Chapter 38, Problem 39PQ

To determine

The distance between the image and the surface 2 and the magnification of the final image.

Answer to Problem 39PQ

The distance of the screen from the surface 2 is $7.5\text{cm}$ and the magnification of the final image is $-0.13$.

Explanation of Solution

Write the expression for the refraction at a spherical surface.

    $\frac{n_{\text{i}}}{d_{\text{o}}}+\frac{n_{\text{t}}}{d_{\text{i}}}=\frac{\left(n_{\text{t}}-n_{\text{i}}\right)}{r}$                                                                           (I)

Here, $n_{\text{i}}$ is the index of refraction of the travelling medium, $n_{\text{t}}$ is the index of refraction of the denser medium, $d_{\text{o}}$ is the object distance, $d_{\text{t}}$ is the image distance and $r$ is the radius of curvature.

Rearrange the equation (I) for $d_{\text{i}}$.

    $\begin{array}{c}\frac{n_{\text{t}}}{d_{\text{i}}}=\frac{\left(n_{\text{t}}-n_{\text{i}}\right)}{r}-\frac{n_{\text{i}}}{d_{\text{o}}}\\ d_{\text{i}}=\frac{n_{\text{t}}}{\frac{\left(n_{\text{t}}-n_{\text{i}}\right)}{r}-\frac{n_{\text{i}}}{d_{\text{o}}}}\end{array}$                                                                          (II)

Write the expression for the magnification for a refractive surface.

    $m=-\frac{n_{\text{i}}{d}_{\text{i}}}{n_{\text{t}}{d}_{\text{o}}}$                                                                                  (III)

Here, $m$ is the magnification for a refractive surface.

Write the expression for the total magnification for a thick lens.

    $M=m_1{m}_2$                                                                                          (IV)

Here, $M$ is the total magnification, $m_1$ is the magnification at surface 1 and $m_2$ is the magnification at surface 2.

Conclusion:

Substitute $1.333$ for $n_{\text{t}}$, $1$ for $n_{\text{i}}$, $6.0\text{cm}$ for $r$ and $100.0\text{cm}$ for $d_{\text{o}}$ in equation (II) to find $d_{\text{i}}$.

    $\begin{array}{c}{d}_{\text{i}}=\frac{1.333}{\frac{\left(1.333-1\right)}{6.0\text{cm}}-\frac{1}{100.0\text{cm}}}\\ =\frac{1.333}{0.0555-0.01}\text{cm}\\ =29.3\text{cm}\end{array}$

The positive sign indicates that the image is formed on the right side of surface 1.

Substitute $1.333$ for $n_{\text{t}}$, $1$ for $n_{\text{i}}$, $100.0\text{cm}$ for $d_{\text{o}}$ and $29.3\text{cm}$ for $d_{\text{i}}$ in equation (III) to find $m_1$.

    $\begin{array}{c}{m}_1=-\frac{\left(1\right)\left(29.3\text{cm}\right)}{\left(1.333\right)\left(100.0\text{cm}\right)}\\ =-0.22\end{array}$

The image formed by surface 1 acts as the object for the surface 2. Thus, the object distance for surface 2 is,

    $\begin{array}{c}{d}_{\text{o}}=12.0\text{cm}-29.3\text{cm}\\ =-17.3\text{cm}\end{array}$

Substitute $1.333$ for $n_{\text{i}}$, $1$ for $n_{\text{t}}$, $-6.0\text{cm}$ for $r$ and $-17.3\text{cm}$ for $d_{\text{o}}$ in equation (II) to find $d_{\text{i}}$.

    $\begin{array}{c}{d}_{\text{i}}=\frac{1}{\frac{\left(1-1.333\right)}{-6.0\text{cm}}-\frac{1.333}{-17.3\text{cm}}}\\ =\frac{1}{0.0555+0.077}\text{cm}\\ =7.5\text{cm}\end{array}$

The positive sign indicates that the image is formed behind the surface 2.

Substitute $1.333$ for $n_{\text{t}}$, $1$ for $n_{\text{i}}$, $-17.3\text{cm}$ for $d_{\text{o}}$ and $7.5\text{cm}$ for $d_{\text{i}}$ in equation (III) to find $m_2$.

    $\begin{array}{c}{m}_2=-\frac{\left(1.333\right)\left(7.5\text{cm}\right)}{\left(1\right)\left(-17.3\text{cm}\right)}\\ =0.58\end{array}$

Substitute $-0.22$ for $m_1$ and $0.58$ for $m_2$ in equation (IV) to find $M$.

    $\begin{array}{c}M=\left(-0.22\right)\left(0.58\right)\\ =-0.13\end{array}$

Therefore, the distance of the screen from the surface 2 is $7.5\text{cm}$ and the magnification of the final image is $-0.13$.