A telescope has an objective lens of focal length `200cm` and an eye piece with focal length `2cm`. If this telescope is used to see a 50 meter tall building at a distance of `2km,` what is the height of the image of the building formed by the objective lens?

Options

A. 5 cm
B. 10 cm
C. 1 cm
D. 2 cm

Answer

To find the height of the image of the building formed by the objective lens of the telescope, we can follow these steps:

### Step 1: Identify the given values
- Focal length of the objective lens, \( f = 200 \, \text{cm} = 2 \, \text{m} \)
- Object distance, \( u = 2 \, \text{km} = 2000 \, \text{m} \)
- Height of the object (building), \( h_o = 50 \, \text{m} \)

### Step 2: Use the lens formula to find the image distance
Since the object is very far away (2 km), we can consider the rays coming from the object to be parallel. In this case, the image will be formed at the focal point of the lens.

Thus, the image distance \( v \) can be approximated as:
\[ v = f = 2 \, \text{m} \]

### Step 3: Calculate the magnification
The magnification \( m \) of a lens is given by the formula:
\[ m = \frac{h_i}{h_o} = \frac{v}{u} \]
Where:
- \( h_i \) is the height of the image
- \( h_o \) is the height of the object
- \( v \) is the image distance
- \( u \) is the object distance

### Step 4: Substitute the known values into the magnification formula
We can rearrange the magnification formula to find the height of the image \( h_i \):
\[ h_i = m \cdot h_o \]
Substituting for \( m \):
\[ h_i = \frac{v}{u} \cdot h_o \]
Substituting the values:
\[ h_i = \frac{2 \, \text{m}}{2000 \, \text{m}} \cdot 50 \, \text{m} \]

### Step 5: Calculate \( h_i \)
Calculating \( h_i \):
\[ h_i = \frac{2}{2000} \cdot 50 = \frac{100}{2000} = 0.05 \, \text{m} = 5 \, \text{cm} \]

### Conclusion
The height of the image of the building formed by the objective lens is \( 5 \, \text{cm} \).

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