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The length of a tube of microscope is $10cm$. The lengths of the objective and the eye lenses are $0.5cm$ and $1cm$ respectively. The magnifying power of the microscope when the images at far point is about
(A) 5
(B) 23
(C) 166
(D) 500

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Last updated date: 13th Jun 2024
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Answer
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Hint: This question is based on the magnifying power of a compound microscope. To solve this we just need to use the formula for normal adjustment which is given by, $m = \dfrac{{{v_o}}}{{{u_o}}}\dfrac{D}{{{f_e}}}$ and then we need to substitute the values of the objective and the eye lenses, length of a tube of microscope from the question and use the value of $D$ as $25cm$.
Formula Used:
$m = \dfrac{{{v_o}}}{{{u_o}}}\dfrac{D}{{{f_e}}}$
where $m$ is magnification
${v_o}$ is image distance
\[{u_o}\] is object distance
$D$ is the distance for distinct vision
${f_e}$ is the focal length of the eye-piece lens.

Complete Step by Step Solution:
From the question, we can see that,
The length of the tube of the microscope, that is the distance between the objective lens and the eye lens
$L = 10cm$
The focal length of the objective lens is ${f_o} = 0.5cm$
and the focal length of the eye lens is ${f_e} = 1.0cm$
It is said in the question that we need to find the magnifying power of the microscope when the image is formed at a faraway point. So here we can use the formula for normal adjustment, which is given by,
$m = \dfrac{{{v_o}}}{{{u_o}}}\dfrac{D}{{{f_e}}}$
Here for the value of $D$ we can take $25cm$as $D$is the distance for distinct vision.
Now in this case we can take ${v_o}$, which is the image distance as similar to $L$ and the value of ${u_o}$, which is the object distance will be equal to the focal length of the objective lens, that is ${f_o}$.
Therefore by putting these in the equation we get,
$m = \dfrac{L}{{{f_o}}}\dfrac{D}{{{f_e}}}$
Now all these values are given in the question. So substituting these values in the equation we get,
$m = \dfrac{{10}}{{0.5}} \times \dfrac{{25}}{1}$
$ \Rightarrow m = \dfrac{{2500}}{5}$
By calculating this we get the value of magnification as,
$m = 500$

So the correct value of $m$ is $500$. And the correct answer is option D.

Note: Here we have used the value of the distance for distinct vision as $D = 25cm$. This is the average value that is considered for a human adult. For children, this value may vary to be $15cm$and in old people, the value might increase up to an extent of $40cm$. But for the sake of calculation, we take the average as $25cm$.