## Section47.10Image Formation Bootcamp

### Subsection47.10.8Miscellaneous

A small light bulb is at the bottom of a tank that has a layer of water and a layer of oil as shown in Figure 47.10.18. Let thickess of the oil layer be $0.4\text{ cm}$ and the depth of bulb from the surface of water be $2\text{ cm}\text{.}$ Determine the apparent depth of the bulb as seen from near normal. Use a ruler to measure the distance(s).

Hint

Work on layer at a time.

$1.84\text{ cm}\text{.}$

Solution

We can work one interface at a time. Suppose we have only Oil/Water interface. That is the detector is inside oil. Then, we will find that the object will appear at the point $\text{Q}_1$ with the following relation between the image distance and the object distance from the interface at $\text{O}_1\text{.}$

\begin{equation*} \text{O}_1\text{Q}_1 = \dfrac{n_2}{n_1}\: \text{O}_1\text{P}. \ \ \ (1) \end{equation*}

Now, we take the image $\text{Q}_1$ as the object for the interface at $\text{O}_2$O. The refraction will now be between the media with refractive indices $n_2$ and $n_3\text{.}$ Therefore, the final image at $\text{Q}_2$ will have the following distance.

\begin{equation*} \text{O}_2\text{Q}_2 = \dfrac{n_3}{n_2}\: \text{O}_2\text{Q}_1. \ \ \ (2) \end{equation*} From (1), (2) and the figure it is readily seen that

\begin{equation*} \text{O}_2\text{Q}_2 = \dfrac{n_3}{n_2}\: \left[ \text{O}_2\text{O}_1 + \left( \dfrac{n_2}{n_1} \right) \: \text{O}_1\text{P} \right]. \end{equation*}

Now, we put in the numbers to obtain, the depth of the final image from the surface of oil.

\begin{equation*} \text{O}_2\text{Q}_2 = \frac{1}{1.2} \left( 0.4 + \frac{1.2}{1.33}\times 2 \right) = 1.84\text{ cm}. \end{equation*}

That is, the bulb appears at only $1.84\text{ cm}$ from the top of the oil rather than $2.4\text{ cm}\text{,}$ the actual depth.

A concave mirror of radius of curvature $10\text{ cm}$ is placed $30\text{ cm}$ from a thin convex lens of focal length $15\text{ cm}$ as shown in Figure 47.10.20.

Using algebraic method, find the location and magnification of a small bulb sitting $50\text{ cm}$ from the lens.

Hint

Work out image from lens only first and then use that image as object for the mirror.

$12\text{ cm}$ to the left of the mirror, $m = \frac{3}{5}\text{.}$

Solution

We first find the intermediate image due to the lens. Let $q_1$ be the image distance from the lens. We have

\begin{equation*} f_1 = 15\:\text{cm},\ \ p_1 = 50\:\text{cm},\ \ \Longrightarrow\ \ q_1 = \dfrac{150}{7}\:\text{cm} \approx 21.7\text{ cm}. \end{equation*}

Since $q_1 \gt 0$ the intermediate image is to the right of the lens and since $q_1 \lt \text{ separation }\text{,}$ the image is to the left of the vertex of the mirror. Now, this intermediate image will serve as the object for the mirror. By using the separation of the mirror and the lens we find the object distance to the mirror to be

\begin{equation*} p_2 = 30\:\text{cm} - \dfrac{150}{7}\:\text{cm} = \dfrac{60}{7}\:\text{cm}. \end{equation*}

The focal length of the mirror is $f_2 = R/2 = 5\:\text{cm}\text{;}$ it is positive since we have a concave mirror (recall the sign convention). Therefore, the image distance from the mirror will be

\begin{equation*} \dfrac{1}{q_2} = \dfrac{1}{5}-\dfrac{7}{60},\ \ \Longrightarrow\ \ q_2 = 12\:\text{cm}. \end{equation*}

Now, sign convenstion for curved mirrors tells us that $q_2 \gt 0$ for a mirror means that the image is in front of the mirror. Therefore, the final image at $q_2\text{,}$ will be in front of the mirror at a distance of $12\text{ cm}$ from the vertex of the mirror.

This image will be real image since it forms in front of the mirror as opposed to image behind the mirror.

The net lateral magnification of the image will come from the product of the magnifications in the lens and the mirror.

\begin{equation*} m = m_1\times m_2 = \left(- \dfrac{q_1}{p_1} \right)\:\left(- \dfrac{q_2}{p_2} \right) = + \dfrac{3}{5}. \end{equation*}

Since $m\gt 0$ the image will be upright, and, since $|m| \lt 1$ the image will be smaller than the object.

An object of height $2\text{ cm}$ is placed at $50\text{ cm}$ in front of a diverging lens of focal length $40\text{ cm}$ as shown in Figure 47.10.22. Behind the lens there is a convex mirror of focal length $15\text{ cm}$ placed $30\text{ cm}$ from the converging lens. Find the location, orientation and size of the final image.

