First slide

Ray optics

Question

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

Moderate

A

50 cm
B

54 cm
C

37.3 cm
D

46 cm

Solution

$Here f_{o} = 40 cm, f_{e} = 4 cm$
Tube length L= Distance between lenses = $v_{o} + f_{e}$
For objective lens,
$\begin{array}{l} u_{o} = - 200 cm, v_{o} = ? \\ \frac{1}{v_{o}} - \frac{1}{u_{0}} = \frac{1}{f_{o}} or \frac{1}{v_{o}} - \frac{1}{- 200} = \frac{1}{40} \\ or \frac{1}{v_{o}} = \frac{1}{40} - \frac{1}{200} = \frac{4}{200} ∴ v_{o} = 50 cm \\ ∴ L = 50 + 4 = 54 cm \end{array}$

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