Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eyepiece is found to be 20 cm. The focal length of lenses are(A) 18 cm, 2 cm(B) 11 cm, 9 cm(C) 10 cm, 10 cm(D) 15 cm, 5 cm

Last updated date: 20th Jun 2024
Total views: 394.2k
Views today: 4.94k
Verified
394.2k+ views
Hint
The magnification of a telescope is directly related to the focal length of the objective lens and inversely related to the focal length of the eye piece. The sum of these focal lengths is the distance between the lenses when adjusted for parallel rays.
Formula used: $M = \dfrac{{{f_0}}}{{{f_e}}}$, where $M$ is the magnifying power, ${f_0}$ is the objective lens focal length, ${f_e}$ is the eyepiece focal length.
$L = {f_0} + {f_e}$, where $L$ is the length between the objective piece and the eyepiece.

We start with using the formula for magnification or magnifying power of a telescope, and it is given as
$M = \dfrac{{{f_0}}}{{{f_e}}}$, ${f_0}$ is the objective lens focal length, ${f_e}$ is the eyepiece focal length.
According to the question, $M$ is 9.
Thus,
$M = \dfrac{{{f_0}}}{{{f_e}}} = 9$
Cross-multiplying, we have
${f_0} = 9{f_e}$
The distance between the objective lens and the eyepiece $L$ is given as 20 cm.
When a telescope is adjusted for parallel rays, the objective and eyepiece are configured in such a way that the image focal point (focal point at the side of the lens where images are formed) of the objective lens coincides with the object focal point (focal point at the side of the lens closer to the objective lens) of the eye piece.
Therefore, the distance between the lenses $L$ is
$L = {f_0} + {f_e}$
Substituting ${f_0} = 9{f_e}$ into $L = {f_0} + {f_e}$ gives
$L = 9{f_e} + {f_e} = 10{f_e}$
Since $L = 20$,
$10{f_e} = 20$
Dividing both sides by 10 we get,
${f_e} = 2$
Since ${f_0} = 9{f_e}$
$\Rightarrow {f_0} = 9 \times 2 = 18$
Thus,
${f_0} = 18$
$\therefore {f_e} = 2cm$ and ${f_0} = 18cm$
Hence, the correct option is A.

Note
Alternately, from the equation ${f_0} = 9{f_e}$ we can conclude our answer from the set of options given by testing the options against this equation as done below
For option A (18 cm, 2 cm):
$18cm = 2cm \times 9$ which corresponds to the equation ${f_0} = 9{f_e}$. Before concluding, let’s ensure it’s the only one.
For option B (11 cm, 9 cm):
$11cm \ne 9cm \times 9$
Thus, it doesn’t correspond.
For option C (10 cm, 10cm):
$10cm \ne 10cm \times 9$, which again doesn’t correspond.
Finally, for option D (15 cm, 5 cm):
$15cm \ne 5cm \times 9$ also defies the equation ${f_0} = 9{f_e}$
Therefore, we can conclude that $18cm = 2cm \times 9$ is the only equation that obeys.
Hence, option A is the correct answer.