Chapter 38, Problem 40PQ

To determine

The magnification of the vase and see whether the vase acts like a magnifying lens.

Answer to Problem 40PQ

The magnification produced by the vase is $0.517$ and the vase will not act like a magnifying lens as the total magnification is less than $1$.

Explanation of Solution

Write the expression for spherical refracting surface.

    $\frac{n_{\text{i}}}{d_0}+\frac{n_{\text{t}}}{d_{\text{i}}}=\frac{n_{\text{t}}-n_{\text{i}}}{R}$

Here, $d_0$ is the distance of the object, $d_{\text{i}}$ is the distance of the image, $R$ is the radius, $n_{\text{t}}$ is the refractive index of the light in refracted medium and $n_{\text{i}}$ is the refractive index of the light in incident medium.

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

    $\begin{array}{c}\frac{n_{\text{t}}}{d_{\text{i}}}=\frac{n_{\text{t}}-n_{\text{i}}}{R}-\frac{n_{\text{i}}}{d_0}\\ d_{\text{i}}=\frac{n_{\text{t}}}{\frac{n_{\text{t}}-n_{\text{i}}}{R}-\frac{n_{\text{i}}}{d_0}}\end{array}$

Write the expression to calculate the distance of the image for the first surface.

    $d_{{\text{i}}_1}=\frac{n_{\text{t}}}{\frac{n_{\text{t}}-n_{\text{i}}}{R}-\frac{n_{\text{i}}}{d_{0_1}}}$                                                                                         (I)

Here, $d_{0_1}$ is the distance of the object for the first surface and $d_{{\text{i}}_1}$ is the distance of the image for the first surface.

Write the expression to calculate the magnification for the first surface.

    $M_1=-\frac{n_{\text{i}}}{n_{\text{t}}}\frac{d_{{\text{i}}_1}}{d_{0_1}}$                                                                                                     (II)

Here, $M_1$ is the magnification for the first surface.

Write the expression to calculate the object distance for the second surface.

    $d_{0_2}=D-d_{{\text{i}}_1}$                                                                                                       (III)

Here, $d_{0_2}$ is the object distance for the second surface and $D$ is the diameter of the vase.

Write the expression to calculate the image for the second surface.

    $d_{{\text{i}}_2}=\frac{n_{\text{t}}}{\frac{n_{\text{t}}-n_{\text{i}}}{R}-\frac{n_{\text{i}}}{d_{0_2}}}$                                                                                             (IV)

Here, $d_{{\text{i}}_2}$ is the distance of the image for the second surface.

Write the expression to calculate the magnification for the second surface.

    $M_2=-\frac{n_{\text{i}}}{n_{\text{t}}}\frac{d_{{\text{i}}_2}}{d_{0_2}}$                                                                                                     (V)

Here, $M_2$ is the magnification for the second surface.

Write the expression to calculate the total magnification.

    $M=M_1\times M_2$                                                                                                     (VI)

Conclusion:

Substitute $1.3$ for $n_{\text{i}}$, $1$ for $n_{\text{t}}$, $3.0\text{cm}$ for $d_{0_1}$ and $6.0\text{cm}$ for $R$ in equation (I) to find $d_{{\text{i}}_1}$.

    $\begin{array}{c}{d}_{{\text{i}}_1}=\frac{1}{\frac{1-1.3}{6.0\text{cm}}-\frac{1.3}{3.0\text{cm}}}\\ =-2.07\text{cm}\end{array}$

Substitute $1$ for $n_{\text{t}}$, $1.3$ for $n_{\text{i}}$, $3.0\text{cm}$ for $d_{0_1}$ and $-2.07\text{cm}$ for $d_{{\text{i}}_1}$ in equation (II) to find $M_1$.

    $\begin{array}{c}{M}_1=-\frac{1.3}{1}\left(\frac{-2.07\text{cm}}{3.0\text{cm}}\right)\\ =0.897\end{array}$

Substitute $12\text{cm}$ for $D$ and $-2.07\text{cm}$ for $d_{{\text{i}}_1}$ in equation (III) to find $d_{0_2}$.

    $\begin{array}{c}{d}_{0_2}=12\text{cm}+2.07\text{cm}\\ =\text{14}\text{.07}\text{cm}\end{array}$

Substitute $1$ for $n_{\text{i}}$, $1.33$ for $n_{\text{t}}$, $3.0\text{cm}$ for $d_{0_2}$ and $-6.0\text{cm}$ for $R$ in equation (IV) to find $d_{{\text{i}}_2}$

    $\begin{array}{c}{d}_{{\text{i}}_2}=\frac{1.33}{\frac{1.33-1}{-6\text{cm}}-\frac{1}{14.07\text{cm}}}\\ =-10.55\text{cm}\end{array}$

Substitute $1$ for $n_{\text{i}}$, $1.33$ for $n_{\text{t}}$, $14.07\text{cm}$ for $d_{0_2}$ and $-10.55\text{cm}$ for $d_{{\text{i}}_2}$ in equation (V) to find $M_2$.

    $\begin{array}{c}{M}_2=-\frac{1}{1.3}\left(\frac{-10.55\text{cm}}{14.07\text{cm}}\right)\\ =0.577\end{array}$

Substitute $0.897$ for $M_1$ and $0.577$ for $M_2$ in equation (VI) to find $M$.

    $\begin{array}{c}M=0.897\times 0.577\\ =0.517\end{array}$

Therefore, the magnification produced by the vase is $0.517$ and the vase will not act like a magnifying lens as the total magnification is less than $1$.