Chapter 25, Problem 65GP

(a)

To determine

The maximum magnification.

Answer to Problem 65GP

The magnification power is $\text{-2}\text{.25}\text{x}$ .

Explanation of Solution

Given info:

Power of lens $\text{+2}\text{.0}\text{D}$ , $\text{+4}\text{.5}\text{D}$ .

Formula used:

It is known that $\text{M=}\text{-}\frac{{\text{f}}_{\text{o}}}{{\text{f}}_{\text{e}}}\text{=}\text{-}\frac{{\text{P}}_{\text{e}}}{{\text{P}}_{\text{o}}}$

  ${\text{f}}_{\text{o}}$ is the focal length of object, ${\text{f}}_{\text{e}}$ focal length of eyepiece, ${\text{P}}_{\text{o}}$ is the power of object ${\text{P}}_{\text{e}}$ power of eyepiece, and $\text{M}$ is the magnifying power.

Calculation:

Substituting the given values in the formula $\text{M=}\text{-}\frac{{\text{f}}_{\text{o}}}{{\text{f}}_{\text{e}}}\text{=}\text{-}\frac{{\text{P}}_{\text{e}}}{{\text{P}}_{\text{o}}}$

  $\text{M=}\text{-}\frac{{\text{P}}_{\text{e}}}{{\text{P}}_{\text{o}}}\text{=}\frac{\text{4}\text{.5}\text{D}}{\text{2}\text{.0}\text{D}}\text{=}\text{-2}\text{.25}\text{x}$

(b)

To determine

The lens used as the eyepiece.

Answer to Problem 65GP

The concave lenses are used for telescope.

Explanation of Solution

Given info:

Power of lens $\text{+2}\text{.0}\text{D}$ , $\text{+4}\text{.5}\text{D}$ .

The magnification power is $\text{-2}\text{.25}\text{x}$ .

Formula used:

It is known that $\text{M=}\text{-}\frac{{\text{f}}_{\text{o}}}{{\text{f}}_{\text{e}}}\text{=}\text{-}\frac{{\text{P}}_{\text{e}}}{{\text{P}}_{\text{o}}}$

  ${\text{f}}_{\text{o}}$ is the focal length of object, ${\text{f}}_{\text{e}}$ focal length of eyepiece, ${\text{P}}_{\text{o}}$ is the power of object ${\text{P}}_{\text{e}}$ power of eyepiece, and $\text{M}$ is the magnifying power.

Calculation:

To get a magnification greater than $\text{1}$ , for the eyepiece we use the lens with the smaller focal length, or greater power, $\text{+4}\text{.5}\text{D}$ . So, the used lens is concave.