Answer
430.2k+ views
Hint: In a Galilean telescope, the focal length of the objective lens is equal to the difference between the length between the two lenses and the focal length of the eye-piece, ${f_o} = L + {f_e}$, where ${f_o},{f_e}$ are the focal lengths of objective lens and eye-piece, L is the distance between them. Substitute the values of L and focal length of the objective lens to find the value of focal length of the eye-piece. Magnification is the negative ratio of focal length of the objective lens and focal length of the eye-piece.
Complete step by step answer:
We are given that a Galilean telescope measures 9 cm from the objective to the eye-piece and the focal length of the objective is 15 cm.
We have to find its magnifying power.
As we can see in the above figure, the objective is a convex lens and the eye-piece is a concave lens.
So the focal length is negative for the concave (eye-piece) lens.
As we can see, the focal length of the objective lens is the sum of the focal length of the eye-piece and the distance between them.
${f_o} = L + {f_e}$, where ${f_o},{f_e}$ are the focal lengths of objective lens and eye-piece, L is the distance between them and ${f_e}$ should be taken negatively.
$
{f_o} = L + \left( { - {f_e}} \right) \\
\implies {f_o} = L - {f_e} \\
\implies {f_e} = L - {f_o} \\
$
Substitute the values of the L and ${f_o}$ as they are given in the question.
$
{f_o} = 15cm,L = 9cm \\
\implies {f_e} = L - {f_o} \\
= 9 - 15 \\
\therefore {f_e} = - 6cm \\
$
Magnification is the negative ratio of focal length of the objective lens and focal length of the eye-piece.
$
m = \dfrac{{ - {f_o}}}{{{f_e}}} \\
{f_o} = 15cm,{f_e} = - 6cm \\
m = \dfrac{{ - 15}}{{ - 6}} \\
\implies m = \dfrac{5}{2} \\
\therefore m = + 2.5 \\
$
The magnifying power of the given Galilean telescope is 2.5.
So, the correct answer is “Option A”.
Note:
The objective lens is concave, which is a spherical lens and focuses the light rays passed through it to a specific point. Eye-piece is a convex lens, which is a spherical lens and diverges the light rays refracted from it. The lenses used in objective and eye-piece are not the same in a Galilean telescope. Be careful in considering the lenses.
Complete step by step answer:
![seo images](https://www.vedantu.com/question-sets/6d65b193-5cee-40ad-bd7f-e58bdacc57961488974755165576376.png)
We have to find its magnifying power.
As we can see in the above figure, the objective is a convex lens and the eye-piece is a concave lens.
So the focal length is negative for the concave (eye-piece) lens.
As we can see, the focal length of the objective lens is the sum of the focal length of the eye-piece and the distance between them.
${f_o} = L + {f_e}$, where ${f_o},{f_e}$ are the focal lengths of objective lens and eye-piece, L is the distance between them and ${f_e}$ should be taken negatively.
$
{f_o} = L + \left( { - {f_e}} \right) \\
\implies {f_o} = L - {f_e} \\
\implies {f_e} = L - {f_o} \\
$
Substitute the values of the L and ${f_o}$ as they are given in the question.
$
{f_o} = 15cm,L = 9cm \\
\implies {f_e} = L - {f_o} \\
= 9 - 15 \\
\therefore {f_e} = - 6cm \\
$
Magnification is the negative ratio of focal length of the objective lens and focal length of the eye-piece.
$
m = \dfrac{{ - {f_o}}}{{{f_e}}} \\
{f_o} = 15cm,{f_e} = - 6cm \\
m = \dfrac{{ - 15}}{{ - 6}} \\
\implies m = \dfrac{5}{2} \\
\therefore m = + 2.5 \\
$
The magnifying power of the given Galilean telescope is 2.5.
So, the correct answer is “Option A”.
Note:
The objective lens is concave, which is a spherical lens and focuses the light rays passed through it to a specific point. Eye-piece is a convex lens, which is a spherical lens and diverges the light rays refracted from it. The lenses used in objective and eye-piece are not the same in a Galilean telescope. Be careful in considering the lenses.
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