A simple microscope is rated $5X$ for a normal relaxed eye. What will be its magnifying power for a relaxed farsighted eye whose near point is $40$?
Answer
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Hint: To solve this question, we need to use the formula for the magnifying power of a simple microscope. Substituting the value of the magnification and the near point given in the question, we will get the focal length of the convex lens. Then substituting the value of the near point for a normal eye and the focal length, we will get the final answer.
Formula used: The formula used to solve this question is given by
$m = \dfrac{D}{f}$, here $m$ is the magnification of a simple microscope whose convex lens has a focal length of $f$ for an eye having a near point of $D$.
Complete step-by-step solution:
Let the focal length of the convex lens of the given simple microscope be $f$.
We know that the magnifying power of a simple microscope is given by the below formula
$m = \dfrac{D}{f}$.........(1)
According to the question, the magnification of the simple microscope is equal to $5X$ for a normal relaxed eye. We know that the near point of a normal relaxed eye is equal to $25cm$. Therefore substituting $m = 5X$ and $D = 25cm$ in (1) we get
$5X = \dfrac{{25}}{f}$
$ \Rightarrow f = \dfrac{5}{X}$....(2)
Now, we are asked about the magnification for the relaxed far sighted eye, which has a near point of $40cm$. Let the magnification be $M$. Therefore substituting $m = M$ and $D = 40cm$ in the above equation we get
$M = \dfrac{{40}}{f}$
Substituting the value of $f$ from (3) in the above equation, we get
\[M = \dfrac{{40X}}{5}\]
$ \Rightarrow M = 8X$
Hence, the magnifying power of the given simple microscope for the given relaxed far sighted eye is equal to $8X$.
Note: The formula for the magnifying power which we have used in the above solution is for the case when the image is formed by the microscope at the infinity. Do not use the formula for the magnifying power for the case when the image is formed at the near point. This is because the eye is relaxed, which happens when the image is formed at the infinity.
Formula used: The formula used to solve this question is given by
$m = \dfrac{D}{f}$, here $m$ is the magnification of a simple microscope whose convex lens has a focal length of $f$ for an eye having a near point of $D$.
Complete step-by-step solution:
Let the focal length of the convex lens of the given simple microscope be $f$.
We know that the magnifying power of a simple microscope is given by the below formula
$m = \dfrac{D}{f}$.........(1)
According to the question, the magnification of the simple microscope is equal to $5X$ for a normal relaxed eye. We know that the near point of a normal relaxed eye is equal to $25cm$. Therefore substituting $m = 5X$ and $D = 25cm$ in (1) we get
$5X = \dfrac{{25}}{f}$
$ \Rightarrow f = \dfrac{5}{X}$....(2)
Now, we are asked about the magnification for the relaxed far sighted eye, which has a near point of $40cm$. Let the magnification be $M$. Therefore substituting $m = M$ and $D = 40cm$ in the above equation we get
$M = \dfrac{{40}}{f}$
Substituting the value of $f$ from (3) in the above equation, we get
\[M = \dfrac{{40X}}{5}\]
$ \Rightarrow M = 8X$
Hence, the magnifying power of the given simple microscope for the given relaxed far sighted eye is equal to $8X$.
Note: The formula for the magnifying power which we have used in the above solution is for the case when the image is formed by the microscope at the infinity. Do not use the formula for the magnifying power for the case when the image is formed at the near point. This is because the eye is relaxed, which happens when the image is formed at the infinity.
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