Chapter 24, Problem 64E

(a)

To determine

The focal length of lenses she required.

Answer to Problem 64E

The focal length lenses she required is $\text{0}\text{.5}\text{m}$.

Explanation of Solution

Write the expression for the focal length of the lenses.

  $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$        (I)

Here, $f$ is the focal length, $v$ is the image distance and $u$ is the object distance from the lenses.

Conclusion:

Substitute $0.5\text{m}$ for $v$ and $\infty$ for $u$ in equation (I).

  $\begin{array}{c}\frac{1}{f}=\frac{1}{\left(0.5\text{m}\right)}+\frac{1}{\infty }\\ =\frac{1}{\left(0.5\text{m}\right)}+0\\ =2{\text{m}}^{-1}\end{array}$

Simplify for the focal length.

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

Thus, the focal length lenses she required is $\text{0}\text{.5}\text{m}$.

(b)

To determine

The near point without the glasses.

Answer to Problem 64E

The near point without the glasses is $0.167\text{m}$.

Explanation of Solution

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

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

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

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

  $P_{\text{n}}=P_{\text{f}}+A$        (III)

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

Write the expression for the near point of the eye.

  $x_{\text{n}}=\frac{D}{D\left(P_{\text{n}}\right)-1}$        (IV)

Here, $x_{\text{n}}$ is the near point.

Conclusion:

Substitute $0.5\text{m}$ for $x_{\text{f}}$ and $0.02\text{m}$ for $D$ in equation (II).

  $\begin{array}{c}{P}_{\text{f}}=\frac{1}{\left(0.5\text{m}\right)}+\frac{1}{\left(0.02\text{m}\right)}\\ =\left(2+50\right)\text{diaptors}\\ =52\text{diaptors}\end{array}$

Substitute $52\text{diaptors}$ for $P_{\text{f}}$ and $4\text{diaptors}$ for $A$ in equation (III).

  $\begin{array}{c}{P}_{\text{n}}=\left(52\text{diaptors}\right)+\left(4\text{diaptors}\right)\\ =56\text{diaptors}\end{array}$

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

  $\begin{array}{c}{x}_{\text{n}}=\frac{\left(0.02\text{m}\right)}{\left(0.02\text{m}\right)\left(56\text{diaptors}\right)-1}\\ =0.167\text{m}\end{array}$

The near point without the glasses is $0.167\text{m}$.

(c)

To determine

The near point with the glasses.

Answer to Problem 64E

The near point with the glasses is $0.25\text{m}$.

Explanation of Solution

Write the expression for the power of eye at infinity.

  ${P^{\prime }}_{\text{f}}=\frac{1}{\infty }+\frac{1}{D}$        (V)

Here, ${P^{\prime }}_{\text{f}}$ is the power of eye at infinity and $D$ is the image distance of the eye.

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

  ${P^{\prime }}_{\text{n}}={P^{\prime }}_{\text{f}}+A$        (VI)

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

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

  ${x^{\prime }}_{\text{n}}=\frac{D}{D\left({P^{\prime }}_{\text{n}}\right)-1}$        (VII)

Here, ${x^{\prime }}_{\text{n}}$ is the near point.

Conclusion:

Substitute $0.02\text{m}$ for $D$ in equation (V).

  $\begin{array}{c}{P^{\prime }}_{\text{f}}=0+\frac{1}{\left(0.02\text{m}\right)}\\ =50\text{diaptors}\end{array}$

Substitute $50\text{diaptors}$ for ${P^{\prime }}_{\text{f}}$ and $4\text{diaptors}$ for $A$ in equation (VI).

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

Substitute $54\text{diaptors}$ for ${P^{\prime }}_{\text{n}}$ and $0.02\text{m}$ for $D$ in equation (IV).

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

The near point with the glasses is $0.25\text{m}$.