The focal length of the objective of an astronomical telescope is 1.0 m. If the magnifying power of telescope is 20, then what is the length of the telescope for relaxed eye ?

Question

The focal length of the objective of an astronomical telescope is 1.0 m. If the magnifying power of telescope is 20, then what is the length of the telescope for relaxed eye ? (A) 85 cm (B) 95 cm (C) 105 cm (D) 115 cm

Tardigrade Admin · Accepted Answer

f0=1 m m=20 for relaxed eye final image is formed at infinity. m=(f0/f e) 20=(1/fe) f e=(1/20)=5 cm length of telescope L= f0+fe 100+5 105 cm

Q. The focal length of the objective of an astronomical telescope is 1.0 m. If the magnifying power of telescope is 20, then what is the length of the telescope for relaxed eye ?

85 cm

95 cm

105 cm

115 cm

$f_{0} = 1 m$
$m = 20$
for relaxed eye final image is formed at infinity.
$m = \frac{f _{0}}{f e}$
$20 = \frac{1}{f _{e}}$
$f e = \frac{1}{20} = 5 c m$
length of telescope
$L = f_{0} + f_{e}$
$100 + 5$
$105 c m$

Q. The focal length of the objective of an astronomical telescope is 1.0 m. If the magnifying power of telescope is 20, then what is the length of the telescope for relaxed eye ?

85 cm

95 cm

105 cm

115 cm

f0​=1m m=20 for relaxed eye final image is formed at infinity. m=fef0​​ 20=fe​1​ fe=201​=5cm length of telescope L=f0​+fe​ 100+5 105cm

$f_{0} = 1 m$
$m = 20$
for relaxed eye final image is formed at infinity.
$m = \frac{f _{0}}{f e}$
$20 = \frac{1}{f _{e}}$
$f e = \frac{1}{20} = 5 c m$
length of telescope
$L = f_{0} + f_{e}$
$100 + 5$
$105 c m$