Question

In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube length is 25 cm. What is the magnitude of the overall magnification of the microscope?

Answer 1

Answer:

The  value  is   $m \approx 310$

Explanation:

From the question we are told that

     The  focal length of the objective is  $f_o = 1.0 \ cm$

    The  focal length of the eyepiece is  $f_e = 2.0 \ cm$

    The  tube length is  $L = 25 \ cm$

Generally the magnitude of the overall magnification is mathematically represented as

            $m = m_o * m_e$

Where  $m_o$ is the objective magnification which is mathematically represented as

        $m_o = \frac{L}{f_o }$

=>      $m_o = \frac{25}{1 }$

=>      $m_o = 25$

$m_e$ is the eyepiece magnification which is mathematically evaluated as

     $m_e = \frac{L }{f_e }$

     $m_e = \frac{25 }{ 2}$

      $m_e = 12.5 \ cm$

So

    $m = 25 * 12.5$

     $m \approx 310$