AIIMS 2005 Physics Telescopes MCQ Question

To solve the problem, we need to calculate the height of the image of the building formed by the objective lens of the telescope. We will use the lens formula and magnification concepts from optics.

Given Data:

• Focal length of the objective lens, $f_o = 200 \, \text{cm}$
• Height of the building, $h = 50 \, \text{m} = 5000 \, \text{cm}$
• Distance from the building, $d = 2 \, \text{km} = 2000 \, \text{m} = 200000 \, \text{cm}$

Step 1: Use the Lens Formula

The lens formula relates the object distance $d_o$, the image distance $d_i$, and the focal length $f$ of the lens:

$$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$

For a telescope, we consider the object distance $d_o$ as positive (the object is real and on the opposite side of the lens). Thus:

$$d_o = 200000 \, \text{cm} \quad \text{and} \quad f = 200 \, \text{cm}$$

Substituting the values into the lens formula:

$$\frac{1}{200} = \frac{1}{200000} + \frac{1}{d_i}$$

Step 2: Solve for $d_i$

To find $d_i$, we rearrange the equation:

$$\frac{1}{d_i} = \frac{1}{200} - \frac{1}{200000}$$

Calculating the right side:

1. Find a common denominator, which is $200000$:

$$\frac{1}{200} = \frac{1000}{200000}$$

1. Now substitute:

$$\frac{1}{d_i} = \frac{1000 - 1}{200000} = \frac{999}{200000}$$

1. Invert to find $d_i$:

$$d_i = \frac{200000}{999} \approx 200.2 \, \text{cm}$$

Step 3: Calculate the Magnification

The magnification $m$ produced by the objective lens is given by the formula:

$$m = -\frac{h_i}{h_o} = \frac{d_i}{d_o}$$

Where:

• $h_i$ is the height of the image,
• $h_o$ is the height of the object (the building).

Rearranging gives:

$$h_i = m \cdot h_o = \frac{d_i}{d_o} \cdot h_o$$

Substituting the values:

$$h_i = \frac{200.2}{200000} \cdot 5000$$

Calculating $h_i$:

1. First, calculate $\frac{200.2 \cdot 5000}{200000}$:

$$h_i = \frac{1001000}{200000} \approx 5.005 \, \text{cm}$$

Conclusion

Since the question provides options rounded to the nearest whole number, we round $5.005 \, \text{cm}$ to $5 \, \text{cm}$, which corresponds to option A (5 cm).

Clarification of Other Options

• Option B (10 cm): This value is too high compared to our calculated image height.
• Option C (1 cm): This value is too low, as the calculations indicate a height much greater than 1 cm.
• Option D (2 cm): This is also too low given the calculated value.

Thus, the correct answer is indeed A) 5 cm.

Summary

The height of the image formed by the objective lens is approximately 5 cm when observing a building 50 meters tall from a distance of 2 km. The use of the lens formula and magnification calculations confirms this result.