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A lens thicker at its edges than at its center is called a diverging lens. It is also called a negative lens or concave lens.

The focal length for a diverging lens is negative.

The thin-lens equation gives a relationship between image $d_i$ and object $d_o$ distances with the focal length $f$ as below \[\frac 1f=\frac 1d_o+\frac 1d_i\] Magnification formula is also provide a relationship between image and object distances or the heights of the image and object as below \[M=-\frac{d_i}{d_o}=\frac{h_i}{h_o}\]

(i) The focal length is always negative.

(ii) For an object in front of the lens the object distance is positive.

(iii) For virtual images, the image distance is negative.

By practicing the following problems, you can understand the notes above better.

**Problem (1): A diverging lens of focal length 10 cm can form an image of an object 20 cm away from it.
(a) What is the location of the image?
(b) Is the image real or virtual?
(c) Is the image upright or inverted? and find its magnification.
(d) Draw a ray diagram and verify the above results.**

**Solution**: Given data is the object distance $d_o=20\,{\rm cm}$ and the focal length of a diverging lens $f=-10\,{\rm cm}$.

Recall that the focal length for a diverging lens is always considered, by convention, negative.

**(a)** The thin-lens equation relates the image distance $d_i$, object distance $d_o$, and focal length $f$ as below \[\frac 1f=\frac 1d_o+\frac 1d_i\] By substituting the given information above into this equation and solving for the unknown image distance, we have \begin{align*} \frac 1f&=\frac 1d_o+\frac 1d_i \\\\ \frac 1{-10}&=\frac{1}{20}+\frac{1}{d_i}\\\\\Rightarrow \frac{1}{d_i}&=\frac{1}{-10}-\frac{1}{20}\\\\&=\frac{-2-1}{20}=\frac{-3}{20} \end{align*} By inverting the above expression, we get $d_i=-6.6\,{\rm cm}$.

Hence, the image is formed about 7 cm away from the diverging lens.

**(b)** According to the sign conventions for thin-lens, a minus sign for image distance indicates that the image is VIRTUAL.

An image is called virtual when it is formed on the same side as where the object is present.

**(c)** To find the orientation of an image formed by thin-lens, we can use the magnification formula below \[M=-\frac{d_i}{d_o}\] A negative magnification indicates an inverted image, and a positive is for an upright image. Placing the image and object distances into the above formula gives \[M=-\frac{-20/3}{20}=+\frac 13\] A positive magnification was obtained so the image is erected (upright).

Note that the absolute value of the magnification indicates how much an object is changed in size. In this case, the image is as small as one-third of the object's size.

**(d)** To sketch a ray-tracing diagram for a diverging lens, it is sufficient to use at least three special rays as below

(i) A ray parallel to the optical axis, bends (refracted) toward the thicker part of the lens as if it came from the focal point on the same side as the rays. (Parallel Ray)

(ii) A ray through the center of the lens, refracted without deflection. This ray is called a central ray.

(iii) A ray through the focal point (on the opposite side the rays are coming in), refracted and emerges parallel to the optical axis. This ray is a focal ray.

Where all the above-refracted rays meet together is the location of the image point.

**Problem (2): A diverging lens has a focal length of 20 cm. An object is placed at different places in front of it. In each case, find the image properties graphically and algebraically.
(a) The object is 40 cm in front of the lens.
(b) The object is 10 cm away from the lens. **

**Solution**: A diverging lens, by convention, has a negative focal length so $f=-20\,{\rm cm}$.

**(a)** In this case, the image object is $d_o=40\,{\rm cm}$. Using thin-lens equation, we can find the location of the image formed by the diverging lens as below \begin{align*} \frac 1f&=\frac 1d_o+\frac 1d_i \\\\ \frac 1{-20}&=\frac{1}{40}+\frac{1}{d_i}\\\\\Rightarrow \frac{1}{d_i}&=\frac{1}{-20}-\frac{1}{40}\\\\&=\frac{-2-1}{40}=\frac{-3}{40} \end{align*} Thus, $d_i=-40/3\,{\rm cm}$. It is negative, so the image is virtual and formed on the same side as the object.

Its size and orientation are also obtained using magnification formula as below \[M=-\frac{d_i}{d_o}=-\frac{-40/3}{40}=+\frac 13\] The sign of magnification indicates its orientation. Here, $M$ is positive, so the image is upright.

Consequently, the image is virtual, upright, and one-third of the size of the object.

All the above results could also be obtained using graph paper as the following image.

**(b)** Now, the object sits 10 cm in front of the lens, so $d_o=10\,{\rm cm}$. Applying thin-lens equation, get the precise location of the image as below \begin{align*} \frac 1f&=\frac 1d_o+\frac 1d_i \\\\ \frac 1{-20}&=\frac{1}{10}+\frac{1}{d_i}\\\\\Rightarrow \frac{1}{d_i}&=\frac{1}{-20}-\frac{1}{10}\\\\&=\frac{-2-1}{20}=\frac{-3}{20} \end{align*} The image is formed at a distance of $d_i=-20/3\,{\rm cm}$ from the lens. The image distance was obtained negative so it is a virtual image.

