \( f = 5 \) cm and \( v = 10u \):
\[ \frac{1}{5} = \frac{1}{10u} - \frac{1}{u} \]

### Step 5: Simplify the equation
To simplify the equation, find a common denominator:
\[ \frac{1}{5} = \frac{1 - 10}{10u} \]
\[ \frac{1}{5} = \frac{-9}{10u} \]

### Step 6: Cross-multiply to solve for \( u \)
Cross-multiplying gives:
\[ 10u = -45 \]
\[ u = \frac{45}{10} = 4.5 \text{ cm} \]

### Step 7: Find \( v \)
Now, substitute \( u \) back into the equation for \( v \):
\[ v = 10u = 10 \times 4.5 = 45 \text{ cm} \]

### Step 8: Calculate the distance between the objective and eyepiece
The total distance \( D \) between the objective and the eyepiece is given by:
\[ D = u + v \]
Substituting the values we found:
\[ D = 4.5 + 45 = 49.5 \text{ cm} \]

### Final Answer
The distance between the objective and the eyepiece is approximately **49.5 cm**.

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