Chapter 17, Problem 10STP

To determine

To draw:The scale-ray diagram that will show the image height.

Answer to Problem 10STP

The scale-ray diagram that will show the image height is shown in Figure 1.

Explanation of Solution

Given:

Height of an object is $h_o=5.0\text{ cm = 5}\times {\text{10}}^{-2}\text{ m}$

Distance between the object and the convex mirror is $d_o=20.0\text{ cm = 0}\text{.2 m}$

Focal length of a convex mirror is $f=-14.0\text{ cm = }-\text{0}\text{.14 m}$

Formula used:

The relationship between the object distance $\left(d_o\right)$ , image distance $\left(d_i\right)$ and the focal length $\left(f\right)$ of a mirror is given by,

  $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$

  $\Rightarrow \frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}$

  $\Rightarrow \frac{1}{d_i}=\frac{d_o-f}{f{d}_o}$

  $\Rightarrow d_i=\frac{f{d}_o}{d_o-f}$ $\left(1\right)$

The magnification $\left(M\right)$ equation gives the relationship between the image height $\left(h_i\right)$ , object height $\left(h_o\right)$ , image distance $\left(d_i\right)$ , and object distance $\left(d_o\right)$ as,

  $M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

  $\Rightarrow \frac{h_i}{h_o}=-\frac{d_i}{d_o}$

  $\Rightarrow h_i=-\frac{d_i}{d_o}{h}_o$ $\left(2\right)$

Calculation:

Image position can be calculated by substituting the numerical values in equation $\left(1\right)$ ,

  $d_i=\frac{\left(-\text{0}\text{.14 m}\right)\left(\text{0}\text{.2 m}\right)}{\left(\text{0}\text{.2 m}\right)-\left(-\text{0}\text{.14 m}\right)}$

  $=-\frac{0.028{\text{ m}}^2}{0.34\text{ m}}=-0.082\text{ m = }-\text{8}\text{.2 cm}$

Image height can be calculated by substituting the numerical values in equation $\left(2\right)$ ,

  $h_i=-\frac{\left(-0.082\text{ m}\right)}{\left(\text{0}\text{.2 m}\right)}\left(\text{5}\times {\text{10}}^{-2}\text{ m}\right)$

  $=2.05\times {\text{10}}^{-2}\text{ m}=2.05\text{ cm}$

Scale-ray diagram showing the image height is shown in figure 1.

  

Figure 1