A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm . If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting , the angle formed by the image of the tower is x , then x is close to ( in degree )

Options

A. a) 30
B. b) 15
C. c) 60
D. d) 1

Explanation

To solve the problem step by step, we will use the concepts of magnifying power of a telescope and the geometry of angles formed by the object and its image.

### Step 1: Understand the given data
- Focal length of the objective lens (f₀) = 150 cm
- Focal length of the eyepiece (fₑ) = 5 cm
- Height of the tower (h) = 50 m = 5000 cm (since we need to keep the units consistent)
- Distance to the tower (d) = 1 km = 100000 cm

### Step 2: Calculate the magnifying power (MP) of the telescope
The magnifying power of a telescope in normal setting is given by the formula:
\[ MP = \frac{f₀}{fₑ} \]

Substituting the values:
\[ MP = \frac{150 \text{ cm}}{5 \text{ cm}} = 30 \]

### Step 3: Calculate the angle subtended by the tower at the objective lens (α)
The angle α subtended by the tower at the objective lens can be calculated using the small angle approximation:
\[ \tan(\alpha) \approx \frac{h}{d} \]

Substituting the values:
\[ \tan(\alpha) = \frac{5000 \text{ cm}}{100000 \text{ cm}} = \frac{1}{20} \]

### Step 4: Calculate the angle subtended by the image at the eyepiece (β)
Using the relationship between the angles and magnifying power:
\[ \tan(\beta) = MP \times \tan(\alpha) \]

Substituting the values:
\[ \tan(\beta) = 30 \times \frac{1}{20} = \frac{30}{20} = 1.5 \]

### Step 5: Calculate the angle β
To find β, we take the inverse tangent:
\[ \beta = \tan^{-1}(1.5) \]

Using a calculator or trigonometric tables:
\[ \beta \approx 56.31^\circ \]

### Step 6: Round to the nearest option
The closest option to 56.31° from the given choices (30°, 15°, 60°, and 1°) is 60°.

### Final Answer:
Thus, the angle formed by the image of the tower (x) is close to **60 degrees**.

---