A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length 6.0 cm. The magnifying power of the telescope for distinct vision adjustment is:

Options

A. 30
B. 29
C. 19
D. 31

Explanation

To find the magnifying power of the telescope, we will use the formula for the magnifying power (M) of a telescope for distinct vision adjustment:

\[ M = \frac{f_o}{f_e} + \frac{f_o}{v} \]

Where:
- \( f_o \) = focal length of the objective lens
- \( f_e \) = focal length of the eyepiece
- \( v \) = distance of distinct vision (which is typically taken as 25 cm)

### Step 1: Identify the given values
- Focal length of the objective lens, \( f_o = 150 \) cm
- Focal length of the eyepiece, \( f_e = 6 \) cm
- Distance of distinct vision, \( v = 25 \) cm

### Step 2: Substitute the values into the formula
Now we will substitute the values into the magnifying power formula:

\[ M = \frac{150}{6} + \frac{150}{25} \]

### Step 3: Calculate each term
1. Calculate \( \frac{150}{6} \):
\[ \frac{150}{6} = 25 \]

2. Calculate \( \frac{150}{25} \):
\[ \frac{150}{25} = 6 \]

### Step 4: Add the results
Now we will add the two results together to find the total magnifying power:

\[ M = 25 + 6 = 31 \]

### Final Answer
The magnifying power of the telescope for distinct vision adjustment is \( M = 31 \).