Diverging Lens Problems with Answers For AP Physics
On this page, we are going to learn some about a diverging lens with a couple of solved examples. All features of a diverging lens, including its physical properties, are discussed through worked examples.
Overview of diverging lenses:
A lens that is 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 of 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 \[\frac 1f=\frac 1d_o+\frac 1d_i\] The magnification formula also provides a relationship between image and object distances, or the heights of the image and object, as below \[M=-\frac{d_i}{d_o}=\frac{h_i}{h_o}\]
Sign conventions for diverging lenses:
(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.
Diverging Lens Solved Problems:
Problem (1): A diverging lens with a focal length of 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 lenses, 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 a thin lens equation, 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 has 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 (refracts) 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 passes through the center of the lens, refracted without deflection. This ray is called a central ray.
(iii) A ray passes through the focal point (on the opposite side the rays are coming in), is 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 in 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 the 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 the magnification formula 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 the size of the object.
(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 negatively, so it is a virtual image.
Substituting the image and object distances into the magnification equation for thin lenses 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 diverging lens always creates a virtual, upright, and smaller in size image from an object in front of it.
Problem (3): A beam of rays strikes parallel to the optical axis of a thin lens and gets refracted as if it came from a point 30 cm from the lens. We want to use this lens to 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 the object be placed? (c) Where is the location of the image?
Solution: the focal point for a thin lens is defined as where the rays, after refracting, meet. If this point is where the exit rays actually converge, 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 distances are $d_o=20\,{\rm cm}$ and $d_i=-12\,{\rm cm}$, respectively.
A minus sign was placed for the image distance, as the problem states 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 the negative of the ratio of the image distance to the object distance. Therefore, we have \[M=-\frac{d_i}{d_o}=-\frac{-12}{20}=+0.6\] Positive magnification always indicates an upright image. Consequently, the image is formed smaller than the object, or precisely, 0.6 times the size of the object.
Overall, 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 can those images 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 always formed within the focal length on the same side of the lens as the object.
Author: Ali Nemati
Date Published: 5/29/2021
