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Astronomical telescope – object at finite distance
Concepts tested here
- telescope-finite-object-tube-length
All Questions
2016 1 question
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A astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance
Tube length = $v_o + f_e$ in normal adjustment.
For the objective: $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o} \Rightarrow \frac{1}{v_o} - \frac{1}{(-200)} = \frac{1}{40}$
$\frac{1}{v_o} = \frac{1}{40} - \frac{1}{200} = \frac{4}{200} \Rightarrow v_o = 50$ cm
For normal adjustment, distance between lenses $l = v_o + f_e = 50 + 4 = 54$ cm