Chapter 24, Problem 66E

(a)

To determine

The far point of the man.

Answer to Problem 66E

The far point of the man is $\text{0}\text{.0714}\text{m}$.

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).

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

Substitute $\text{60}\text{diaptors}$ for $P_{\text{n}}$ and $4\text{diaptors}$ for $A$ equation (II).

  $\begin{array}{c}{P}_{\text{f}}=\text{60}\text{diaptors}-\text{4}\text{diaptors}\\ =56\text{diaptors}\end{array}$

Substitute $56\text{diaptors}$ for $P_{\text{f}}$ and $0.02\text{m}$ for $D$ in equation (III).

  $\begin{array}{l}56\text{diaptors}=\frac{1}{x_{\text{f}}}+\frac{1}{\left(0.02\text{m}\right)}\\ 56\text{diaptors}=\frac{1}{x_{\text{f}}}+\text{50}\text{diaptors}\end{array}$

Simplify the above expression as:

  $\begin{array}{c}\frac{1}{x_{\text{f}}}=56\text{diaptors}-\text{50}\text{diaptors}\\ =\text{6}\text{diaptors}\end{array}$

Simplify the above expression for the far point as:

  $\begin{array}{c}{x}_{\text{f}}=\frac{1}{\text{6}}\text{m}\\ =\text{0}\text{.1667}\text{m}\end{array}$

Thus, the far point of the man is $\text{0}\text{.1667}\text{m}$.

(b)

To determine

The power of lens he needed.

Answer to Problem 66E

The power of lens he needed is $\text{60}\text{diaptors}$.

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.

Conclusion:

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

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

Thus, the power of lens he needed is $\text{60}\text{diaptors}$.

(c)

To determine

The near point with the glasses.

Answer to Problem 66E

The near point with the glasses is $\text{0}\text{.1}\text{m}$.

Explanation of Solution

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

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

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

Conclusion:

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

  $\text{60}\text{diaptors}=\frac{1}{{x^{\prime }}_{\text{n}}}+\frac{1}{\left(0.02\text{m}\right)}$

Simplify the above expression for the near point.

  $\begin{array}{c}{x^{\prime }}_{\text{n}}=\frac{1}{\left(10+50\right)\text{diaptors}}\\ =\text{0}\text{.1}\text{m}\end{array}$

Thus, the near point with the glasses is $\text{0}\text{.1}\text{m}$.