Substituting the image and object distances into the magnification equation for thin-lens get \[M=-\frac{d_i}{d_o}=-\frac{-20/3}{10}=+\frac 23\] $M$ is positive and larger than one, so the image is upright and two-thirds the object size.

The ray diagram for this configuration is as below

The lesson we learn from these example problems is that a *converging lens always creates a virtual, upright, and smaller in size image from an object in front of it. *

**Problem (3): A beam of rays strike parallel to the optical axis of thin-lens and get refracted as if they came from a point 30 cm from the lens. We want to use this lens and form an upright and virtual image that is one-third of the object's size. (a) What is the type of lens? (b) Where should be the object placed? (c) Where is the location of the image?**

**Solution**: the focal point for thin-lens is defined where the rays, after refracting, meet. If this point where the exit rays actually converge at it, then that lens is a converging lens with a positive focal length (by convention).

But if the rays appear to diverge from a point on the optical axis, then that lens is a diverging lens with a negative focal length.

**(a)** In this problem, we want an upright, virtual, and small image. All these are the properties of an image formed by a diverging lens.

Therefore, this is a diverging lens with a focal length of $f=-30\,{\rm cm}$.

**(b)** The object and image distances are unknown but the magnification equation, $M=-\frac{d_i}{d_o}$, relates them together. By convention, we know that a positive magnification indicates an upright image so \[M=-\frac{d_i}{d_o}=+\frac 13\] Thus, we have $d_i=-\frac 13 d_o$.

Putting $d_i$ and $f$ into the thin-lens equation, and solving for the unknown object distance we have \begin{align*} \frac 1f&=\frac 1d_o+\frac 1d_i \\\\ \frac 1{-30}&=\frac{1}{d_o}+\frac{1}{-1/3 d_o}\\\\ \frac{1}{-30}&=\frac{1-3}{d_o} \end{align*} Rearranging the above expression, and solving for $d_o$ we have $d_o=60\,{\rm cm}$. Thus, the object should be 60 cm from the lens.

**(c)** Using the thin-lens equation again and solving for the unknown image distance, we get \begin{align*} \frac 1f&=\frac 1d_o+\frac 1d_i \\\\ \frac 1{-30}&=\frac{1}{60}+\frac{1}{d_i}\\\\\Rightarrow \frac{1}{d_i}&=\frac{1}{-30}-\frac{1}{60}\\\\&=\frac{-2-1}{60}\end{align*} Thus, $d_i=-20\,{\rm cm}$. The image distance is negative,as expected for a virtual image. Consequently, the image is formed on the same side as the object.

Recall that the image formed by a diverging lens is always virtual so, by convention, $d_i<0$.

A ray-tracing diagram for this arrangement is also drawn as follows:

**Problem (4): An object is located 20 cm to the left of an unknown lens. Its image is also formed 12 cm on the same side as the object. (a) Find the focal length of the lens? (b) Is the lens converging or diverging? (c) Determine the magnification of the lens.**

**Solution**: The object and image distance are $d_o=20\,{\rm cm}$ and $d_i=-12\,{\rm cm}$, respectively.

A minus sign was placed for the image distance as in the problems said that the image is formed on the same side of the lens as the object so, by convention, the image is virtual.

**(a)** Substituting the known values into the thin-lens equation, we have \begin{align*} \frac 1f&=\frac 1d_o+\frac 1d_i \\\\ \frac 1{f}&=\frac{1}{20}+\frac{1}{-12}\\\\&=\frac{1}{-20}-\frac{1}{-12}\\\\ \Rightarrow d_i&=-7.5\quad {\rm cm}\end{align*} Therefore, the focal length is 7.5 cm.

**(b)** Because the focal length was obtained negatively, the lens is a diverging or concave lens.

**(c)** How much an object is enlarged by a lens is called the magnification whose formula is negative of the ratio of the image distance to the object distance. Therefore, we have \[M=-\frac{d_i}{d_o}=-\frac{-12}{20}=+0.6\] A positive magnification always indicates an upright image. Consequently, the image is formed smaller than the object or precisely, 0.6 the size of the object.

As a summary, we pose a question about the diverging lens as follows:

Question: for a diverging or concave lens, what kind of images can be formed, and under what conditions those images can be formed?

Answer: in all of the problems above, we saw that

(i) a diverging lens can form only virtual images from a real object in front of it, regardless of the location of the object.

(ii) the images are smaller in size than the object.

(iii) the image is formed always within the focal length on the same side of the lens as the object.

**Author**: Ali Nemati

**Date Published**: 5/29/2021

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