Chapter 25, Problem 58GP

(a)

To determine

The focal length of Person S’s glasses.

Answer to Problem 58GP

  $\text{28}\text{.6}\text{cm}$

Explanation of Solution

Given:

Power of lens, $\text{P}\text{=}\text{+3}\text{.50}\text{D}$

Formula used:

It is known that 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

Calculation:

Substituting the given values in the formula $\text{f =}\frac{\text{1}}{\text{P}}$

  $\text{f = }\frac{\text{1}}{\text{3}\text{.50}\text{D}}\text{= 0}\text{.286}\text{m =28}\text{.6}\text{cm}$

(b)

To determine

Person S’s near point.

Answer to Problem 58GP

  $\text{118}\text{cm}$

Explanation of Solution

Given:

Power of lens, $\text{P}\text{=}\text{+3}\text{.50}\text{D}$

Distance from the source to lens, ${\text{d}}_{\text{o}}\text{=}\text{25}\text{cm}\text{=}\text{0}\text{.25}\text{m}$

Formula used:

It is known that the Lens maker’s equation

  $\frac{\text{1}}{{\text{d}}_{\text{o}}}\text{ + }\frac{\text{1}}{{\text{d}}_{\text{i}}}\text{ = }\frac{\text{1}}{\text{f}}\text{ = P}$

Where,

  $\text{f}$ is a focal length

  $\text{P}$ is a power

  ${\text{d}}_{\text{o}}$ is distance from the source to lens 

  ${\text{d}}_{\text{i}}$ is distance from the screen to the lens

Calculation:

Distance is always positive. Take absolute value and add the $\text{2}\text{cm}$ distance to the eyes.

So, $\frac{\text{1}}{\text{0}\text{.23}\text{cm}}\text{ + }\frac{\text{1}}{{\text{d}}_{\text{i}}}\text{ = 3}\text{.50}\text{D}$

  $\text{i}\text{=}\text{-1}\text{.18}\text{m}$

Conclusion:

So, Person S’s near point is $\text{118}\text{cm}$ .

(c)

To determine

Person P’s near point.

Answer to Problem 58GP

  $\text{13}\text{cm}$

Explanation of Solution

Given:

Power of lens, $\text{P}\text{=}\text{+3}\text{.50}\text{D}$

Distance from the source to lens, ${\text{d}}_{\text{o}}\text{=}\text{25}\text{cm}\text{=}\text{0}\text{.25}\text{m}$

Formula used:

It is known that the Lens maker's equation

  $\frac{\text{1}}{{\text{d}}_{\text{o}}}\text{ + }\frac{\text{1}}{{\text{d}}_{\text{i}}}\text{ = }\frac{\text{1}}{\text{f}}\text{ = P}$

Where,

  $\text{f}$ is a focal length

  $\text{P}$ is a power

  ${\text{d}}_{\text{o}}$ is distance from the source to lens 

  ${\text{d}}_{\text{i}}$ is distance from the screen to the lens

Calculation:

Distance is always positive. Take absolute value and add the $\text{2}\text{cm}$ distance to the eyes.

So, $\frac{\text{1}}{{\text{d}}_{\text{o}}}\text{ + }\frac{\text{1}}{\text{-0}\text{.23}}\text{ = 3}\text{.50}\text{D}$

  $\text{o}\text{=}\text{-0}\text{.13}\text{m}$

Conclusion:

So, Person S’s near point is $\text{13}\text{cm}$ .