Chapter 18, Problem 66A

(a)

To determine

The focal length of the lens.

Answer to Problem 66A

The focal length of the lens is $6.0\text{ cm}$ .

Explanation of Solution

Given:

The height of the object is $h_{\text{o}}=8.0\text{ cm}$ .

The position of the object is $d_{\text{o}}=15.0\text{ cm}$ .

The position of the image is $d_{\text{i}}=10.0\text{ cm}$ .

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 focal length is,

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ f=\frac{d_{\text{i}}{d}_{\text{o}}}{d_{\text{o}}+d_{\text{i}}}\\ f=\frac{\left(10.0\text{ cm}\right)\left(15.0\text{ cm}\right)}{\left(15.0\text{ cm}\right)+\left(10.0\text{ cm}\right)}\\ f=6.0\text{ cm}\end{array}$

Conclusion:

Thus, the focal length of the lens is $6.0\text{ cm}$ .

(b)

To determine

The location, size, and orientation of the image for the given changes in lens.

Answer to Problem 66A

The location of the image is $60.0\text{ cm}$ .

The size of the image is $-32.0\text{ cm}$ .

The image is real and inverted.

Explanation of Solution

Given:

The height of the object is $h_{\text{o}}=8.0\text{ cm}$ .

The position of the object is $d_{\text{o}}=15.0\text{ cm}$ .

The position of the image is $d_{\text{i}}=10.0\text{ cm}$ .

Focal length of lens is doubled.

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.

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:

From part (a), the focal length of the lens is $6.0\text{ cm}$ .

The new focal length of the lens as follows:

  $\begin{array}{l}{f}_{\text{new}}=2f\\ f_{\text{new}}=2\times 6.0\text{ cm}\\ f_{\text{new}}=12.0\text{ cm}\end{array}$

The new location of the image is,

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\ d_{\text{i}}=\frac{f{d}_{\text{o}}}{d_{\text{o}}-f}\\ d_{\text{i}}=\frac{\left(12.0\text{ cm}\right)\left(15.0\text{ cm}\right)}{\left(15.0\text{ cm}\right)-\left(12.0\text{ cm}\right)}\end{array}$

  $d_{\text{i}}=60.0\text{ cm}$

The new height of the image is calculated using magnification formula.

  $\begin{array}{l}m=\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}\\ h_{\text{i}}=-\frac{d_{\text{i}}{h}_{\text{o}}}{d_{\text{o}}}\\ h_{\text{i}}=-\frac{\left(60.\text{ cm}\right)\left(\text{8}\text{.0 cm}\right)}{\left(15.0\text{ cm}\right)}\\ h_{\text{i}}=-32.0\text{ cm}\end{array}$

Hence, the image is real and inverted.

Conclusion:

Thus, the location of the image is $60.0\text{ cm}$ , the size of the image is $-32.0\text{ cm}$ and the image is real and inverted.