Chapter 25, Problem 10P

A patient can’t see objects closer than 40.0 cm and wishes to clearly see objects that are 20.0 cm from his eye. (a) Is the patient nearsighted or farsighted? (b) If the eye-lens distance is 2.00 cm, what is the minimum object distance p from the lens? (c) What image position with respect to the lens will allow the patient to see the object? (d) Is the image real or virtual? Is the image distance q positive or negative? (e) Calculate the required focal length. (f) Find the power of the lens in diopters. (g) If a contact lens is to be prescribed instead, find p, q, and f and the power of the lens.

To determine

Is the person nearsighted or farsighted.

Answer to Problem 10P

The person is farsighted.

Explanation of Solution

The person is able to see distant objects but unable to focus on objects at the normal near point for the human eye.

Conclusion:

Thus, the person is farsighted.

To determine

The minimum object distance from the lens.

Answer to Problem 10P

The minimum object distance from the lens is $18.0\text{cm}$.

Explanation of Solution

With the corrective lens $2.00\text{cm}$ in front of the eye, the object distance for an object $20.0\text{cm}$ in front of the eye is,

$\begin{array}{c}p=20.0\text{cm}-2.00\text{cm}\\ =\text{18}\text{.0}\text{cm}\end{array}$

Conclusion:

Thus, the minimum object distance from the lens is $\underset{\_}{18.0\text{cm}}$.

To determine

What image position with respect to the lens will allow the patient to see the object.

Answer to Problem 10P

The image position is $38.0\text{cm}$.

Explanation of Solution

The upright, virtual image formed by the corrective lens will serve as the object for the eye.

The object must be $40.0\text{cm}$ in front of the eye. With the lens $2.00\text{cm}$ in front of the eye, the image distance is,

$\begin{array}{c}\left|q\right|=40.0\text{cm}-2.00\text{cm}\\ =\text{38}\text{.0}\text{cm}\end{array}$

Conclusion:

Thus, the image position is $\underset{\_}{38.0\text{cm}}$.

To determine

Is the image real or virtual. Is the image distance positive or negative.

Answer to Problem 10P

The image is virtual and image distance is negative.

Explanation of Solution

The image must be located in front of the corrective lens, so it is a virtual image. The image distance is negative.

$q=-38.0\text{cm}$

Conclusion:

Thus, the image is virtual and image distance is negative.

To determine

Calculate the required focal length.

Answer to Problem 10P

The required focal length is $34.2\text{cm}$.

Explanation of Solution

The lens equation is,

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

• $f$ is the focal length
• $p$ is the object distance
• $q$ is the image distance

The focal length is,

$f=\frac{pq}{p+q}$

Substitute $18.0\text{cm}$ for $p$ and $-38.0\text{cm}$ for $q$.

$\begin{array}{c}f=\frac{\left(18.0\text{cm}\right)\left(-38.0\text{cm}\right)}{18.0\text{cm}-38.0\text{cm}}\\ =34.2\text{cm}\end{array}$

Conclusion:

Thus, the required focal length is $\underset{\_}{34.2\text{cm}}$.

To determine

The power of the corrective lens.

Answer to Problem 10P

The power of the corrective lens is $2.92\text{diopters}$.

Explanation of Solution

The power of the lens is,

$P=\frac{1}{f_{\text{in meters}}}$

Substitute $0.342\text{m}$ for $f$.

$\begin{array}{c}P=\frac{1}{0.342\text{m}}\\ =2.92\text{diopters}\end{array}$

Conclusion:

Thus, the power of the corrective lens is $\underset{\_}{2.92\text{diopters}}$.

To determine

The focal length of the corrective lens.

Answer to Problem 10P

The required values are $p=20.0\text{cm}$, $q=-40.0\text{cm}$, $f=40.0\text{cm}$ and $P=2.50\text{diopters}$.

Explanation of Solution

With the contact lens, the lens to eye distance would be zero. So, the object distance is,

$p=20.0\text{cm}$

The image distance is,

$q=-40.0\text{cm}$

The focal length is,

$f=\frac{pq}{p+q}$

Substitute $20.0\text{cm}$ for $p$ and $-40.0\text{cm}$ for $q$.

$\begin{array}{c}f=\frac{\left(20.0\text{cm}\right)\left(-40.0\text{cm}\right)}{20.0\text{cm}-40.0\text{cm}}\\ =40.0\text{cm}\end{array}$

The power of the lens is,

$P=\frac{1}{f_{\text{in meters}}}$

Substitute $0.400\text{m}$ for $f$.

$\begin{array}{c}P=\frac{1}{0.400\text{m}}\\ =2.50\text{diopters}\end{array}$

Conclusion:

Thus, the required values are $\underset{\_}{p=20.0\text{cm}}$, $\underset{\_}{q=-40.0\text{cm}}$, $\underset{\_}{f=40.0\text{cm}}$ and $\underset{\_}{P=2.50\text{diopters}}$.