Chapter 23, Problem 16P

To determine

The object distance as multiple of its focal length.

Answer to Problem 16P

Solution:

The object distance as multiple of its focal length $u=2f$

Explanation of Solution

Given Info:

Let height of object be $+h_0$

Height of image $h_i=-h_o$

Let object and image distance be ‘u’ and ‘v’.

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 $m=\frac{h_i}{h_0}=-\frac{v}{u}$

$\begin{array}{l}\frac{-h_o}{+h_0}=-\frac{v}{u}\\ or,v=u\end{array}$

Apply focal length formula for mirror

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

Hence the object distance will be double of its focal length. That is $u=2f$