An astronomical telescope has an objective and eyepiece of focal lengths $40 ext{ cm}$ and $4 ext{ cm}$ resp

Question

An astronomical telescope has an objective and eyepiece of focal lengths $40	ext{ cm}$ and $4	ext{ cm}$ respectively. To view an object $200	ext{ cm}$ away from the objective, the lenses must be separated by a distance of:

Accepted Answer

1.  **Image Formation by Objective:**
    *   Using the lens formula for the objective lens: $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$.
    *   Given: $f_o = +40	ext{ cm}$, $u_o = -200	ext{ cm}$.
    *   Substitution: $\frac{1}{v_o} - \frac{1}{-200} = \frac{1}{40} \implies \frac{1}{v_o} = \frac{1}{40} - \frac{1}{200}$.
    *   Calculation: $\frac{1}{v_o} = \frac{5 - 1}{200} = \frac{4}{200} = \frac{1}{50}$.
    *   Image distance $v_o = 50	ext{ cm}$.
2.  **Separation of Lenses:**
    *   For an astronomical telescope in normal adjustment (final image at infinity), the real image formed by the objective acts as a virtual object for the eyepiece and must lie at the focus of the eyepiece.
    *   Therefore, the distance between the intermediate image and the eyepiece is $u_e = f_e = 4	ext{ cm}$.
    *   The total distance between the lenses (tube length) is the sum of the image distance from the objective and the focal length of the eyepiece: $L = v_o + f_e$.
    *   $L = 50	ext{ cm} + 4	ext{ cm} = 54	ext{ cm}$.

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