Chapter 36, Problem 33SP

It is desired to cast the image of a lamp, magnified 5 times, upon a wall 12 m distant from the lamp. What kind of spherical mirror is required, and what is its position?

To determine

The nature of the mirror required and the position of the image when the mirror magnified the image $5$ times upon a wall that is at a distance of $12\text{ m}$ from the lamp.

Answer to Problem 33SP

Solution:

Concave, radius of $5.0\text{m}$ , and $3.0\text{m}$ distance from the lamp.

Explanation of Solution

Given data:

The distance between thewall and the lampis $12\text{ m}$.

The magnification needed by the mirror is $5$ times.

Formula used:

Themirror equation is written as,

$\frac{1}{f}=\frac{1}{s_o}+\frac{1}{s_i}$

Here, $f$ is the focal length of the mirror, $s_o$ is the object distance from the mirror, and $s_i$ is the image distance from the mirror.

The formula for magnification of the mirror is written as,

$\begin{array}{c}{M}_T=\frac{y_i}{y_o}\\ =-\frac{s_i}{s_o}\end{array}$

Here, $y_i$ is the height of the image and $y_o$ is the height of the object.

The thin mirror equation is written as,

$\frac{1}{f}=-\frac{2}{R}$

Here, $f$ is the focal length of the mirror and $R$ is the radius of curvature of the mirror.

Sign convention:

If R is negative, then the centre of curvature is to the left of the mirror, and the mirror is concave.

If R is positive, then the centre of curvature is to the right of the mirror, and the mirror is convex.

If $s_o$ is positive, then the object is in front (that is to the left) of the mirror.

If $s_i$ is positive, then the image is real (that isin front or to the left of the mirror).

If $s_i$ is negative, then the image is virtual (that isbehind or to the right of the mirror).

If f is positive, then the mirror is concave.

If f is negative, then the mirror is convex.

Explanation:

Recall the expression for themagnification of the mirror.

$M_T=-\frac{s_i}{s_o}$

Understand that the image is formed to the right of the mirror. Hence, the image would be negative and the total distance of image from the mirror will be,

$s_i=-\left(s_o+d\right)$

Here, $d$ is the distance between the wall and the lamp.

Substitute $12\text{ m}$ for $d$

$s_i=-\left(s_o+12\text{ m}\right)$

Substitute $5$ for $M_T$ and $-\left(s_o+12\text{ m}\right)$ for $s_i$

$5=-\frac{-\left(s_o+12\text{ m}\right)}{s_o}$

Solve for $s_o$.

$\begin{array}{c}5s_0=s_o+12\text{ m}\\ 4s_0=12\text{ m}\\ s_0=\frac{12\text{ m}}{4}\\ s_o=3\text{ m}\end{array}$

Understand that the distance of image from the mirror will be the sum of object distance and distance between the wall and the lamp.

$s_i=s_o+d$

Substitute $3\text{ m}$ for $s_o$ and $12\text{ m}$ for $d$.

$\begin{array}{c}{s}_i=3\text{ m}+12\text{ m}\\ =15\text{ m}\end{array}$

Recall the expression formirror equation.

$\frac{1}{f}=\frac{1}{s_o}+\frac{1}{s_i}$

Substitute $15\text{ m}$ for $s_i$ and $\text{3 m}$ for $s_o$

$\begin{array}{c}\frac{1}{f}=\frac{1}{15\text{ m}}+\frac{1}{\text{3 m}}\\ =\frac{18}{45}\text{ m}\end{array}$

Solve for $f$.

$\begin{array}{c}f=\frac{45}{18}\text{ m}\\ =2.5\text{ m}\end{array}$

Recall the expression forthin mirror.

$\frac{1}{f}=-\frac{2}{R}$

Solve for $R$.

$R=-2f$

Substitute $2.5\text{ m}$ for $f$

$\begin{array}{c}R=-2\left(2.5\text{ m}\right)\\ =-5.0\text{ m}\end{array}$

Negative sign indicates that the mirror is concave.

Conclusion:

Therefore, the mirror is c oncave, has a radius of curvature of $5.0\text{m}$ , and $3.0\text{m}$ distance from the lamp.