JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Geometrical Optics

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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✓ Correct
(B)30 cm
(C)10 cm
(D)15 cm

Answer: (A)

Solution

For a telescope in normal adjustment,

f_{0} + f_{e} = 30

cm. The angular magnification is

m = \frac{f _{0}}{f _{e}} = 2

, so

f_{0} = 2 f_{e}

. Substituting:

2 f_{e} + f_{e} = 30

gives

f_{e} = 10

cm and

f_{0} = 20

cm.

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option

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