Hint

Find the image by the lens first and use that as an object for the mirror.

Image 11 cm behind the mirror, $m = 0.1\text{,}$ $h_i = 0.2\text{ cm}\text{.}$

Solution

We first find the intermediate image due to lens. Let $q_1$ be the image distance from the lens. We have

\begin{equation*} f_1 = -40\:\text{cm},\ \ p_1 = 50\:\text{cm},\ \ \Longrightarrow\ \ q_1 = -\dfrac{200}{9}\:\text{cm}. \end{equation*}

Since $q_1\lt 0$ the intermediate image is to the left of the lens. Now, this intermediate image will serve as the object for the mirror. By using the separation of the mirror and the lens we find the object distance to the mirror to be

\begin{equation*} p_2 = 30\:\text{cm} - \left( -\dfrac{200}{9}\:\text{cm}\right) = \dfrac{470}{9}\:\text{cm}. \end{equation*}

The focal length of the mirror is $f_2 = -15\:\text{cm}\text{,}$ which is negative since the mirror is convex. Therefore, the image distance from the mirror will be

\begin{equation*} \dfrac{1}{q_2} = \dfrac{1}{-15}-\dfrac{9}{470},\ \ \Longrightarrow\ \ q_2 = -11.7\:\text{cm}. \end{equation*}

Since $q\lt 0$ for a mirror means that the image is behind the mirror, the final image at $q_2$ will be behind the mirror at a distance of 11.7 cm from the vertex of the mirror. This image will be a virtual image since it forms behind the mirror where real rays do not go. The net magnification of the image will come from the product of the magnifications in the lens and the mirror.

\begin{equation*} m = m_1\times m_2 = \left(- \dfrac{q_1}{p_1} \right)\:\left(- \dfrac{q_2}{p_2} \right) = +0.1. \end{equation*}

Since $m>0$ the image will be upright and since $|m|\lt 1$ the image will be smaller than the object. The height of the image is given by the product of $m$ and the height of the object.

\begin{equation*} h_i = m\:h_o = (0.1)(2\:\text{cm}) = 0.2\:\text{cm}. \end{equation*}

Two identical concave mirrors are placed facing each other with a distance of $d=R/2$ between their vertices as shown in Figure 47.10.24. One of them has a small hole in the middle. When a penny is placed on the bottom mirror, you observe a real image of the penny floating above the hole.

Explain how this happens by showing images in the two mirrors and locate the distance above the mirror with the hole that you will see the floating image. If you like, you can use the following numerical values, $R=25\text{ cm}\text{,}$ $\Delta = 0.3\text{ cm}\text{.}$

Hint

Start with image in $\text{M}_1\text{.}$

$\frac{R\Delta}{ R-2\Delta}\text{,}$ $0.31 \text{ cm}$

Solution

Since the penny is located within a focal distance of mirror $\text{M}_1\text{,}$ a virtual image will be formed at $q_1\text{,}$ which will serve as an object for mirror $\text{M}_2\text{.}$ Now, the object distance $p_2$ from mirror $\text{M}_2$ is larger than the focal length, hence image will be a real image at $q_2\text{.}$ This is illustrated in Figure 47.10.25. Below, I will work out the distances.

With $p=d-\Delta$ and $f=2/R\text{,}$ the image distance $q_1$ will be

\begin{equation*} \frac{1}{q_1} = \frac{2}{R} - \frac{1}{d-\Delta}. \end{equation*}

Therefore,

\begin{equation*} q_1 = \frac{R(d-\Delta)}{2d-2\Delta - R} = - \frac{R(d-\Delta)}{2\Delta}. \end{equation*}

This will be negative. Therefore, we will get object distance to mirror $\text{M}_2$ by adding its absolute value to $d$ and using $d=R/2\text{.}$

\begin{equation*} p_2 = |q_1| + d = \frac{R(d-\Delta)}{2\Delta} + d = \frac{R^2}{4\Delta}. \end{equation*}

Now, we work out $q_2\text{.}$

\begin{equation*} \frac{1}{q_2} = \frac{2}{R} - \frac{4\Delta}{R^2}. \end{equation*}

This gives

\begin{equation*} q_2 = \frac{R^2}{2R-4\Delta}. \end{equation*}

Now, this is a little bit beyond mirror $\text{M}_1\text{.}$ We can get the distance above the hole by subtracting $d$ from it.

\begin{equation*} \text{above hole} = q_2 - d = \frac{R^2}{2R-4\Delta} - \frac{R}{2} = \frac{R\Delta}{ R-2\Delta}. \end{equation*}

Now, with the numerical values given, we get

\begin{equation*} \text{above hole} = 0.31 \text{ cm}. \end{equation*}