Chapter 23, Problem 82GP

To determine

Part (a) To Determine:

The meaning of the limiting case of spherical mirror is plane mirror.

Answer to Problem 82GP

Solution:

Yes, plane mirror is the limiting case of spherical mirror.

Explanation of Solution

Given Info:

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

Let Object distance be $u$ and Image distance be $v$

Let the focal length of mirror be ‘f’

Formula Used:

Apply focal length formula for spherical mirror

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

Here, all alphabets are in their usual meanings.

Calculation:

A plane mirror can be considered a limiting case of a spherical mirror.

This will be only applicable when its radius of curvature increases then its surface becomes flatter. When its radius becomes infinity then its focal length will also be infinity and finally it will behave as plane mirror.

Hence, we can say; plane mirror is the limiting case of spherical mirror.

To determine

Part (b) To Determine:

The equation related to image and object distances in term of the limit of plane mirror.

Answer to Problem 82GP

Solution:

The equation related to image and object distances in term of the limit of plane mirror is $v+u=0$

Explanation of Solution

Given Info:

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

Let Object distance be $u$ and Image distance be $v$

Let the focal length of mirror be ‘f’

Formula Used:

Apply focal length formula for spherical mirror

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

Here, all alphabets are in their usual meanings.

Calculation:

Apply focal length formula for spherical mirror

$\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{\infty }=\frac{1}{v}+\frac{1}{u}----incaseofplanemirror,f=\infty \\ or,v=-u------(i)\\ or,v+u=0\end{array}$

This is the required equation.

To determine

Part (c) To Determine:

The magnification of plane mirror.

Answer to Problem 82GP

Solution:

The magnification of plane mirror ‘m’ = + 1

Explanation of Solution

Given Info:

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

Let Object distance be $u$ and Image distance be $v$

Let the focal length of mirror be ‘f’

Formula Used:

Magnification of spherical mirror $m=\frac{h_i}{h_0}=-\frac{v}{u}$

Here, all alphabets are in their usual meanings.

Calculation:

Apply focal length formula for spherical mirror

$\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{\infty }=\frac{1}{v}+\frac{1}{u}----incaseofplanemirror,f=\infty \\ or,v=-u------(i)\end{array}$

Magnification of spherical mirror $m=\frac{h_i}{h_0}=-\frac{v}{u}$

In case of plane mirror, we use equation (i)

Magnification of mirror $\begin{array}{l}m=-\frac{v}{u}\\ m=-\frac{-u}{+u}\\ or,m=+1------(ii)\end{array}$

To determine

Part (d) To Determine:

Are the above results consistent.

Answer to Problem 82GP

Solution:

Yes, the result of part (b)and part (c)both are the consistence results in terms of plane mirror which is the limiting case of spherical mirror.

Explanation of Solution

Given Info:

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

Let Object distance be $u$ and Image distance be $v$

Let the focal length of mirror be ‘f’

Formula Used:

Apply focal length formula for spherical mirror

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

Magnification of spherical mirror $m=\frac{h_i}{h_0}=-\frac{v}{u}$

Here, all alphabets are in their usual meanings.

Calculation:

Apply focal length formula for spherical mirror

$\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{\infty }=\frac{1}{v}+\frac{1}{u}----incaseofplanemirror,f=\infty \\ or,v=-u------(i)\\ or,v+u=0\end{array}$

This is the required equation.

Magnification of spherical mirror $m=\frac{h_i}{h_0}=-\frac{v}{u}$

In case of plane mirror, we use equation (i)

Magnification of mirror $\begin{array}{l}m=-\frac{v}{u}\\ m=-\frac{-u}{+u}\\ or,m=+1------(ii)\end{array}$

From equation (i)and (ii), we conclude that part (b)and part (c)both are the consistent results in terms of plane mirror which is the limiting case of spherical mirror.