A simple telescope, consisting of an objective of focal length `60cm` and a single eye lens of focal length `5cm` is focused on a distant object in such a way that parallel rays emerge form the eye lens. If the object makes an angle of `2^(@)` at the objective, the angular width if the image is

Options

A. `10^(@)`
B. `24^(@)`
C. `50^(@)`
D. `48^(@)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the setup of the telescope
The telescope consists of two lenses:
- An objective lens with a focal length \( f_o = 60 \, \text{cm} \)
- An eye lens (eyepiece) with a focal length \( f_e = 5 \, \text{cm} \)

The telescope is focused on a distant object, meaning that the rays coming from the object are parallel when they reach the objective lens.

### Step 2: Determine the magnification of the telescope
The magnification \( m \) of a simple telescope is given by the formula:
\[ m = -\frac{f_o}{f_e} \]
Substituting the values:
\[ m = -\frac{60 \, \text{cm}}{5 \, \text{cm}} = -12 \]
This indicates that the image is inverted and magnified by a factor of 12.

### Step 3: Relate the angular width of the object to the angular width of the image
The angular width of the object is given as \( \theta_0 = 2^\circ \). The angular width of the image \( \theta \) can be calculated using the magnification:
\[ \theta = m \cdot \theta_0 \]
Substituting the values:
\[ \theta = -12 \cdot 2^\circ = -24^\circ \]
The negative sign indicates that the image is inverted, but we are interested in the magnitude of the angular width.

### Step 4: Final result
Thus, the angular width of the image is:
\[ \theta = 24^\circ \]

### Conclusion
The angular width of the image formed by the telescope is \( 24^\circ \).

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