Chapter 23, Problem 15P

To determine

The radius of curvature of the mirror.

Answer to Problem 15P

Solution:

The nature of mirror is concave mirror as focal length is positive and its radius of curvature is + 5.7 m.:

Explanation of Solution

Given Info:

Image is upright.

Let height of object be $+h_0$

Height of image $h_i=+3h_o$

Object distance $u=+1.9\text{m}$

Image distance = ‘v’

Nature of image is upright and three times of the object’s height

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

$\begin{array}{l}\frac{h_i}{h_0}=-\frac{v}{u}\\ \frac{+3h_o}{+h_0}=-\frac{v}{+u}\\ or,v=-3u\end{array}$

Apply focal length formula for mirror

$\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{f}=\frac{1}{-3u}+\frac{1}{u}\\ f=\frac{3u}{2}=\frac{3\times \left(\text{+1}\text{.9}\text{m}\right)}{2}\\ f=+2.85\text{m}\end{array}$

Since the radius of curvature R = 2f

So, the radius of curvature R = 2 ×(+2.85 m)= + 5.7 m: