Chapter 26, Problem 72P

Figure P26.72 shows a thin converging lens for which the radii of curvature of its surfaces have magnitudes of 9.00 cm and 11.0 cm. The lens is in front of a concave spherical mirror with the radius of curvature R = 8.00 cm. Assume the focal points F₁ and F₂ of the lens are 5.00 cm from the center of the lens. (a) Determine the index of refraction of the lens material. The lens and mirror are 20.0 cm apart, and an object is placed 8.00 cm to the left of the lens. Determine (b) the position of the final image and (c) its magnification as seen by the eye in the figure. (d) Is the final image inverted or upright? Explain.

To determine

Refractive index of the lens material.

Answer to Problem 72P

Refractive index is $n=\underset{\_}{1.99}$.

Explanation of Solution

Write down the Lens- maker’s equation.

    $\frac{1}{f}=\left(n-1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$        (I)

Here $f$ is the focal length, $n$ is the refractive index, $R_1$ is the radius of curvature of front surface of the lens and $R_2$ is the radius of curvature.

Rearrange (I) in terms of $n$

    $n=\frac{1}{f\left[\frac{1}{R_1}-\frac{1}{R_2}\right]}+1$        (II)

Conclusion:

Substitute $5\text{cm}$ for $f$, $9\text{cm}$ for $R_1$ and $-11\text{cm}$ for $R_2$ in (II)

$\begin{array}{c}n=\frac{1}{5\text{cm}\left[\frac{1}{9\text{cm}}-\frac{1}{-11\text{cm}}\right]}+1\\ =1.99\end{array}$

Refractive index is $n=\underset{\_}{1.99}$.

To determine

Position of the final image.

Answer to Problem 72P

The final image is real

The image is $\underset{\_}{10\text{cm}}$ left to the lens.

Explanation of Solution

Write the thin lens equation

  $\frac{1}{f}=\frac{1}{p_1}+\frac{1}{q_1}$        (III)

Here $f$ is the focal length, $p_1$ is the object distance and $q_1$ is the image distance.

Rewrite (III) in terms of $q_1$.

    $q_1=\frac{p_{1}f}{p_1-f}$        (IV)

Write the equation for image magnification.

    $M=\frac{-q_1}{p_1}$        (V)

The image becomes the object for the concave mirror.

Then,

    $p_M=d-q_1$        (VI)

Here $p_M$ is the object distance of the mirror and $d$ is the distance between mirror and lens.

Write the equation for focal length

    $f^{\prime }=\frac{R}{2}$        (VII)

Here $R$ is the radius of curvature of the mirror.

Write the equation for $q_M$ from mirror equation

    $q_M=\frac{p_M{f}^{\prime }}{p_M-f^{\prime }}$        (VIII)

Write the equation for image magnification

    $M^{\prime }=\frac{-q_M}{p_M}$        (IX)

The image formed by the mirror serves as real object for the lens on the second pass.

    $p=d-q_M$        (X)

Then the image distance will be

    $q=\frac{pf}{p-f}$        (XI)

Write the equation for image magnification for this case

    $M^{″}=\frac{-q}{p}$        (XII)

Conclusion:

Substitute $8\text{cm}$ for $p_1$ and $5\text{cm}$ for $f$ in (IV)

    $\begin{array}{c}{q}_1=\frac{\left(8\text{cm}\right)\left(5\text{cm}\right)}{8\text{cm}-5\text{cm}}\\ =13.3\text{cm}\end{array}$

Substitute $8\text{cm}$ for $p_1$ and $13.3\text{cm}$ for $q_1$ in (V)

    $\begin{array}{c}M=\frac{-13.3\text{cm}}{8\text{cm}}\\ =-1.67\end{array}$

Substitute $20\text{cm}$ for $d$ and $13.3\text{cm}$ for $q_1$ in (VI)

    $\begin{array}{c}{p}_M=20\text{cm}-13.3\text{cm}\\ \text{=6}\text{.67}\text{cm}\end{array}$

Substitute $8\text{cm}$ for $R$ in (VII)

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

Substitute $\text{6}\text{.67}\text{cm}$ for $p_M$ and $4\text{cm}$ for $f^{\prime }$ in (VIII)

    $\begin{array}{c}{q}_M=\frac{\left(\text{6}\text{.67}\text{cm}\right)\left(4\text{cm}\right)}{\text{6}\text{.67}\text{cm}-4\text{cm}}\\ =10\text{cm}\end{array}$

Substitute $\text{6}\text{.67}\text{cm}$ for $p_M$ and $10\text{cm}$ for $q_M$ in (IX)

    $\begin{array}{c}{M}^{\prime }=\frac{-10\text{cm}}{\text{6}\text{.67}\text{cm}}\\ =-1.5\end{array}$

Substitute $20\text{cm}$ for $d$ and $10\text{cm}$ for $q_M$ in (X)

    $\begin{array}{c}p=20\text{cm}-10\text{cm}\\ \text{=10}\text{cm}\end{array}$

Substitute $\text{10}\text{cm}$ for $p$ and $5\text{cm}$ for $f$ in (XI)

    $\begin{array}{c}q=\frac{\left(\text{10}\text{cm}\right)\left(5\text{cm}\right)}{\text{10}\text{cm}-5\text{cm}}\\ =10\text{cm}\end{array}$

Substitute $10\text{cm}$ for $q$ and $10\text{cm}$ for $p$ in (XII)

    $\begin{array}{c}{M}^{″}=\frac{-10\text{cm}}{10\text{cm}}\\ =-1\end{array}$

The final image is real

The image is $\underset{\_}{10\text{cm}}$ left to the lens.

To determine

Overall magnification of the image

Answer to Problem 72P

Overall magnification is $\underset{\_}{-2.5}$.

Explanation of Solution

Write the equation for overall magnification

    $M_F=M{M}^{\prime }{M}^{″}$        (XIII)

Here $M_F$ is the overall magnification of the final image, $M$ is the magnification of the first image, $M^{\prime }$ is the magnification of the second image and $M^{″}$ is the magnification of the third image.

Conclusion:

Substitute $-1.67$ for $M$, $-1.5$ for $M^{\prime }$ and $-1$ for $M^{″}$ in (XIII)

    $\begin{array}{c}{M}_F=\left(-1.67\right)\left(-1.5\right)\left(-1\right)\\ =-2.5\end{array}$

To determine

Whether the image is upright or inverted.

Answer to Problem 72P

The image is inverted.

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 s inverted

Conclusion:

As the overall magnification is $-2.5$ , the image is inverted.