Chapter 18, Problem 74A

(a)

To determine

The location of the image.

Answer to Problem 74A

The location of the image is $d_{\text{i}}=51\text{mm}$ .

Explanation of Solution

Given:

The distance of the object from the lens is $d_{\text{o}}=3.0\text{m}$ .

The focal length of the lens is $f=50.0\text{mm}$ .

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:

Consider the location of the image is denoted by $d_{\text{i}}$ . The location of the image is,

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\ d_{\text{i}}=\frac{f{d}_{\text{o}}}{d_{\text{o}}-f}\\ d_{\text{i}}=\frac{\left(\text{50 mm}\right)\left(\text{3}\text{.0}\text{m}\times \frac{1000\text{mm}}{1\text{m}}\right)}{\left(\text{3}\text{.0}\text{m}\times \frac{1000\text{mm}}{1\text{m}}\right)-\left(\text{50 mm}\right)}\\ d_{\text{i}}=\left(\frac{\text{50 mm}\times 3000\text{mm}}{3000\text{mm}-\text{50 mm}}\right)\\ d_{\text{i}}=51\text{mm}\end{array}$

Conclusion:

Thus, the location of the image is $d_{\text{i}}=51\text{mm}$ .

(b)

To determine

The location of the image for $1000\text{mm}$ lens.

Answer to Problem 74A

The location of the image is $d_{\text{i}}=1.01\times {10}^3\text{mm}$ .

Explanation of Solution

Given:

The distance of the object from the lens is $d_{\text{o}}=125.0\text{m}$ .

The focal length of the lens is $f=1000.0\text{mm}$ .

Consider the location of the image is denoted by $d_{\text{i}}$ .

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 location of the image is,

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\ d_{\text{i}}=\frac{f{d}_{\text{o}}}{d_{\text{o}}-f}\\ d_{\text{i}}=\frac{\left(\text{1000 mm}\right)\left(\text{125}\text{.0}\text{m}\times \frac{1000\text{mm}}{1\text{m}}\right)}{\left(\text{125}\text{.0}\text{m}\times \frac{1000\text{mm}}{1\text{m}}\right)-\left(\text{1000 mm}\right)}\\ d_{\text{i}}=\left(\frac{\text{1000 mm}\times 125000\text{mm}}{125000\text{mm}-100\text{0 mm}}\right)\\ d_{\text{i}}=1.01\times {10}^3\text{mm}\end{array}$

Conclusion:

Thus, the location of the image is $d_{\text{i}}=1.01\times {10}^3\text{mm}$ .