A telescope, consisting of an objective of focal length 60 cm and a single eye lens of focal length 15 cm is focused on a distant object in such a way that parallel rays comes out from the eye lens. If the object subtends an angle 2° at the objective, the angular width of Che image is

Options

A. `10^(@)`
B. `8^(@)`
C. `5^(@)`
D. `1//6^(@)`

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Identify the given values
- Focal length of the objective lens, \( f_o = 60 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 15 \, \text{cm} \)
- Angle subtended by the object at the objective lens, \( \alpha = 2^\circ \)

### Step 2: Understand the relationship between angular width and magnification
The magnification \( m \) of a telescope can be expressed in terms of the angles subtended by the object and the image:
\[ m = \frac{\beta}{\alpha} \]
where:
- \( \beta \) is the angle subtended by the image at the eyepiece,
- \( \alpha \) is the angle subtended by the object at the objective.

### Step 3: Relate magnification to focal lengths
The magnification can also be expressed in terms of the focal lengths of the objective and the eyepiece:
\[ m = \frac{f_o}{f_e} \]

### Step 4: Substitute the values into the magnification equation
From the two expressions for magnification, we can set them equal to each other:
\[ \frac{\beta}{\alpha} = \frac{f_o}{f_e} \]
Substituting the known values:
\[ \frac{\beta}{2^\circ} = \frac{60 \, \text{cm}}{15 \, \text{cm}} \]

### Step 5: Calculate the magnification
Calculating the right side:
\[ \frac{60}{15} = 4 \]
Thus, we have:
\[ \frac{\beta}{2^\circ} = 4 \]

### Step 6: Solve for \( \beta \)
Now, we can solve for \( \beta \):
\[ \beta = 4 \times 2^\circ = 8^\circ \]

### Step 7: Conclusion
The angular width of the image \( \beta \) is \( 8^\circ \).

### Final Answer:
The angular width of the image is \( 8^\circ \).

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