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## Learning Objectives

By the end of this section, you will be able to:

• Describe image formation by spherical mirrors.
• Use ray diagrams and the mirror equation to calculate the properties of an image in a spherical mirror.

The image in a plane mirror has the same size as the object, is upright, and is the same distance behind the mirror as the object is in front of the mirror. A curved mirror, on the other hand, can form images that may be larger or smaller than the object and may form either in front of the mirror or behind it. In general, any curved surface will form an image, although some images make be so distorted as to be unrecognizable (think of fun house mirrors). Because curved mirrors can create such a rich variety of images, they are used in many optical devices that find many uses. We will concentrate on spherical mirrors for the most part, because they are easier to manufacture than mirrors such as parabolic mirrors and so are more common.

## Curved Mirrors

We can define two general types of spherical mirrors. If the reflecting surface is the outer side of the sphere, the mirror is called a convex mirror . If the inside surface is the reflecting surface, it is called a concave mirror .

Symmetry is one of the major hallmarks of many optical devices, including mirrors and lenses. The symmetry axis of such optical elements is often called the principal axis or optical axis. For a spherical mirror, the optical axis passes through the mirror’s center of curvature and the mirror’s vertex, as shown in Figure $$\PageIndex{1}$$.

Consider rays that are parallel to the optical axis of a parabolic mirror, as shown in Figure $$\PageIndex{2a}$$. Following the law of reflection, these rays are reflected so that they converge at a point, called the focal point . Figure $$\PageIndex{2b}$$ shows a spherical mirror that is large compared with its radius of curvature. For this mirror, the reflected rays do not cross at the same point, so the mirror does not have a well-defined focal point. This is called spherical aberration and results in a blurred image of an extended object. Figure $$\PageIndex{2c}$$ shows a spherical mirror that is small compared to its radius of curvature. This mirror is a good approximation of a parabolic mirror, so rays that arrive parallel to the optical axis are reflected to a well-defined focal point. The distance along the optical axis from the mirror to the focal point is called the focal length of the mirror.

A convex spherical mirror also has a focal point, as shown in Figure $$\PageIndex{3}$$. Incident rays parallel to the optical axis are reflected from the mirror and seem to originate from point $$F$$ at focal length $$f$$ behind the mirror. Thus, the focal point is virtual because no real rays actually pass through it; they only appear to originate from it.

How does the focal length of a mirror relate to the mirror’s radius of curvature? Figure $$\PageIndex{4}$$ shows a single ray that is reflected by a spherical concave mirror. The incident ray is parallel to the optical axis. The point at which the reflected ray crosses the optical axis is the focal point. Note that all incident rays that are parallel to the optical axis are reflected through the focal point—we only show one ray for simplicity. We want to find how the focal length $$FP$$ (denoted by $$f$$) relates to the radius of curvature of the mirror, $$R$$, whose length is

$R=CF+FP. \label{eq31}$

The law of reflection tells us that angles $$\angle OXC$$ and $$\angle CXF$$ are the same, and because the incident ray is parallel to the optical axis, angles $$\angle OXC$$ and $$\angle XCP$$ are also the same. Thus, triangle $$CXF$$ is an isosceles triangle with $$CF=FX$$. If the angle $$θ$$ is small then

$\sin θ≈ θ \label{sma}$

which is called the “ small-angle approximation ”), then $$FX≈FP$$ or $$CF≈FP$$. Inserting this into Equation \ref{eq31} for the radius $$R$$, we get

\begin{align} R &=CF+FP \nonumber \\[4pt] &=FP+FP \nonumber \\[4pt] &=2FP\nonumber \\[4pt] &=2f \end{align} \nonumber

In other words, in the small-angle approximation, the focal length $$f$$ of a concave spherical mirror is half of its radius of curvature, $$R$$:

$f=\dfrac{R}{2}. \nonumber$

In this chapter, we assume that the small-angle approximation (also called the paraxial approximation ) is always valid. In this approximation, all rays are paraxial rays, which means that they make a small angle with the optical axis and are at a distance much less than the radius of curvature from the optical axis. In this case, their angles $$θ$$ of reflection are small angles, so

