Chapter 24, Problem 63P

(a)

To determine

The uncorrected far point.

Answer to Problem 63P

The uncorrected far point is $1.602\text{m}$.

Explanation of Solution

Write the equation to find the distance between the glass and book.

  $p=d_{book}-d_{glass}$                                                                                                       (I)

Here, $p$ is the distance between the glass and book, $d_{book}$ is the distance from relaxed eye to book, and $d_{glass}$ is the separation between glass and eye.

Write the lens formula.

  $\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$                                                                                                              (II)

Here, $q$ is the uncorrected far point, $p$ is the object distance, $f$ is the focal length, and $\frac{1}{f}$ is the power.

Rewrite the above relation in terms of $q$.

    $q={\left(\frac{1}{f}-\frac{1}{p}\right)}^{-1}$                                                                                                          (III)

Write the equation to find the corrected far point.

    $q^{\prime }=q+d_{glass}$                                                                                                           (IV)

Here, $q^{\prime }$ is the corrected far point.

Conclusion:

Substitute $40.0\text{cm}$ for $d_{book}$ and $2.0\text{cm}$ for $d_{glass}$ in equation (I) to find $p$.

    $\begin{array}{c}p=40.0\text{cm}-2.0\text{cm}\\ \text{=38}\text{.0}\text{cm}\end{array}$

Substitute $2.0\text{D}$ for $\frac{1}{f}$ and $\text{38}\text{.0}\text{cm}$ for $p$ in equation (III) to find $q$.

  $\begin{array}{c}q={\left(2.0\text{D}-\frac{1}{\left(\text{38}\text{.0}\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}\right)}^{-1}\\ =\frac{1}{0.625{\text{m}}^{-1}}\\ =-1.6\text{m}\end{array}$

Substitute $1.6\text{m}$ for $q$ and $2.0\text{cm}$ for $d_{glass}$ in equation (IV) to find $q^{\prime }$.

    $\begin{array}{c}{q}^{\prime }=1.6\text{m}+2.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ \text{=1}\text{.602}\text{m}\end{array}$

Therefore, the uncorrected far point is $1.602\text{m}$.

(b)

To determine

Power of lens required for distance vision.

Answer to Problem 63P

Power of lens required for distance vision is $-0.63\text{D}$.

Explanation of Solution

Object distance is taken as infinity for distant objects.

Write the equation to find $q$ in the given situation.

  $q=q^{\prime }+d_{glass}$                                                                                                           (V)

Write the relation between $f$ and power of lens.

    $P=\frac{1}{f}$                                                                                                                       (VI)

Here,$P$ is the power of lens.

Conclusion:

Substitute $-1.602$ for $q^{\prime }$ and $2.0\text{cm}$ for $d_{glass}$ in equation (V) to find $q$.

  $\begin{array}{c}q=-1.602\text{m}+2.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =-1.582\text{m}\end{array}$

Substitute $\infty$ for $p$ and $-1.582\text{m}$ for $q$ in equation (II) to find $\frac{1}{f}$.

    $\begin{array}{c}\frac{1}{\infty }+\frac{1}{-1.582\text{m}}=\frac{1}{f}\\ 0-0.63{\text{m}}^{-1}=\frac{1}{f}\end{array}$

Substitute $0.63{\text{m}}^{-1}$ for $\frac{1}{f}$ in equation (VI) to find $P$.

  $\begin{array}{c}P=-0.63{\text{m}}^{-1}\left(\frac{1\text{D}}{{\text{1m}}^{-1}}\right)\\ =-0.63\text{D}\end{array}$

Therefore, the power of lens required for distance vision is $-0.63\text{D}$.

(c)

To determine

Power of two lenses required for clear vision from $25\text{cm}$ to infinity.

Answer to Problem 63P

Power of two lenses required for clear vision from $25\text{cm}$ to infinity are $-0.63\text{D}$ and $3.3\text{D}$.

Explanation of Solution

Object distance is taken as infinity for distant objects.

Write the equation to find $p$.

    $p=d-d_{glass}$                                                                                                  (VII)

Here, $d$ is the near point of eye.

Write the equation to find $q$.

  $q=-d_{un}+d_{glass}$                                                                                         (VIII)

Conclusion:

Substitute $25\text{cm}$ for $d$ and $2.0\text{cm}$ for $d_{glass}$ in equation (VII) to find $p$.

  $\begin{array}{c}p=25\text{cm}-2.0\text{cm}\\ \text{=23}\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =0.23\text{m}\end{array}$

Substitute $1.0\text{m}$ for $d$ and $2.0\text{cm}$ for $d_{glass}$ in equation (VIII) to find $q$.

    $\begin{array}{c}q=-1.0\text{m}+2.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =-0.98\text{m}\end{array}$

Substitute $0.23\text{m}$$\infty$ for $p$ and $-0.98\text{m}$ for $q$ in the lens formula to find $\frac{1}{f}$.

    $\begin{array}{l}\frac{1}{0.23\text{m}}+\frac{1}{-0.98\text{m}}=\frac{1}{f}\\ 3.3{\text{m}}^{-1}=\frac{1}{f}\end{array}$

Substitute $3.3{\text{m}}^{-1}$ for $\frac{1}{f}$ in equation (VI) to find $P$.

  $\begin{array}{c}P=3.3{\text{m}}^{-1}\left(\frac{1\text{D}}{1{\text{m}}^{-1}}\right)\\ =3.3\text{D}\end{array}$

Therefore, Power of two lenses required for clear vision from $25\text{cm}$ to infinity are $-0.63\text{D}$ and $3.3\text{D}$.