Chapter 25, Problem 69GP

To determine

The maximum magnification.

Answer to Problem 69GP

The magnification power is $\text{-19}\text{x}$ .

Explanation of Solution

Given info:

Power of lens $\text{+23}\text{D}$ .

Distance between the objective lens and the eyepiece of telescope is $\text{85}\text{cm}$ .

Formula used:

The focal length is the inverse of power. So,

  $\text{f =}\frac{\text{1}}{\text{P}}$

Where,

  $\text{f}$ is a focal length.

  $\text{P}$ is a power.

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

Where,

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

  ${\text{f}}_{\text{e}}$ is the focal length of eyepiece.

  $\text{M}$ is the magnifying power.

Calculation:

The focal length of the eyepiece is ${\text{f}}_{\text{e}}\text{=}\frac{\text{1}}{{\text{P}}_{\text{e}}}$

  ${\text{f}}_{\text{e}}\text{=}\frac{\text{1}}{\text{23}\text{D}}$

  ${\text{f}}_{\text{e}}\text{=}\text{4}\text{.3×}{\text{10}}^{\text{-2}}\text{m}\text{=}\text{4}\text{.3}\text{cm}$

For both object as well as image far away, the focal length of the objective from separation of the lenses is $\text{L=}{\text{f}}_{\text{o}}{\text{+f}}_{\text{e}}$

  $\text{85}\text{cm}\text{=}{\text{f}}_{\text{o}}\text{+}\text{4}\text{.3}\text{cm}$

So, focus length of object is $\text{80}\text{.7}\text{cm}$

The magnification of the telescope is given by $\text{M=}\text{-}\frac{{\text{f}}_{\text{o}}}{{\text{f}}_{\text{e}}}$

  $\text{M=}\text{-}\frac{\left(\text{80}\text{.7}\text{cm}\right)}{\left(\text{4}\text{.3}\text{cm}\right)}\text{= -19}\text{x}$

Conclusion: The required value is $\text{ -19}\text{x}$ .