Solveeit Logo

Question

Question: A Galilean telescope measures 9 cm from the objective to the eye-piece. The focal length of the obje...

A Galilean telescope measures 9 cm from the objective to the eye-piece. The focal length of the objective is 15 cm. Its magnifying power is
A. 2.52.5
B. 25\dfrac{2}{5}
C. 53\dfrac{5}{3}
D. 35\dfrac{3}{5}

Explanation

Solution

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, fo=L+fe{f_o} = L + {f_e}, where fo,fe{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.
fo=L+fe{f_o} = L + {f_e}, where fo,fe{f_o},{f_e} are the focal lengths of objective lens and eye-piece, L is the distance between them and fe{f_e} should be taken negatively.
fo=L+(fe)     fo=Lfe     fe=Lfo  {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 fo{f_o} as they are given in the question.
fo=15cm,L=9cm     fe=Lfo =915 fe=6cm  {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=fofe fo=15cm,fe=6cm m=156     m=52 m=+2.5  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.