You are given following three lenses- Which two lenses will you see as an eyepiece and as an objective to construct an astronomical telescope-LensesPowerApertureL-1-3D8 cmL-2-6D1 cmL-3-10D1 cm

Question

Toppr Admin · Accepted Answer

In an astronomical telescope-The eyepiece has greater power while a smaller aperture- and- the objective has lower power with a longer aperture so- by using 4 as objective and L3 are eyepiece- the best possible magnification-m-PeP0-103-3-33 can be achieved-

Lenses	Power	Aperture
$L_{1}$	$3 D$	$8 c m$
$L_{2}$	$6 D$	$1 c m$
$L_{3}$	$10 D$	$1 c m$

Lenses	Power	Aperture
$L_{1}$	$3 D$	$8 c m$
$L_{2}$	$6 D$	$1 c m$
$L_{3}$	$10 D$	$1 c m$

Lenses	Power	Aperture
$L_{1}$	$3 D$	$8 cm$
$L_{2}$	$6 D$	$1 cm$
$L_{3}$	$10 D$	$1 cm$

Lenses	Power	Aperture
$L_{1}$	$3 D$	$8 cm$
$L_{2}$	$6 D$	$1 cm$
$L_{3}$	$10 D$	$1 cm$

You are given following three lenses. Which two lenses will you see as an eyepiece and as an objective to construct an astronomical telescope?
Lenses Power Aperture
$L_{1}$ $3 D$ $8 c m$
$L_{2}$ $6 D$ $1 c m$
$L_{3}$ $10 D$ $1 c m$

In an astronomical telescope,
The eyepiece has greater power while a smaller aperture. and, the objective has lower power with a longer aperture so, by using $4$ as objective and $L_{3}$ are eyepiece, the best possible magnification,
$m = \frac{P_{e}}{P_{0}} = \frac{10}{3} = 3.33$ can be achieved.

You are given following three lenses. Which two lenses will you see as an eyepiece and as an objective to construct an astronomical telescope?LensesPowerApertureL13D8 cmL26D1 cmL310D1 cm

In an astronomical telescope,The eyepiece has greater power while a smaller aperture. and, the objective has lower power with a longer aperture so, by using 4 as objective and L3 are eyepiece, the best possible magnification,m=PeP0=103=3.33 can be achieved.

You are given following three lenses. Which two lenses will you see as an eyepiece and as an objective to construct an astronomical telescope?
Lenses Power Aperture
$L_{1}$ $3 D$ $8 c m$
$L_{2}$ $6 D$ $1 c m$
$L_{3}$ $10 D$ $1 c m$

In an astronomical telescope,
The eyepiece has greater power while a smaller aperture. and, the objective has lower power with a longer aperture so, by using $4$ as objective and $L_{3}$ are eyepiece, the best possible magnification,
$m = \frac{P_{e}}{P_{0}} = \frac{10}{3} = 3.33$ can be achieved.