The magnifying power of an astronomical telescope for normal adjustment is 10 and the length of the telescope is 110cm..Find the magnifying power of the telescope when the image is formed at the least distance of distinct vision for normal eye.

Options

A. 14
B. 48
C. 28
D. 52

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the given data
- The magnifying power of the telescope for normal adjustment (m) = 10
- Length of the telescope (L) = 110 cm
- Least distance of distinct vision (d) = 25 cm

### Step 2: Relate the magnifying power to the focal lengths
For a telescope in normal adjustment, the magnifying power (m) is given by the formula:
\[ m = \frac{f_o}{f_e} \]
Where:
- \( f_o \) = focal length of the objective lens
- \( f_e \) = focal length of the eyepiece lens

Given that \( m = 10 \), we can write:
\[ 10 = \frac{f_o}{f_e} \]
This implies:
\[ f_o = 10 \cdot f_e \]

### Step 3: Use the length of the telescope
The length of the telescope is the sum of the focal lengths of the objective and eyepiece:
\[ L = f_o + f_e \]
Substituting the expression for \( f_o \):
\[ 110 = 10 \cdot f_e + f_e \]
This simplifies to:
\[ 110 = 11 \cdot f_e \]

### Step 4: Solve for \( f_e \)
Now, we can solve for \( f_e \):
\[ f_e = \frac{110}{11} = 10 \, \text{cm} \]

### Step 5: Find \( f_o \)
Using the value of \( f_e \) to find \( f_o \):
\[ f_o = 10 \cdot f_e = 10 \cdot 10 = 100 \, \text{cm} \]

### Step 6: Calculate the new magnifying power when the image is formed at the least distance of distinct vision
When the image is formed at the least distance of distinct vision, the new magnifying power \( m_2 \) is given by:
\[ m_2 = -f_o \left( \frac{1}{d} + \frac{1}{f_e} \right) \]
Substituting the values:
\[ m_2 = -100 \left( \frac{1}{25} + \frac{1}{10} \right) \]

### Step 7: Calculate the terms inside the parentheses
Calculating the fractions:
\[ \frac{1}{25} = 0.04 \]
\[ \frac{1}{10} = 0.1 \]
Thus:
\[ \frac{1}{25} + \frac{1}{10} = 0.04 + 0.1 = 0.14 \]

### Step 8: Calculate \( m_2 \)
Now substituting back into the equation for \( m_2 \):
\[ m_2 = -100 \cdot 0.14 = -14 \]

### Final Answer
The magnifying power of the telescope when the image is formed at the least distance of distinct vision is:
\[ m_2 = -14 \]