Chapter 18.2, Problem 30SSC

(a)

To determine

The position of the image from the lens.

To identify: Whether the image is real or virtual.

Answer to Problem 30SSC

The position of the image from the lens is $30\text{ cm}$ .

The image is a real image.

Explanation of Solution

Given:

The height of the salt shaker is $h_{\text{o}}=6.5\text{ cm}$ .

The position of the salt shaker is $d_{\text{o}}=6.0\text{ cm}$ .

The focal length is $f=5.0\text{ cm}$

Formula used:

The expression for the mirror equation is,

  $\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}$

Here, $d_{\text{o}}$ is the position of the object, $d_{\text{i}}$ is the position of the image, and $f$ is the focal length.

Calculation:

The position of the image is,

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\ d_{\text{i}}=\frac{f{d}_{\text{o}}}{d_{\text{o}}-f}\\ d_{\text{i}}=\frac{\left(5.0\text{ cm}\right)\left(6.0\text{ cm}\right)}{\left(6.0\text{ cm}\right)-\left(5.0\text{ cm}\right)}\end{array}$

  $d_{\text{i}}=30.0\text{ cm}$

Hence, the image is a real image.

Conclusion:

Thus, the position of the image from the lens is $30.0\text{ cm}$ .

Thus, the image is a real image.

(b)

To determine

The magnification.

To identify: Whether the image is smaller or larger than the object.

Answer to Problem 30SSC

The magnification is $-5.0$ .

The image is an enlarged image.

Explanation of Solution

Given:

The height of the salt shaker is $h_{\text{o}}=6.5\text{ cm}$ .

The position of the salt shaker is $d_{\text{o}}=6.0\text{ cm}$ .

The focal length is $f=5.0\text{ cm}$

Formula used:

The expression for the magnification equation is,

  $m=\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}$

Here, $h_{\text{i}}$ is the height of the image and $h_{\text{o}}$ is the height of the object.

Calculation:

From part (a), the position of the image from the lens is $d_{\text{i}}=30.0\text{ cm}$ .

The magnification is,

  $\begin{array}{l}m=-\frac{d_{\text{i}}}{d_{\text{o}}}\\ m=-\frac{30.0\text{ cm}}{6.0\text{ cm}}\\ m=-5.0\end{array}$

Hence, the image is an enlarged image.

Conclusion:

Thus, the magnification is $m=-5.0$ .

Thus, the image is an enlarged image.

(c)

To determine

The distance of the image from the lens and the magnification.

To identify: Whether the image is smaller or larger than the object.

Answer to Problem 30SSC

The distance of the image from the lens is $-20.0\text{ cm}$ .

The magnification is $-5.0$ .

Explanation of Solution

Given:

The height of the salt shaker is $h_{\text{o}}=6.5\text{ cm}$ .

The position of the salt shaker is $d_{\text{o}}=4.0\text{ cm}$ .

The focal length is $f=5.0\text{ cm}$

Formula used:

Show the expression for the mirror equation as follows:

  $\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}$

Here, $d_{\text{o}}$ is the position of the object, $d_{\text{i}}$ is the position of the image, and $f$ is the focal length.

Show the expression for the magnification equation as follows:

  $m=\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}$

Here, $h_{\text{i}}$ is the height of the image and $h_{\text{o}}$ is the height of the object.

Calculation:

The position of the image is,

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\ d_{\text{i}}=\frac{f{d}_{\text{o}}}{d_{\text{o}}-f}\\ d_{\text{i}}=\frac{\left(5.0\text{ cm}\right)\left(4.0\text{ cm}\right)}{\left(4.0\text{ cm}\right)-\left(5.0\text{ cm}\right)}\end{array}$

  $d_{\text{i}}=-20.0\text{ cm}$

Hence, the image is the virtual image.

The magnification is,

  $\begin{array}{l}m=-\frac{d_{\text{i}}}{d_{\text{o}}}\\ m=-\frac{20.0\text{ cm}}{4.0\text{ cm}}\\ m=-5.0\end{array}$

Thus, the image is an enlarged image.

Conclusion:

Thus, the position of the image from the lens is $d_{\text{i}}=-20.0\text{ cm}$ .

Thus, the magnification is $m=-5.0$ .

Thus, the image is an enlarged image.