The length (L) of the astronomical telescope, for normal adjustment is

Options

A. `((f_(o)+f_(e))/(2))`
B. `((f_(o)-f_(e))/(2))`
C. `f_(o) xx f_(e)`
D. `(f_(o) xx f_(e))`

Answer

To find the length (L) of the astronomical telescope for normal adjustment, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Components of the Telescope**:
An astronomical telescope consists of two lenses: the objective lens (F₀) and the eyepiece lens (Fₑ). The length of the telescope is the sum of the focal lengths of these two lenses when the telescope is in normal adjustment.

2. **Define Normal Adjustment**:
Normal adjustment of an astronomical telescope occurs when the final image is formed at infinity. This means that the light rays coming from the object are parallel when they enter the eyepiece.

3. **Apply the Lens Formula**:
The lens formula is given by:
\[ \frac{1}{V} - \frac{1}{U} = \frac{1}{F} \]
Where:
- V = image distance
- U = object distance
- F = focal length of the lens

4. **Set Up the Equation for the Eyepiece**:
For the eyepiece lens, when the final image is at infinity:
\[ \frac{1}{V} = 0 \quad (\text{since } V = \infty) \]
Therefore, the equation simplifies to:
\[ -\frac{1}{Uₑ} = \frac{1}{Fₑ} \]
Rearranging gives:
\[ Uₑ = Fₑ \]

5. **Calculate the Length of the Telescope**:
The total length of the telescope (L) is the sum of the focal lengths of the objective and the eyepiece:
\[ L = F₀ + Uₑ \]
Substituting \( Uₑ = Fₑ \) into the equation:
\[ L = F₀ + Fₑ \]

6. **Final Result**:
Thus, the length of the astronomical telescope for normal adjustment is:
\[ L = F₀ + Fₑ \]