The focal length of the objective of an astronomical telescope is 75 cm and that of the eye-piece is 5 cm. If the final image is formed at the least distance of distinct vision from the eye, calculate the magnify- ing power of the telescope.

Answer

To calculate the magnifying power of the astronomical telescope, we can follow these steps:

### Step 1: Understand the Parameters
We are given:
- Focal length of the objective lens (Fo) = 75 cm
- Focal length of the eyepiece lens (Fe) = 5 cm
- Final image distance (D) = 25 cm (least distance of distinct vision)

### Step 2: Use the Formula for Magnifying Power
The magnifying power (M) of an astronomical telescope when the final image is at the least distance of distinct vision can be calculated using the formula:

\[ M = -\frac{F_o}{F_e} \left(1 + \frac{F_e}{D}\right) \]

### Step 3: Substitute the Values
Now, we substitute the values into the formula:
- \(F_o = 75 \, \text{cm}\)
- \(F_e = 5 \, \text{cm}\)
- \(D = 25 \, \text{cm}\)

Substituting these values into the formula:

\[ M = -\frac{75}{5} \left(1 + \frac{5}{25}\right) \]

### Step 4: Calculate Each Component
First, calculate \(-\frac{75}{5}\):

\[ -\frac{75}{5} = -15 \]

Next, calculate \(1 + \frac{5}{25}\):

\[ \frac{5}{25} = 0.2 \quad \Rightarrow \quad 1 + 0.2 = 1.2 \]

### Step 5: Combine the Results
Now, multiply the two results together:

\[ M = -15 \times 1.2 = -18 \]

### Step 6: Interpret the Result
The magnifying power of the telescope is \(-18\). The negative sign indicates that the final image is inverted. Therefore, the image is 18 times larger than the object.

### Final Answer
The magnifying power of the telescope is **18** (inverted).

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