Answer
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Hint: We have two equations for the magnifying power of a microscope and telescope. The magnifying power is related to the focal length of the objective lens. By analyzing their relationship, we can solve the question.
Formula used:
\[m=\dfrac{LD}{{{f}_{o}}\times {{f}_{e}}}\]
\[m=\dfrac{{{f}_{o}}}{{{f}_{e}}}\]
Complete step-by-step solution:
For a microscope, magnifying power is given by,
\[m=\dfrac{LD}{{{f}_{o}}\times {{f}_{e}}}\]
\[L\] is the tube length
\[D\] is the least distance of distinct vision.
\[{{f}_{o}}\] is the focal length of the objective lens
\[{{f}_{e}}\] is the focal length of the eye lens
From the above equation, we can see that \[m\] is inversely proportional to \[{{f}_{o}}\]. Hence if we increase the focal length of the objective lens, the magnifying power of the microscope will decrease. For a telescope, magnifying power is given by,
\[m=\dfrac{{{f}_{o}}}{{{f}_{e}}}\]
Where,
\[{{f}_{o}}\] is the focal length of the objective lens
\[{{f}_{e}}\] is the focal length of the eye lens
From the equation, we can see that, \[m\] is directly proportional to \[{{f}_{o}}\]
Hence, if we increase the focal length of the objective lens, the magnifying power of the telescope will increase.
Therefore, the answer is option D.
Note: For a telescope, magnifying power and focal length of eye lens are inversely related. Hence increasing the focal length of the eye lens will decrease the magnifying power. But in the case of a microscope, magnifying power is inversely related to the focal lens of the eye lens. The magnification produced by a simple microscope is small. It can only be increased by decreasing the focal length of the lens. Due to the practical limit in increasing the focal length, we use a compound microscope for large magnification in which magnification is obtained in two stages by using two convex lenses.
Formula used:
\[m=\dfrac{LD}{{{f}_{o}}\times {{f}_{e}}}\]
\[m=\dfrac{{{f}_{o}}}{{{f}_{e}}}\]
Complete step-by-step solution:
For a microscope, magnifying power is given by,
\[m=\dfrac{LD}{{{f}_{o}}\times {{f}_{e}}}\]
\[L\] is the tube length
\[D\] is the least distance of distinct vision.
\[{{f}_{o}}\] is the focal length of the objective lens
\[{{f}_{e}}\] is the focal length of the eye lens
From the above equation, we can see that \[m\] is inversely proportional to \[{{f}_{o}}\]. Hence if we increase the focal length of the objective lens, the magnifying power of the microscope will decrease. For a telescope, magnifying power is given by,
\[m=\dfrac{{{f}_{o}}}{{{f}_{e}}}\]
Where,
\[{{f}_{o}}\] is the focal length of the objective lens
\[{{f}_{e}}\] is the focal length of the eye lens
From the equation, we can see that, \[m\] is directly proportional to \[{{f}_{o}}\]
Hence, if we increase the focal length of the objective lens, the magnifying power of the telescope will increase.
Therefore, the answer is option D.
Note: For a telescope, magnifying power and focal length of eye lens are inversely related. Hence increasing the focal length of the eye lens will decrease the magnifying power. But in the case of a microscope, magnifying power is inversely related to the focal lens of the eye lens. The magnification produced by a simple microscope is small. It can only be increased by decreasing the focal length of the lens. Due to the practical limit in increasing the focal length, we use a compound microscope for large magnification in which magnification is obtained in two stages by using two convex lenses.
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