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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 $25cm$.

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