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Hint: Astronomical telescope is a system of lenses which produces images of the Object which are very far from earth.. It follows lens law $\dfrac{1}{u_{o}}+\dfrac{1}{v_{o}}=\dfrac{1}{f_{o}}$ and magnification is given by $M=\dfrac{f_{o}}{f_{e}\prime}$, or $m=\dfrac{f_{o}}{f_{e}}\left(1+\dfrac{f_{e}}{D}\right)$ where $D=25cm$ is the distance till which the relaxed eyes can see .
Formula used:
$M=\dfrac{f_{o}}{f_{e}}$
Complete answer:
Clearly, since a telescope is made up of a combination of lenses, namely the objective lens and eyepiece lens. For a telescope since the final image is formed at infinity the magnification is given as $M=\dfrac{f_{o}}{f_{e}}$ where $f_{o}$ and $f_{e}$ is the focal length of the objective lens and the eyepiece.
Here it is given that $M=10$
$\implies 10=\dfrac{f_{o}}{f_{e}}$
$\implies 10f_{e}=f_{o}$
We also know that the length of the telescope is nothing but the sum of the focal length of the lenses. Then we have;
$L=f_{o}+f_{e}$
$\implies 1.1=f_{o}+f_{e}$
$\implies 1.1=11 f_{e}$
$\therefore f_{e}=0.1\;m=10cm$
Then, the focal length of the objective lens is $10f_{e}=f_{o}$
$\implies f_{o}=100\;cm=1\;m$
If the image is at a distance $25 cm=0.25 m$ from the eye piece and the object is at a distance $f_{e
}$ then we can calculate the apparent focal length of the eyepiece $f_{e}\prime$
Then from the lens formula we have
$\dfrac{1}{f_e \prime}=-1\left[\dfrac{1}{D}+\dfrac{1}{f_{e}}\right]$
Then the magnification is given as $m=\dfrac{f_{o}}{f_{e}\prime}$
We also know that the magnification at the least distance of vision is given as $m=\dfrac{f_{o}}{f_{e}}\left(1+\dfrac{f_{e}}{D}\right)$
Substituting we get, $m=10\left(1+\dfrac{10}{25}\right)=10\left(\dfrac{35}{25}\right)=14$
Thus the correct answer is option B.14 .
Note:
A telescope is an optical instrument which uses lenses and mirrors to observe the objects which are at a distance from the earth, either by reflecting the light rays or by the emission or absorption of the rays emitted by the distant objects. Earlier they were made of two lenses, but in the present days a combination of lens and mirror to avoid distortion and catch all the possible light rays.
Formula used:
$M=\dfrac{f_{o}}{f_{e}}$
Complete answer:
Clearly, since a telescope is made up of a combination of lenses, namely the objective lens and eyepiece lens. For a telescope since the final image is formed at infinity the magnification is given as $M=\dfrac{f_{o}}{f_{e}}$ where $f_{o}$ and $f_{e}$ is the focal length of the objective lens and the eyepiece.
Here it is given that $M=10$
$\implies 10=\dfrac{f_{o}}{f_{e}}$
$\implies 10f_{e}=f_{o}$
We also know that the length of the telescope is nothing but the sum of the focal length of the lenses. Then we have;
$L=f_{o}+f_{e}$
$\implies 1.1=f_{o}+f_{e}$
$\implies 1.1=11 f_{e}$
$\therefore f_{e}=0.1\;m=10cm$
Then, the focal length of the objective lens is $10f_{e}=f_{o}$
$\implies f_{o}=100\;cm=1\;m$
If the image is at a distance $25 cm=0.25 m$ from the eye piece and the object is at a distance $f_{e
}$ then we can calculate the apparent focal length of the eyepiece $f_{e}\prime$
Then from the lens formula we have
$\dfrac{1}{f_e \prime}=-1\left[\dfrac{1}{D}+\dfrac{1}{f_{e}}\right]$
Then the magnification is given as $m=\dfrac{f_{o}}{f_{e}\prime}$
We also know that the magnification at the least distance of vision is given as $m=\dfrac{f_{o}}{f_{e}}\left(1+\dfrac{f_{e}}{D}\right)$
Substituting we get, $m=10\left(1+\dfrac{10}{25}\right)=10\left(\dfrac{35}{25}\right)=14$
Thus the correct answer is option B.14 .
Note:
A telescope is an optical instrument which uses lenses and mirrors to observe the objects which are at a distance from the earth, either by reflecting the light rays or by the emission or absorption of the rays emitted by the distant objects. Earlier they were made of two lenses, but in the present days a combination of lens and mirror to avoid distortion and catch all the possible light rays.
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