When a telescope is in normal adjustment, the distance of the objective from the eyepiece is found to be 100 cm. If the magnifying power of the telescope, at normal adjustment, is 24, the focal lengths of the lenses are

Question

When a telescope is in normal adjustment, the distance of the objective from the eyepiece is found to be 100 cm. If the magnifying power of the telescope, at normal adjustment, is 24, the focal lengths of the lenses are (A) 96 cm, 4 cm (B) 48 cm, 2 cm (C) 50 cm, 50 cm (D) 80 cm, 20 cm

Tardigrade Admin · Accepted Answer

In normal adjustment of telescope, magnifying power is

M = \frac{f _{0}}{f _{e}}

where

f_{0}

and

f_{e}

be the focal lengths of the objective lens and eye-piece respectively.
In normal adjustment of telescope, distance between objective lens and eye piece is

L = f_{0} + f_{e}

According to the problem

24 = \frac{f _{0}}{f _{e}} ... (i)

100 = f_{0} + f_{e} ... (ii)

Solving

(i)

and

(ii)

, we get

f_{0} = 96 c m, f_{e} = 4 c m

Q. When a telescope is in normal adjustment, the distance of the objective from the eyepiece is found to be $100 c m$ . If the magnifying power of the telescope, at normal adjustment, is $24$ , the focal lengths of the lenses are

$96 c m, 4 c m$

$48 c m, 2 c m$

$50 c m, 50 c m$

$80 c m, 20 c m$

Q. When a telescope is in normal adjustment, the distance of the objective from the eyepiece is found to be 100cm. If the magnifying power of the telescope, at normal adjustment, is 24, the focal lengths of the lenses are

96cm,4cm

48cm,2cm

50cm,50cm

80cm,20cm

Q. When a telescope is in normal adjustment, the distance of the objective from the eyepiece is found to be $100 c m$ . If the magnifying power of the telescope, at normal adjustment, is $24$ , the focal lengths of the lenses are

$96 c m, 4 c m$

$48 c m, 2 c m$

$50 c m, 50 c m$

$80 c m, 20 c m$