Chapter 23, Problem 20P

To determine

Part(a)To Determine:

The object’s distance

Answer to Problem 20P

Solution:

The object’s distance will be it radius of curvature.

Explanation of Solution

Given Info:

Image is formed at the same location of the object.

Nature of mirror is concave

Let height of object be $+h_0$ and height of image be $h_i$

Object distance $u$= Image distance $v$

Let the focal length of mirror be ‘f’

Formula Used:

Apply focal length formula for mirror

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

Here, all alphabets are in their usual meanings.

Calculation:

Apply focal length formula for mirror

$\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{f}=\frac{1}{u}+\frac{1}{u}\\ f=\frac{u}{2}\\ or,u=2f\end{array}$

Hence, the object should be placed at the centre of the curvature of concave mirror.

To determine

Part(b)To Determine:

Is the image real or virtual?

Answer to Problem 20P

Solution:

The image will be real.:

Explanation of Solution

Given Info:

Image is formed at the same location of the object.

Nature of mirror is concave

Let height of object be $+h_0$ and height of image be $h_i$

Object distance $u$= Image distance $v$

Let the focal length of mirror be ‘f’

Formula Used:

Apply focal length formula for mirror

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

Here, all alphabets are in their usual meanings.

Calculation:

Apply focal length formula for mirror

$\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{f}=\frac{1}{u}+\frac{1}{u}\\ f=\frac{u}{2}\\ or,u=2f\end{array}$

Since, Object distance $u$= Image distance $v$ and both are lie in front of the concave mirror. Hence the image will be real.:

To determine

Part(c)To Determine:

Is the image inverted or upright?

Answer to Problem 20P

Solution:

The image will be inverted.:

Explanation of Solution

Given Info:

Image is formed at the same location of the object.

Nature of mirror is concave

Let height of object be $+h_0$ and height of image be $h_i$

Object distance $u$= Image distance $v$

Let the focal length of mirror be ‘f’

Formula Used:

(i)Apply focal length formula for mirror

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

(ii)Magnification of mirror $m=\frac{h_i}{h_0}=-\frac{v}{u}$

Here, all alphabets are in their usual meanings.

Calculation:

Apply focal length formula for mirror

$\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{f}=\frac{1}{u}+\frac{1}{u}\\ f=\frac{u}{2}\\ or,u=2f\end{array}$

Apply magnification of mirror $m=\frac{h_i}{h_0}=-\frac{v}{u}$

$\begin{array}{l}\frac{h_i}{h_0}=-\frac{+v}{+u}\\ h_i=-h_o------(i)\end{array}$

Since $h_i=-h_o$ so the image will be inverted.:

To determine

Part(d)To Determine:

The magnification of the image

Answer to Problem 20P

Solution:

The magnification of the image will be – 1:

Explanation of Solution

Given Info:

Image is formed at the same location of the object.

Nature of mirror is concave

Let height of object be $+h_0$ and height of image be $h_i$

Object distance $u$= Image distance $v$

Let the focal length of mirror be ‘f’

Formula Used:

(i)Apply focal length formula for mirror

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

(ii)Magnification of mirror $m=\frac{h_i}{h_0}=-\frac{v}{u}$

Here, all alphabets are in their usual meanings.

Calculation:

Apply magnification of mirror $m=\frac{h_i}{h_0}=-\frac{v}{u}$

$\begin{array}{l}\frac{h_i}{h_0}=-\frac{+v}{+u}\\ h_i=-h_o------(i)\end{array}$

Magnification of mirror $m=\frac{h_i}{h_0}=\frac{-h_0}{h_0}=-1$