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The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal lengths of lenses are
A
11 cm, 9 cm
B
10 cm, 10 cm
C
15 cm, 5 cm
D
18 cm, 2 cm
Detailed Solution
In normal adjustment: $f_o + f_e = 20$ cm ...(i)
Magnifying power: $\frac{f_o}{f_e} = 9 \Rightarrow f_o = 9f_e$ ...(ii)
From (i) and (ii): $10f_e = 20 \Rightarrow f_e = 2$ cm
$f_o = 18$ cm
Magnifying power: $\frac{f_o}{f_e} = 9 \Rightarrow f_o = 9f_e$ ...(ii)
From (i) and (ii): $10f_e = 20 \Rightarrow f_e = 2$ cm
$f_o = 18$ cm