Chapter 23, Problem 48P

To determine

To Find: The position, type and height of the image.

Answer to Problem 48P

The position of the image is $\mathrm{v}=-7.9\text{ cm}$ and the image is virtual. Also, the height of the image is ${\mathrm{h}}_{\mathrm{i}}=0.23\text{ cm}$

Explanation of Solution

Given info:

Power of lens, $\mathrm{P}=-5.5\text{ D}$

Object distance, $\mathrm{u}=14.0\text{ cm}=0.14\text{ m}$

Height of object, ${\mathrm{h}}_{\mathrm{o}}=4.0\text{ mm}=4\times {10}^{-3}\text{m}$

Formula used:

Lens formula can be given as:

  $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}$

Here,

  $\mathrm{f}$ is the focal length.

  $\mathrm{u}$ is the object distance.

  $\mathrm{v}$ is the image distance.

Calculation:

Now, power can be written as:

  $\mathrm{P}=\frac{1}{\mathrm{f}}=-5.5\text{ D}$

So, putting all the values in the formula, we get:

  $\begin{array}{l}-5.5\text{ D}=\frac{1}{0.14\text{ m}}+\frac{1}{\mathrm{v}}\\ \mathrm{v}=-0.079\text{ m}\end{array}$

Image will be virtual.

Now, the height of the image can be calculated as follows

  $\begin{array}{l}\mathrm{m}=\frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{h}}_{\mathrm{o}}}=-\frac{\mathrm{v}}{\mathrm{u}}\\ \frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{h}}_{\mathrm{o}}}=-\frac{\mathrm{v}}{\mathrm{u}}\\ {\mathrm{h}}_{\mathrm{i}}=-\frac{\mathrm{v}{\mathrm{h}}_{\mathrm{o}}}{\mathrm{u}}\\ {\mathrm{h}}_{\mathrm{i}}=-\frac{\left(-0.079\text{ m}\right)\left(4\times {10}^{-3}\text{m}\right)}{0.14\text{ m}}\\ {\mathrm{h}}_{\mathrm{i}}=0.0023\text{ m}\\ {\mathrm{h}}_{\mathrm{i}}=0.23\text{ cm}\end{array}$

Conclusion:

Thus, the position of the image is $\mathrm{v}=-7.9\text{ cm}$ and the image is virtual. Also, the height of the image is ${\mathrm{h}}_{\mathrm{i}}=0.23\text{ cm}$ .