Chapter 32, Problem 27P

(a)

To determine

The image distances for given object distances.

Explanation of Solution

Given:

The object distances are, $55\text{cm}$ , $24\text{cm}$ , $12\text{cm}$ , $8.0\text{cm}$ and $1.0\text{cm}$ .

The radius of curvature is $24\text{cm}$ .

Formula used:

Write expression for mirror equation.

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

Here, $v$ is image distance, $u$ is object distance and $f$ is focal length.

Substitute $\frac{r}{2}$ for $f$ in above expression.

  $\frac{1}{v}+\frac{1}{u}=\frac{2}{r}$

Solve above expression for $v$ .

  $v=\frac{ru}{2u-r}$  ....... (1)

Calculation:

Substitute $v_1$ for $v$ , $-24\text{cm}$ for $r$ and $55\text{cm}$ for $u$ in equation (1).

  $\begin{array}{l}{v}_1=\frac{\left(55\text{cm}\right)\left(-24\text{cm}\right)}{2\left(55\text{cm}\right)-\left(-24\text{cm}\right)}\\ v_1=-9.85\text{cm}\end{array}$

Substitute $v_2$ for $v$ , $-24\text{cm}$ for $r$ and $24\text{cm}$ for $u$ in equation (1).

  $\begin{array}{l}{v}_2=\frac{\left(-24\text{cm}\right)\left(24\text{cm}\right)}{2\left(24\text{cm}\right)-\left(-24\text{cm}\right)}\\ v_2=-8\text{cm}\end{array}$

Substitute $v_3$ for $v$ , $-24\text{cm}$ for $r$ and $12\text{cm}$ for $u$ in equation (1).

  $\begin{array}{l}{v}_3=\frac{\left(-24\text{cm}\right)\left(12\text{cm}\right)}{2\left(12\text{cm}\right)-\left(-24\text{cm}\right)}\\ v_3=-6\text{cm}\end{array}$

Substitute $v_4$ for $v$ , $-24\text{cm}$ for $r$ and $8\text{cm}$ for $u$ in equation (1).

  $\begin{array}{l}{v}_4=\frac{\left(-24\text{cm}\right)\left(8\text{cm}\right)}{2\left(8\text{cm}\right)-\left(-24\text{cm}\right)}\\ v_4=-4.8\text{cm}\end{array}$

Substitute $v_5$ for $v$ , $-24\text{cm}$ for $r$ and $1.0\text{cm}$ for $u$ in equation (1).

  $\begin{array}{l}{v}_5=\frac{\left(-24\text{cm}\right)\left(1\text{cm}\right)}{2\left(1\text{cm}\right)-\left(-24\text{cm}\right)}\\ v_5=-0.92\text{cm}\end{array}$

Conclusion:

Thus, the image distance is $-9.85\text{cm}$ for $55\text{cm}$ ,the image formation for object at $24\text{cm}$ is $-8\text{cm}$ , the image formation for object at $12\text{cm}$ is $-6\text{cm}$ , the image formation for object at $8\text{cm}$ is $-4.8\text{cm}$ and the image formation for object at $1.0\text{cm}$ is at $-0.92\text{cm}$ .

(b)

To determine

The magnification of each given object distance.

Explanation of Solution

Given:

The object distances are, $55\text{cm}$ , $24\text{cm}$ , $12\text{cm}$ , $8.0\text{cm}$ and $1.0\text{cm}$ .

The radius of curvature is $24\text{cm}$ .

Formula used:

Write expression for magnification for mirror.

  $m=\frac{-v}{u}$  ....... (1)

Calculation:

Substitute $m_1$ for $m$ , $55\text{cm}$ for $u$ and $-9.85\text{cm}$ for $v$ in equation (1).

  $\begin{array}{l}{m}_1= \frac{-\left(-9.85\text{cm}\right)}{55}\\ m_1=0.18\end{array}$

Substitute $m_2$ for $m$ , $24\text{cm}$ for $u$ and $-8\text{cm}$ for $v$ in equation (1).

  $\begin{array}{l}{m}_2=\frac{-\left(-8\text{cm}\right)}{24\text{cm}}\\ m_2=0.33\end{array}$

Substitute $m_3$ for $m$ , $12\text{cm}$ for $u$ and $-6\text{cm}$ for $v$ in equation (1).

  $\begin{array}{l}{m}_3=\frac{-\left(-6\text{cm}\right)}{12}\\ m_3=0.5\end{array}$

Substitute $m_4$ for $m$ , $8.0\text{cm}$ for $u$ and $-4.8\text{cm}$ for $v$ in equation (1).

  $\begin{array}{l}{m}_4=\frac{-\left(-4.8\text{cm}\right)}{8.0}\\ m_4=0.60\end{array}$

Substitute $m_5$ for $m$ , $1.0\text{cm}$ for $u$ and $-0.92\text{cm}$ for $v$ in equation (1).

  $\begin{array}{l}{m}_5=\frac{-\left(-0.92\right)}{1.0}\\ m_5=0.92\end{array}$

Conclusion:

Thus, the magnification for first case is $0.18$ , for second case is $0.33$ , for third case is $0.50$ , for fourth case is $0.60$ and for fifth case is $0.92$ .