In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30 cm. The focal length of the objective, when the angular magnification of the telescope is 2, will be: (a) 20 cm (b) 30 cm (c) 10 cm (d) 15 cm

Answer: (a). fo + fe = 30 cm and (fo)/(fe) = 2. 3fe = 30 ⇒ fe = 10 cm, fo = 20 cm So the answer is (a) 20 cm.

In normal adjustment, for a refracting telescope, the distance… | JEE Main 2022 (28 Jul, Shift 1)

Q20JEE Main 2022 · 28 JUL · SHIFT 1Easy

Optical Instruments & the Eye

In normal adjustment, for a refracting telescope, the distance between objective and eye piece is $30\text{ cm}$ . The focal length of the objective, when the angular magnification of the telescope is 2, will be:

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Answer: (a)

$f_o + f_e = 30\,\text{cm}$ and $\dfrac{f_o}{f_e} = 2$ .
$3f_e = 30 \Rightarrow f_e = 10\,\text{cm},\ f_o = 20\,\text{cm}$
So the answer is (a) 20 cm.

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