Chapter 37, Problem 55PQ

(a)

To determine

The location of the image.

Answer to Problem 55PQ

The location of the image is $-3.27\text{cm}$.

Explanation of Solution

In case of convex mirror, no matter where the object is placed. The image is always erect and virtual.

Write the expression for focal length.

    $f=-\frac{R}{2}$                                                                                                                   (I)

Here, $R$ is radius of mirror.

Write the expression for mirror equation.

    $\frac{1}{f}=\frac{1}{d_0}+\frac{1}{d_{\text{i}}}$

Here, $f$ is the focal length, $d_0$ is object distance and $d_{\text{i}}$ is image distance.

Solve the above equation for $d_{\text{i}}$.

    $\begin{array}{c}\frac{1}{f}=\frac{1}{d_0}+\frac{1}{d_{\text{i}}}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_0}\\ =\frac{d_0-f}{f{d}_0}\end{array}$

Simplify the above equation.

    $d_{\text{i}}=\frac{f{d}_0}{d_0-f}$                                                                                                             (II)

Conclusion:

Substitute $8\text{cm}$ for $R$ in equation (I) to find $f$.

    $\begin{array}{c}f=-\frac{8\text{cm}}{2}\\ =-4\text{cm}\end{array}$

Substitute, $-4\text{cm}$ for $f$ and $18\text{cm}$ for $d_{\text{o}}$ in equation (II) to find $d_{\text{i}}$.

    $\begin{array}{c}{d}_{\text{i}}=\frac{\left(-4\text{cm}\right)\times \left(18\text{cm}\right)}{18\text{cm}-\left(-4\text{cm}\right)}\\ =-3.27\text{cm}\end{array}$

Therefore, the location of the image is $-3.27\text{cm}$.

(b)

To determine

The magnification of the image.

Answer to Problem 55PQ

The magnification of the image is $0.182$.

Explanation of Solution

Write the expression for magnification in case of mirror.

  $M=-\frac{d_{\text{i}}}{d_0}$

Conclusion:

Substitute, $-3.27\text{cm}$ for $d_{\text{i}}$ and $18\text{cm}$ for $d_0$ in the above equation to find $M$.

    $\begin{array}{c}M=-\frac{\left(-3.27\text{cm}\right)}{18\text{cm}}\\ =0.182\end{array}$

Therefore, the magnification of the image is $0.182$.

(c)

To determine

The image is inverted or upright.

Answer to Problem 55PQ

The image is upright.

Explanation of Solution

The image is upright because the magnification value is positive as calculated in the part (b). And as the image distance is negative as calculated in part (a), it implies the image is virtual.

Hence, the image is virtual and upright.

Therefore, the image is upright.

(d)

To determine

The height of the image.

Answer to Problem 55PQ

The height of the image is $0.636\text{cm}$.

Explanation of Solution

Write the expression for magnification in case of mirror.

  $\begin{array}{c}M=-\frac{d_{\text{i}}}{d_0}\\ =\frac{h_{\text{i}}}{h_0}\end{array}$

Here, $h_{\text{i}}$ is height of image, $h_{\text{o}}$ is the height of the object, $d_{\text{i}}$ is the distance of the image and $d_{\text{o}}$ is the distance of the object.

Solve the above equation for $h_{\text{i}}$.

    $\begin{array}{c}-\frac{d_{\text{i}}}{d_0}=\frac{h_{\text{i}}}{h_0}\\ h_{\text{i}}=h_0\left(-\frac{d_{\text{i}}}{d_0}\right)\end{array}$

Conclusion:

Substitute $3.5\text{cm}$ for $h_0$, $-3.27\text{cm}$ for $d_{\text{i}}$ and $18\text{cm}$ for $d_0$ in the above equation to find $h_{\text{i}}$.

    $\begin{array}{c}{h}_{\text{i}}=\left(3.5\text{cm}\right)\left(-\frac{\left(-3.27\text{cm}\right)}{18\text{cm}}\right)\\ =0.636\text{cm}\end{array}$

Therefore, the height of the image is $0.636\text{cm}$.