Chapter 37, Problem 39PQ

(a)

To determine

The location, magnification and nature of the image formed by the mirror.

Answer to Problem 39PQ

A real and inverted image is formed at a distance of $23.7\text{cm}$. The magnification of the image is $-0.395$.

Explanation of Solution

Write the equation for the focal length of the mirror.

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

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

Write the equation for the concave mirror.

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_0}+\frac{1}{d_{\text{i}}}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_0}\end{array}$                                                               (II)

Here, $d_0$ is the distance of object from the mirror and $d_{\text{i}}$ is the distance of image from the mirror.

Write the equation for magnification produced by the mirror.

  $M=-\frac{d_{\text{i}}}{d_0}$                                                                    (III)

Conclusion:

Substitute $34.0\text{cm}$ for $R$ in equation (I) to find $f$.

  $\begin{array}{c}f=\frac{34.0\text{cm}}{2}\\ =17.0\text{cm}\end{array}$

Substitute $17.0\text{cm}$ for $f$ and $60.0\text{cm}$ for $d_0$ in equation (II) to find $d_{\text{i}}$.

  $\begin{array}{c}\frac{1}{d_{\text{i}}}=\frac{1}{17.0\text{cm}}-\frac{1}{60.0\text{cm}}\\ =\frac{\left(60.0\text{cm}-17.0\text{cm}\right)}{\left(17.0\text{cm}\right)\left(60.0\text{cm}\right)}\\ =\frac{43}{1020.0\text{cm}}\\ =23.7\text{cm}\end{array}$

The distance of image is positive so the image distance will be real.

Substitute $23.7\text{cm}$ for $d_{\text{i}}$ and $60.0\text{cm}$ for $d_0$ in equation (III) to find $M$.

  $\begin{array}{c}M=-\frac{\text{23}\text{.7}\text{cm}}{\text{60}\text{.0}\text{cm}}\\ =-0.395\end{array}$

The magnification is negative so the image will be inverted.

Therefore, a real and inverted image is formed at a distance of $23.7\text{cm}$. The magnification of the image is $-0.395$.

(b)

To determine

The location, magnification and nature of the image formed by the mirror.

Answer to Problem 39PQ

A real and inverted image is formed at a distance of $34.0\text{cm}$. The magnification of the image is $-1.00$.

Explanation of Solution

Consider equation(II) and equation (III) to solve the problem.

Conclusion:

Substitute $17.0\text{cm}$ for $f$ and $34.0\text{cm}$ for $d_0$ in equation (II) to find $d_{\text{i}}$.

  $\begin{array}{c}\frac{1}{d_{\text{i}}}=\frac{1}{17.0\text{cm}}-\frac{1}{34.0\text{cm}}\\ =\frac{\left(34.0\text{cm}-17.0\text{cm}\right)}{\left(17.0\text{cm}\right)\left(34.0\text{cm}\right)}\\ =\frac{17.0}{\left(17.0\times 34.0\right)\text{cm}}\\ d_{\text{i}}=34.0\text{cm}\end{array}$

The distance of image is positive so the image distance will be real.

Substitute $34.0\text{cm}$ for $d_{\text{i}}$ and $34.0\text{cm}$ for $d_0$ in equation (III) to find $M$.

  $\begin{array}{c}M=-\frac{34.0\text{cm}}{34.0\text{cm}}\\ =-1.00\end{array}$

The magnification is negative so the image will be inverted.

Therefore, a real and inverted image is formed at a distance of $34.0\text{cm}$. The magnification of the image is $-1.00$.