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3 years ago

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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 le

ngth is 25 cm. What is the magnitude of the overall magnification of the microscope?

1 answer:

expeople1 [14]3 years ago

6 0

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

$m = 25 * 12.5$

$m \approx 310$

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