A telescope `(f_0=140 cm,f_e=5cm)` is used to view a distant tower 100 m high situated at a distance 3 km. The tower is made of iron channels consisting rectangular grades of approximate size `2 xx2` m each in the image formed by telescope's objective , each square channel will appear to be of area

Options

A. `100 cm^2`
B. `86 cm^2`
C. `70 cm^2`
D. `60 cm^2`

Answer

To solve the problem step by step, we will follow the outlined procedure to find the area of the image formed by the telescope's objective.

### Step 1: Understand the given parameters
- Focal length of the objective lens, \( f_0 = 140 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 5 \, \text{cm} \)
- Height of the tower, \( h = 100 \, \text{m} \)
- Distance to the tower, \( d = 3 \, \text{km} = 3000 \, \text{m} \)

### Step 2: Calculate the magnification of the telescope
The magnification \( M \) of the telescope can be calculated using the formula:
\[ M = \frac{f_0}{f_e} \]
Substituting the values:
\[ M = \frac{140 \, \text{cm}}{5 \, \text{cm}} = 28 \]

### Step 3: Calculate the angle \( \alpha \)
To find the angle \( \alpha \), we can use the tangent function:
\[ \tan \alpha = \frac{\text{height of the tower}}{\text{distance to the tower}} = \frac{100 \, \text{m}}{3000 \, \text{m}} = \frac{1}{30} \]

### Step 4: Calculate the height of the image \( h' \)
Using the magnification, we can express the height of the image \( h' \) formed by the telescope's objective:
\[ h' = M \cdot \tan \alpha \]
Substituting the values:
\[ h' = 28 \cdot \tan \alpha = 28 \cdot \frac{100}{3000} = 28 \cdot \frac{1}{30} \cdot 100 = \frac{2800}{30} \approx 93.33 \, \text{cm} \]

### Step 5: Calculate the area of the image
The area of the image formed by each square channel can be calculated as:
\[ \text{Area} = (h')^2 \]
Substituting the value of \( h' \):
\[ \text{Area} = (9.33 \, \text{cm})^2 \approx 87.56 \, \text{cm}^2 \]

### Final Answer
The area of each square channel in the image formed by the telescope's objective is approximately \( 87.56 \, \text{cm}^2 \).

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