Magnification produced by astronominal telescope for normal adjustment is 10 and length of telescope is 1.1m. The magnification when the image is formed at least distance of distinct vision `(D = 25cm)` is-

Options

A. 6
B. 14
C. 16
D. 18

Explanation

To solve the problem, we need to find the magnification produced by an astronomical telescope when the image is formed at the least distance of distinct vision (D = 25 cm).

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- Magnification (M) for normal adjustment = 10
- Length of the telescope (L) = 1.1 m = 110 cm
- Least distance of distinct vision (D) = 25 cm

2. **Relating Focal Lengths:**
- The magnification for a telescope in normal adjustment is given by:
\[ M = \frac{f_o}{f_e} \]
where \( f_o \) is the focal length of the objective and \( f_e \) is the focal length of the eyepiece.
- From the information given, we can write:
\[ f_o = 10 f_e \quad (1) \]

3. **Using the Length of the Telescope:**
- The total length of the telescope is the sum of the focal lengths:
\[ L = f_o + f_e \]
- Substituting the values, we have:
\[ 110 = f_o + f_e \quad (2) \]

4. **Substituting Equation (1) into Equation (2):**
- Replacing \( f_o \) in equation (2) with \( 10 f_e \):
\[ 110 = 10 f_e + f_e \]
\[ 110 = 11 f_e \]
- Solving for \( f_e \):
\[ f_e = \frac{110}{11} = 10 \text{ cm} \]

5. **Finding \( f_o \):**
- Now substituting \( f_e \) back into equation (1):
\[ f_o = 10 f_e = 10 \times 10 = 100 \text{ cm} \]

6. **Calculating Magnification for D = 25 cm:**
- The magnification when the image is formed at the least distance of distinct vision is given by:
\[ M' = \frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right) \]
- Substituting the values:
\[ M' = \frac{100}{10} \left(1 + \frac{10}{25}\right) \]
\[ M' = 10 \left(1 + 0.4\right) = 10 \times 1.4 = 14 \]

7. **Final Answer:**
- The magnification when the image is formed at the least distance of distinct vision is **14**.