Chapter 26, Problem 53P

(a)

To determine

Position of the final image.

Answer to Problem 53P

The image is $\underset{\_}{25.3\text{cm}}$ to the right of the mirror.

Explanation of Solution

Consider the mirror and write the equation for radius of curvature

  $\frac{2}{R}=\frac{1}{p_1}+\frac{1}{q_1}$        (I)

Here, $R$ is the radius of curvature of the mirror, $p_1$ is the object distance and $q_1$ is the image distance.

Rearrange (I) in terms of $q_1$.

    $\frac{1}{q_1}=\frac{2}{R}-\frac{1}{p_1}$        (II)

The image of the mirror serves as the object for the lens.

The object distance for the lens would be

    $p_2=d-p_1$        (III)

Write the mirror equation.

    $\frac{1}{f_2}=\frac{1}{p_2}+\frac{1}{q_2}$        (IV)

Here $f_2$ is the focal length of the lens, $p_2$ is the object distance and $q_2$ is the image distance.

Rewrite (III) in terms of $q_2$.

    $\frac{1}{q_2}=\frac{1}{f_2}-\frac{1}{p_2}$        (V)

The final image will be at a distance $q_2-d$

Conclusion:

Substitute $20\text{cm}$ for $R$ and $12.5\text{cm}$ for $p_1$ in (II)

    $\begin{array}{c}\frac{1}{q_1}=\frac{2}{20\text{cm}}-\frac{1}{12.5\text{cm}}\\ q_1=50\text{cm}\end{array}$

Substitute $25\text{cm}$ for $d$ and $50\text{cm}$ for $q_1$ in (III)

    $\begin{array}{c}{p}_2\text{=25}\text{cm-50}\text{cm}\\ =-25\text{cm}\end{array}$

Substitute $-25\text{cm}$ for $p_2$ and $-16.7\text{cm}$ for $f_2$ in (V)

    $\begin{array}{l}\frac{1}{q_2}=\frac{1}{\left(-16.7\text{cm}\right)}-\frac{1}{\left(-25\text{cm}\right)}\\ q_2=-50.3\text{cm}\end{array}$

Thus the final image will be at $50.3\text{cm}-25\text{cm}=25.3\text{cm}$ to the right

(b)

To determine

Whether the image is real or virtual

Answer to Problem 53P

The image is virtual.

Explanation of Solution

The sign of image distance decides the nature of the image.\

The positive sign indicates that the image is real. Negative sign suggests that the image is virtual

Conclusion:

As the image distance is $-50.3\text{cm}$, the image is vertical.

(c)

To determine

Whether the image is upright or inverted..

Answer to Problem 53P

The image is upright.

Explanation of Solution

Sign of magnification decides whether the image is upright or inverted.

If magnification is positive, image is upright. If magnification is negative the image is inverted,

Write the equation for magnification mirror and lens

    $M_1=\frac{-q_1}{p_1}$        (VI)

    $M_2=\frac{-q_2}{p_2}$        (VII)

Write the equation for overall magnification

    $M=M_1{M}_2$        (VIII)

Conclusion:

Substitute $50\text{cm}$ for $q_1$ and $12.5\text{cm}$ for $p_1$ in (VI)

    $\begin{array}{c}{M}_1=\frac{-\left(50\text{cm}\right)}{12.5\text{cm}}\\ =-4\end{array}$

Substitute $-50.3\text{cm}$ for $q_2$ and $-25\text{cm}$ for $p_2$ in (VII)

    $\begin{array}{c}{M}_2=\frac{-\left(-50.3\text{cm}\right)}{-25\text{cm}}\\ =-2.01\end{array}$

Substitute $-4$ for $M_1$ and $-2$ for $M_2$ in (VIII)

    $\begin{array}{c}M=\left(-4\right)\left(-2\right)\\ =8.05\end{array}$

As the magnification is positive, the image is upright.

(d)

To determine

Overall magnification of the image.

Answer to Problem 53P

Magnification is $\underset{\_}{8.05}$.

Explanation of Solution

Refer sub part (c) and write the equation for overall magnification

  $M=M_1{M}_2$

Conclusion:

Substitute $-4$ for $M_1$ and $-2$ for $M_2$ in (VIII)

    $\begin{array}{c}M=\left(-4\right)\left(-2\right)\\ =8.05\end{array}$

Magnification is $\underset{\_}{8.05}$.