Chapter 24, Problem 65E

(a)

To determine

The power of eyeglasses required.

Answer to Problem 65E

The power of eyeglasses required is $3.5\text{diaptors}$.

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

Write the expression for the power of eye at a new 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 the eye at a new near point and ${x^{\prime }}_{\text{n}}$ is the new near point.

Write the expression for the power of eyeglasses required.

  $P={P^{\prime }}_{\text{n}}-P_{\text{n}}$        (III)

Here, $P$ is the power of eyeglasses required.

Conclusion:

Substitute $2\text{m}$ for $x_{\text{n}}$ and $0.02\text{m}$ for $D$ in equation (I).

  $\begin{array}{c}{P}_{\text{n}}=\frac{1}{\left(2\text{m}\right)}+\frac{1}{\left(0.02\text{m}\right)}\\ =\left(0.5+50\right)\text{diaptors}\\ =50.5\text{diaptors}\end{array}$

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

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

Substitute $50.5\text{diaptors}$ for  $P_{\text{n}}$  and $54\text{diaptors}$ for ${P^{\prime }}_{\text{n}}$ in equation (iii).

  $\begin{array}{c}P=\left(54\text{diaptors}\right)-\left(50.5\text{diaptors}\right)\\ =3.5\text{diaptors}\end{array}$

Thus, the power of eyeglasses required is $3.5\text{diaptors}$.

(b)

To determine

The far point with the eyeglasses.

Answer to Problem 65E

The far point with the eyeglasses is $1\text{m}$.

Explanation of Solution

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

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

Here, ${P^{\prime }}_{\text{n}}$ is the power of the eye at a new near point, $P_{\text{f}}$ is the power of eye at the far point and $A$ is the power of accommodation.

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

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

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

Conclusion:

Substitute $54\text{diaptors}$ for  ${P^{\prime }}_{\text{n}}$  and $3\text{diaptors}$ for $A$ in equation (IV).

  $\begin{array}{c}{P}_{\text{f}}=\left(54\text{diaptors}\right)-\left(3\text{diaptors}\right)\\ =51\text{diaptors}\end{array}$

Substitute $51\text{diaptors}$ for $P_{\text{f}}$  and $0.02\text{m}$ for $D$ in equation (V).

  $\begin{array}{l}51\text{diaptors}=\frac{1}{x_{\text{f}}}+\frac{1}{\left(0.02\text{m}\right)}\\ 51\text{diaptors}=\frac{1}{x_{\text{f}}}+50\text{diaptors}\end{array}$

Simplify the above expression as:

  $\begin{array}{c}\frac{1}{x_{\text{f}}}=51\text{diaptors}-50\text{diaptors}\\ =1\text{diaptor}\end{array}$

Simplify the above expression for far point.

  $\begin{array}{c}{x}_{\text{f}}=\frac{1}{1\text{diaptor}}\\ =1\text{m}\end{array}$

Thus, the far point with the eyeglasses is $1\text{m}$.