The focal lengths of objective lens and eye lens of a Galilean telescope are respectively 30 cm and 3.0 cm. telescope produces virtual, erect image of an object situated far away from it at least distance of distinct vision from the eye lens. In this condition, the magnifying power of the Galilean telescope should be:

Options

A. `+11.2`
B. `-11.2`
C. `-8.8`
D. `+8.8`

Explanation

To find the magnifying power of the Galilean telescope, we will follow these steps:

### Step 1: Understand the parameters given
We have the following information:
- Focal length of the objective lens (F_o) = 30 cm
- Focal length of the eyepiece lens (F_e) = 3 cm
- The least distance of distinct vision (D) = 25 cm

### Step 2: Write the formula for magnifying power (M)
The formula for the magnifying power (M) of a Galilean telescope is given by:
\[ M = \frac{F_o}{F_e} \]

### Step 3: Substitute the values into the formula
Now, substituting the values of the focal lengths into the formula:
\[ M = \frac{F_o}{F_e} = \frac{30 \, \text{cm}}{3 \, \text{cm}} \]

### Step 4: Calculate the magnifying power
Calculating the above expression:
\[ M = 10 \]

### Step 5: Adjust for the least distance of distinct vision
Since the telescope produces a virtual and erect image, we need to consider the adjustment for the least distance of distinct vision (D):
\[ M = \frac{F_o}{F_e} \cdot \left(1 - \frac{F_e}{D}\right) \]

Substituting the values:
\[ M = 10 \cdot \left(1 - \frac{3}{25}\right) \]

### Step 6: Simplify the expression
Calculating the term in the parentheses:
\[ 1 - \frac{3}{25} = \frac{25 - 3}{25} = \frac{22}{25} \]

Now substituting back into the magnifying power equation:
\[ M = 10 \cdot \frac{22}{25} \]

### Step 7: Final calculation
Calculating the final value:
\[ M = \frac{220}{25} = 8.8 \]

### Conclusion
Thus, the magnifying power of the Galilean telescope is:
\[ M = 8.8 \]

---