Chapter 25, Problem 72GP

To determine

The focal length of each lens.

Answer to Problem 72GP

  $\text{3}\text{.1}\text{cm,}\text{24}\text{.9}\text{cm}$

Explanation of Solution

Given info:

Magnification is $\text{8}\text{.0}$ .

Sum of focal lengths is $\text{28}\text{cm}$ .

Formula used:

  $\text{M=}\text{-}\frac{{\text{f}}_{\text{o}}}{{\text{f}}_{\text{e}}}$

Where,

  ${\text{f}}_{\text{o}}$ is the focal length of object,

  ${\text{f}}_{\text{e}}$ focal length of eyepiece, and

  $\text{M}$ is the magnifying power.

Calculation:

Write the magnification equation and also, the given equation for the sum of the focal lengths.

Calculate the system of equations.

  $\text{M=}\text{-}\frac{{\text{f}}_{\text{o}}}{{\text{f}}_{\text{e}}}$

  $\text{M=}\text{-}\frac{{\text{f}}_{\text{o}}}{{\text{f}}_{\text{e}}}=-8.0$

Therefore, ${\text{f}}_{\text{o}}=8.0{\text{f}}_{\text{e}}$ ,

  $\text{8}\text{.0}{\text{f}}_{\text{e}}{\text{+ f}}_{\text{e}}\text{=9}\text{.0 cm}\text{=}\text{28 cm}$

Solving ${\text{f}}_{\text{e}}\text{=}\frac{\text{28}\text{cm}}{\text{9}\text{.0}}\text{=3}\text{.1}\text{and}{\text{f}}_{\text{e}}\text{=24}\text{.9}\text{cm}$

Conclusion: The required values are $\text{3}\text{.1}\text{cm,}\text{24}\text{.9}\text{cm}$ .