Chapter 23, Problem 22P

(a)

To determine

To find: Type of the mirror.

Explanation of Solution

Introduction:

A mirror that has a reflecting surface inwards is called a concave mirror. It is also called a converging mirror.

According to the given statement, the image is enlarged. So, when the object is placed between the optical centre and centre of curvature of the mirror, only then the image produced is enlarged in case of a concave mirror. Hence, the mirror is concave.

Conclusion:

Concave mirror.

(b)

To determine

To Explain: The type of the image.

Explanation of Solution

Introduction:

A mirror that has a reflecting surface inwards is called a concave mirror. It is also called a converging mirror.

When the object is placed closer to the concave mirror, then the image formed will be upright, virtual and enlarged.

Conclusion:

The image is virtual, upright and enlarged.

(c)

To determine

To find: Required radius of curvature of the mirror.

Answer to Problem 22P

The required radius of curvature is $\mathrm{R}=1.6\text{ m}$

Explanation of Solution

Given:

Magnification, $\mathrm{m}=1.33$

Object distance, $\mathrm{u}=20.0\text{ cm}$

Formula used:

Magnification can be given as: $\mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}$

Mirror formula is given by:

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

Here,

  $\mathrm{f}$ is the focal length

  $\mathrm{u}$ is the object distance

  $\mathrm{v}$ is the image distance.

Calculation:

Firstly, the image distance is given by:

  $\begin{array}{l}\mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}\\ \mathrm{v}=-\mathrm{m}\mathrm{u}\\ \mathrm{v}=-\left(1.33\right)\left(20.0\text{ cm}\right)\\ \mathrm{v}=26.6\text{ cm}\end{array}$

Now, the focal length can be calculated as follows:

  $\begin{array}{l}\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}\\ \mathrm{f}=\frac{\mathrm{u}\mathrm{v}}{\mathrm{u}+\mathrm{v}}\\ \mathrm{f}=\frac{\left(20.0\text{ cm}\right)\left(-26.6\text{ cm}\right)}{\left(20.0\text{ cm}\right)+\left(-26.6\text{ cm}\right)}\\ \mathrm{f}=80.6\text{ cm}\end{array}$

Lastly, the radius of curvature is:

  $\begin{array}{l}\mathrm{R}=2\mathrm{f}\\ \mathrm{R}=2\left(80.6\text{ cm}\right)\\ \mathrm{R}=1.6\text{ m}\end{array}$

Conclusion:

The radius of curvature is $\mathrm{R}=1.6\text{ m}$