Chapter 24, Problem 75E

(a)

To determine

The far point of the man without glasses.

Answer to Problem 75E

The far point of the man without glasses is $\text{0}\text{.125}\text{m}$.

Explanation of Solution

Write the expression for the power of eye at near point.

  $P_{\text{n}}=\frac{1}{x_{\text{n}}}+\frac{1}{D}$        (I)

Here, $P_{\text{n}}$ is the power of eye at near point, $x_{\text{n}}$ is the near point of the eye and $D$ is the image distance of the eye.

Write the expression for the power of the eye at far point.

  $P_{\text{f}}=P_{\text{n}}-A$        (II)

Here, $P_{\text{f}}$ is the power of the eye at far point and $A$ is power of accommodation of the eye.

Write the expression for the power of the eye at far point.

  $P_{\text{f}}=\frac{1}{x_{\text{f}}}+\frac{1}{D}$        (III)

Here, $x_{\text{f}}$ is the far point of the eye.

Conclusion:

Substitute $0.1\text{m}$ for $x_{\text{n}}$ and $0.02\text{m}$ for $D$ in equation (I).

  $\begin{array}{c}{P}_{\text{n}}=\frac{1}{0.1\text{m}}+\frac{1}{0.02\text{m}}\\ =\left(10+50\right)\text{diaptors}\\ =\text{60}\text{diaptors}\end{array}$

Substitute $\text{60}\text{diaptors}$ for $P_{\text{n}}$ and $2\text{diaptors}$ for $A$ in equation (II).

  $\begin{array}{c}{P}_{\text{f}}=\text{60}\text{diaptors}-2\text{diaptors}\\ =\text{58}\text{diaptors}\end{array}$

Substitute $\text{58}\text{diaptors}$ for  $P_{\text{f}}$ and $0.02\text{m}$ for $D$ in equation (III).

  $\begin{array}{l}\text{58}\text{diaptors}=\frac{1}{x_{\text{f}}}+\frac{1}{\left(0.02\text{m}\right)}\\ \text{58}\text{diaptors}=\frac{1}{x_{\text{f}}}+\text{50}\text{diaptors}\end{array}$

Simplify the above expression.

  $\begin{array}{c}\frac{1}{x_{\text{f}}}=\text{58}\text{diaptors}-\text{50}\text{diaptors}\\ =\text{8}\text{diaptors}\end{array}$

Simplify the above expression for the far point.

  $\begin{array}{c}{x}_{\text{f}}=\frac{1}{\text{8}}\text{m}\\ =\text{0}\text{.125}\text{m}\end{array}$

Thus, the far point of the man without glasses is $\text{0}\text{.125}\text{m}$.

(b)

To determine

The far point of the man with the correct glasses.

Answer to Problem 75E

The far point of the man with the correct glasses is $\text{0}\text{.5}\text{m}$.

Explanation of Solution

Write the expression for the power of eye at near point with the glasses.

  ${P^{\prime }}_{\text{n}}=\frac{1}{{x^{\prime }}_{\text{n}}}+\frac{1}{D}$        (IV)

Here, ${P^{\prime }}_{\text{n}}$ is the power of eye at near point with the glasses, $x_{\text{n}}{}^{\prime }$ is the near point of the eye with the glasses and $D$ is the image distance of the eye.

Write the expression for the power of lens in the glasses.

  $P_{\text{f}}={P^{\prime }}_{\text{n}}-A$        (V)

Write the expression for the far point.

  $P_{\text{f}}=\frac{1}{x_{\text{f}}}+\frac{1}{D}$        (VI)

Conclusion:

Substitute $0.25\text{m}$ for ${x^{\prime }}_{\text{n}}$ and $0.02\text{m}$ for $D$ in equation (IV).

  $\begin{array}{c}{P^{\prime }}_{\text{n}}=\frac{1}{\left(0.25\text{m}\right)}+\frac{1}{\left(0.02\text{m}\right)}\\ =\left(4+50\right)\text{diaptors}\\ =\text{54}\text{diaptors}\end{array}$

Substitute $\text{54}\text{diaptors}$ for ${P^{\prime }}_{\text{n}}$ and $2\text{diaptors}$ for $A$ equation (V).

  $\begin{array}{c}{P}_{\text{f}}=\text{54}\text{diaptors}-\text{2}\text{diaptors}\\ =\text{52}\text{diaptors}\end{array}$

Substitute $\text{52}\text{diaptors}$ for  $P_{\text{f}}$ and $0.02\text{m}$ for $D$ in equation (VI).

  $\begin{array}{l}\text{52}\text{diaptors}=\frac{1}{x_{\text{f}}}+\frac{1}{\left(0.02\text{m}\right)}\\ \text{52}\text{diaptors}=\frac{1}{x_{\text{f}}}+\text{50}\text{diaptors}\end{array}$

Simplify the above expression.

  $\begin{array}{c}\frac{1}{x_{\text{f}}}=\text{52}\text{diaptors}-\text{50}\text{diaptors}\\ \text{=}2\text{diaptors}\end{array}$

Simplify the above expression for the far point as:

  $\begin{array}{c}{x}_{\text{f}}=\frac{1}{\text{2}}\text{m}\\ =\text{0}\text{.5}\text{m}\end{array}$

Thus, the far point of the man with the correct glasses is $\text{0}\text{.5}\text{m}$.