The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eyepiece is found to be 20 cm. The focal lengths of the lenses are

Question

The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eyepiece is found to be 20 cm. The focal lengths of the lenses are (A) 18 cm , 2 cm (B) 11 cm , 9 cm (C) 10 cm , 10 cm (D) 15 cm , 5 cm

Tardigrade Admin · Accepted Answer

As, M=(f o /f e ) 9=(fo/fe) or fo=9 f e Also, L=fo+fe or 20=fo+fe or 20=9 fe+fe or 20=10 fe or f e =2 cm ∴ fo=9 × 2 cm =18 cm

Q. The magnifying power of a telescope is $9$ . When it is adjusted for parallel rays, the distance between the objective and the eyepiece is found to be $20 c m$ . The focal lengths of the lenses are

18 cm , 2 cm

11 cm , 9 cm

10 cm , 10 cm

15 cm , 5 cm

As, $M = \frac{f _{o}}{f _{e}}$ $9 = \frac{f _{o}}{f _{e}}$
or $f_{o} = 9 f_{e}$
Also, $L = f_{o} + f_{e}$
or $20 = f_{o} + f_{e}$
or $20 = 9 f_{e} + f_{e}$
or $20 = 10 f_{e}$
or $f_{e} = 2 c m$
$∴$ $f_{o} = 9 \times 2 c m = 18 c m$

Q. The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eyepiece is found to be 20cm. The focal lengths of the lenses are