Chapter 23, Problem 71P

(a)

To determine

Focal length of each lens.

Answer to Problem 71P

The focal length of each lens is $0.500\text{m}$.

Explanation of Solution

Write an expression for the focal length of each lens.

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

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

Conclusion:

Substitute $2.00{\text{m}}^{-1}$ for $P$ in equation (I) to find $f$.

    $\begin{array}{c}f=\frac{1}{2.00{\text{m}}^{-1}}\\ =0.500\text{m}\end{array}$

Thus, the focal length of each lens is $0.500\text{m}$.

(b)

To determine

The type of lens.

Answer to Problem 71P

The lenses are converging.

Explanation of Solution

Lenses can be categorized in to two. The categorization of lenses is based on the sign of focal length of the lens. Types of lenses according to the focal length are converging and diverging.

Converging lenses have positive focal length. Diverging lenses have negative focal length. Here, the lenses have positive focal length. Thus, the lenses are converging.

(c)

To determine

The image distance.

Answer to Problem 71P

The image is formed $2.00\text{m}$ from the lens on the same side of the lens as the object.

Explanation of Solution

Write an expression for the image distance.

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

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

Conclusion:

Substitute $0.500\text{m}$ for $f$ and $0.400\text{m}$ for $p$ in equation (II) to find $q$.

    $\begin{array}{c}q={\left(\frac{1}{0.500\text{m}}-\frac{1}{0.400\text{m}}\right)}^{-1}\\ ={\left(2-2.5\right)}^{-1}\\ =-2.00\text{m}\end{array}$

Thus, the image is formed $2.00\text{m}$ from the lens on the same side of the lens as the object.

(d)

To determine

Size of image relative to the size of object.

Answer to Problem 71P

The size of the image is 5 times larger than the object.

Explanation of Solution

Size of image relative to the size of object is given by the magnification of the lens.

Write an expression for the magnification of the lens.

    $\frac{h\text{'}}{h}=-\frac{q}{p}$                                                                                                                (III)

Here, $h\text{'}/h$ is the size of image relative to the size of object.

Conclusion:

Substitute $-2.00\text{m}$ for $q$ and $0.400\text{m}$ for $p$ in equation (III) to find $h\text{'}/h$.

    $\begin{array}{c}\frac{h\text{'}}{h}=-\frac{-2.00\text{m}}{0.400\text{m}}\\ =5\end{array}$

Thus, the size of the image is 5 times larger than the object.

(e)

To determine

The characteristics of image.

Answer to Problem 71P

The image is upright.

Explanation of Solution

Write an expression for the magnification

    $m=\frac{h\text{'}}{h}$                                                                                                               (IV)

Here, $m$ is the magnification.

If the magnification is positive, the image is upright. If the magnification is negative, the image is inverted. Here, the magnification is positive. Thus, the image is upright.