Answer
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Hint:A telescope is an optical instrument which is mainly used for providing angular magnification of objects which are distant. For solving this question, we need to use the relation between the magnification of the telescope and the focal lengths of the objective lens and eyepiece.
Formula used:
$m = \dfrac{{{f_o}}}{{{f_e}}}$,
Where, $m$ is the magnifying power of a telescope, \[{f_o}\] is the focal length of the objective lens and \[{f_e}\] is the focal length of the eyepiece.
Complete step by step answer:
Let us first see how the telescope works. As shown in the figure, the telescope has two lenses: objective and eyepiece. It is clearly observed that the objective has a large focal length and a much larger lens than an eyepiece. When a light from a distant object enters the objective lens, a real image is formed in the tube at its second focal point. This image is then magnified by the eyepiece producing a final inverted image.
Now, the magnifying power of the telescope is calculated as the ratio the angle $\beta$ subtended at the eye by the final image to the angle $\alpha $ subtended by the object at the lens or the eye
Therefore, magnification can be given as
$m = \dfrac{\beta }{\alpha } \\
\Rightarrow m = \dfrac{h}{{{f_e}}} \times \dfrac{{{f_o}}}{h} \\
\therefore m = \dfrac{{{f_o}}}{{{f_e}}}$
From this equation, it is clear that magnification is directly proportional to the focal length of the objective and inversely proportional to the focal length of the eyepiece. Therefore, if we want to increase the magnification of the telescope, the objective lens should be of large focal length and the eyepiece should be of small focal length.
Hence, option A is the correct answer.
Note:We have seen that the magnification of the telescope is directly proportional to the focal length of the objective and inversely proportional to the focal length of the eyepiece. Also for a clear image, an objective lens should be of much higher diameter. Sometimes, it is difficult and expensive to make these types of large lenses to form images which are free from any type of chromatic aberration and distortion. Due to this, modern telescopes concave mirror instead of lens as objective.
Formula used:
$m = \dfrac{{{f_o}}}{{{f_e}}}$,
Where, $m$ is the magnifying power of a telescope, \[{f_o}\] is the focal length of the objective lens and \[{f_e}\] is the focal length of the eyepiece.
Complete step by step answer:
Let us first see how the telescope works. As shown in the figure, the telescope has two lenses: objective and eyepiece. It is clearly observed that the objective has a large focal length and a much larger lens than an eyepiece. When a light from a distant object enters the objective lens, a real image is formed in the tube at its second focal point. This image is then magnified by the eyepiece producing a final inverted image.
Now, the magnifying power of the telescope is calculated as the ratio the angle $\beta$ subtended at the eye by the final image to the angle $\alpha $ subtended by the object at the lens or the eye
Therefore, magnification can be given as
$m = \dfrac{\beta }{\alpha } \\
\Rightarrow m = \dfrac{h}{{{f_e}}} \times \dfrac{{{f_o}}}{h} \\
\therefore m = \dfrac{{{f_o}}}{{{f_e}}}$
From this equation, it is clear that magnification is directly proportional to the focal length of the objective and inversely proportional to the focal length of the eyepiece. Therefore, if we want to increase the magnification of the telescope, the objective lens should be of large focal length and the eyepiece should be of small focal length.
Hence, option A is the correct answer.
Note:We have seen that the magnification of the telescope is directly proportional to the focal length of the objective and inversely proportional to the focal length of the eyepiece. Also for a clear image, an objective lens should be of much higher diameter. Sometimes, it is difficult and expensive to make these types of large lenses to form images which are free from any type of chromatic aberration and distortion. Due to this, modern telescopes concave mirror instead of lens as objective.
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