Chapter 24, Problem 30P

(a)

To determine

The refractive power of the corrective eyeglass lens to enable the man to clearly see distant objects.

Answer to Problem 30P

The refractive power of the corrective eyeglass lens to enable the man to clearly see distant objects is $-0.51\text{ D}$.

Explanation of Solution

Refractive power of a lens is the reciprocal of its focal length.

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

Here, $P$ is the refractive power of the lens and $f$ is the focal length of the lens.

Write the lens equation.

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

Here, $p$ is the object distance and $q$ is the image distance

Put equation (II) in equation (I).

  $P=\frac{1}{p}+\frac{1}{q}$                                                                                                              (III)

Conclusion:

To clear distant objects, the object distance of the lens should be infinity.

Given that the man cannot see objects farther than $2.0\text{ m}$ away and the distance between the eyes and the lens is $2.0\text{ cm}$ .

Find the image distance of the lens.

  $\begin{array}{c}q=-2.0\text{ m}+2.0\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\\ =-2.0\text{ m}+0.020\text{ m}\\ =-2.02\text{ m}\end{array}$

Substitute $\infty$  for $p$ and $2.02\text{ m}$ for $q$ in equation (III) to find $P$ .

  $\begin{array}{c}P=\frac{1}{\infty }+\frac{1}{-2.02\text{ m}}\\ =-0.505\text{ D}\\ \approx -0.51\text{ D}\end{array}$

Therefore, the refractive power of the corrective eyeglass lens to enable the man to clearly see distant objects is $-0.51\text{ D}$.

(b)

To determine

The near point of the man can see clearly with and without his glasses.

Answer to Problem 30P

The near point of the man with his glasses is $25\text{ cm}$ and without his glasses is $22\text{ cm}$ .

Explanation of Solution

Use equation (III) to find the refractive power of eye.

  $P_{\text{eye}}=\frac{1}{p}+\frac{1}{q}$

Here, $P_{\text{eye}}$ is the refractive power of the eye.

Substitute $2.0\text{ m}$ for $p$ and $2.0\text{ cm}$ for $q$ in the above equation to find $P_{\text{eye}}$ .

  $\begin{array}{c}{P}_{\text{eye}}=\frac{1}{2.0\text{ m}}+\frac{1}{2.0\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)}\\ =\frac{1}{2.0\text{ m}}+\frac{1}{0.020\text{ m}}\\ =50.5\text{ D}\end{array}$

It is given that the power of accommodation of the eye is $4.0\text{ D}$ .

Find the refractive power of the eye using the accommodation.

  $\begin{array}{c}{P^{\prime }}_{\text{eye}}=50.5\text{ D}+4.0\text{ D}\\ =54.5\text{ D}\end{array}$

Here, ${P^{\prime }}_{\text{eye}}$ is the refractive power of the eye using the accommodation.

Rewrite equation (I) for $p$ .

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

Put equation (I) in the above equation.

  $p={\left(P-\frac{1}{q}\right)}^{-1}$                                                                                                       (IV)

Replace $P$ in equation (IV) by ${P^{\prime }}_{\text{eye}}$ to find the expression for the near point without glasses.

  $p={\left({P^{\prime }}_{\text{eye}}-\frac{1}{q}\right)}^{-1}$                                                                                                       (V)

Write the equation for the refractive power of the eye with the glasses.

  $P_{e-g}=P_{\text{eye}}+P_{\text{glasses}}$

Here, $P_{e-g}$ is the refractive power of the eye with the glasses and $P_{\text{glasses}}$ is the refractive power of the glasses.

Substitute $50.5\text{ D}$ for $P_{\text{eye}}$ and $-0.505\text{ D}$ for $P_{\text{glasses}}$ in the above equation to find $P_{e-g}$ .

  $\begin{array}{c}{P}_{e-g}=50.5\text{ D}+\left(-0.505\text{ D}\right)\\ =50\text{ D}\end{array}$

Find the refractive power of the eye with the glasses using the accommodation.

  $\begin{array}{c}{P^{\prime }}_{e-g}=50\text{ D}+4.0\text{ D}\\ =54\text{ D}\end{array}$

Here, ${P^{\prime }}_{e-g}$ is the refractive power of the eye with the glasses using the accommodation.

Replace $P$ in equation (IV) by ${P^{\prime }}_{e-g}$ to find the expression for the near point with glasses.

  $p={\left({P^{\prime }}_{e-g}-\frac{1}{q}\right)}^{-1}$                                                                                                   (VI)

Conclusion:

Substitute $54.5{\text{ m}}^{-1}$ for ${P^{\prime }}_{\text{eye}}$ and $2.0\text{ cm}$ for $q$ in equation (V) to find the near point of the man without glasses.

  $\begin{array}{c}p={\left(54.5{\text{ m}}^{-1}\frac{1\text{ m}}{100\text{ cm}}-\frac{1}{2.0\text{ cm}}\right)}^{-1}\\ =22\text{ cm}\end{array}$

Substitute $54{\text{ m}}^{-1}$ for ${P^{\prime }}_{e-g}$ and $2.0\text{ cm}$ for $q$ in equation (VI) to find the near point of the man with glasses.

  $\begin{array}{c}p={\left(54{\text{ m}}^{-1}\frac{1\text{ m}}{100\text{ cm}}-\frac{1}{2.0\text{ cm}}\right)}^{-1}\\ =25\text{ cm}\end{array}$

Therefore, the near point of the man with his glasses is $25\text{ cm}$ and without his glasses is $22\text{ cm}$ .