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

The value is
`m \approx 310`

Step-by-step 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 = (L)/(f_o )`

=>
`m_o = (25)/(1 )`

=>
`m_o = 25`

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

`m_e = (L )/(f_e )`

`m_e = (25 )/( 2)`

`m_e = 12.5 \ cm`

So

`m = 25 * 12.5`

`m \approx 310`