$\sin θ≈ \tan θ≈ θ. \label{smallangle}$

## Using Ray Tracing to Locate Images

To find the location of an image formed by a spherical mirror, we first use ray tracing, which is the technique of drawing rays and using the law of reflection to determine the reflected rays (later, for lenses, we use the law of refraction to determine refracted rays). Combined with some basic geometry, we can use ray tracing to find the focal point, the image location, and other information about how a mirror manipulates light. In fact, we already used ray tracing above to locate the focal point of spherical mirrors, or the image distance of flat mirrors. To locate the image of an object, you must locate at least two points of the image. Locating each point requires drawing at least two rays from a point on the object and constructing their reflected rays. The point at which the reflected rays intersect, either in real space or in virtual space, is where the corresponding point of the image is located. To make ray tracing easier, we concentrate on four “principal” rays whose reflections are easy to construct.

Figure $$\PageIndex{5}$$ shows a concave mirror and a convex mirror, each with an arrow-shaped object in front of it. These are the objects whose images we want to locate by ray tracing. To do so, we draw rays from point $$Q$$ that is on the object but not on the optical axis. We choose to draw our ray from the tip of the object. Principal ray 1 goes from point $$Q$$ and travels parallel to the optical axis. The reflection of this ray must pass through the focal point, as discussed above. Thus, for the concave mirror, the reflection of principal ray 1 goes through focal point $$F$$, as shown in Figure $$\PageIndex{5b}$$. For the convex mirror, the backward extension of the reflection of principal ray 1 goes through the focal point (i.e., a virtual focus). Principal ray 2 travels first on the line going through the focal point and then is reflected back along a line parallel to the optical axis. Principal ray 3 travels toward the center of curvature of the mirror, so it strikes the mirror at normal incidence and is reflected back along the line from which it came. Finally, principal ray 4 strikes the vertex of the mirror and is reflected symmetrically about the optical axis.

The four principal rays intersect at point $$Q′$$, which is where the image of point $$Q$$ is located. To locate point $$Q′$$, drawing any two of these principle rays would suffice. We are thus free to choose whichever of the principal rays we desire to locate the image. Drawing more than two principal rays is sometimes useful to verify that the ray tracing is correct.

To completely locate the extended image, we need to locate a second point in the image, so that we know how the image is oriented. To do this, we trace the principal rays from the base of the object. In this case, all four principal rays run along the optical axis, reflect from the mirror, and then run back along the optical axis. The difficulty is that, because these rays are collinear, we cannot determine a unique point where they intersect. All we know is that the base of the image is on the optical axis. However, because the mirror is symmetrical from top to bottom, it does not change the vertical orientation of the object. Thus, because the object is vertical, the image must be vertical. Therefore, the image of the base of the object is on the optical axis directly above the image of the tip, as drawn in the figure.

For the concave mirror, the extended image in this case forms between the focal point and the center of curvature of the mirror. It is inverted with respect to the object, is a real image, and is smaller than the object. Were we to move the object closer to or farther from the mirror, the characteristics of the image would change. For example, we show, as a later exercise, that an object placed between a concave mirror and its focal point leads to a virtual image that is upright and larger than the object. For the convex mirror, the extended image forms between the focal point and the mirror. It is upright with respect to the object, is a virtual image, and is smaller than the object.

## Ray-Tracing Rules

Ray tracing is very useful for mirrors. The rules for ray tracing are summarized here for reference:

• A ray traveling parallel to the optical axis of a spherical mirror is reflected along a line that goes through the focal point of the mirror (ray 1 in Figure $$\PageIndex{5}$$).
• A ray traveling along a line that goes through the focal point of a spherical mirror is reflected along a line parallel to the optical axis of the mirror (ray 2 in Figure $$\PageIndex{5}$$).
• A ray traveling along a line that goes through the center of curvature of a spherical mirror is reflected back along the same line (ray 3 in Figure $$\PageIndex{5}$$).
• A ray that strikes the vertex of a spherical mirror is reflected symmetrically about the optical axis of the mirror (ray 4 in Figure $$\PageIndex{5}$$).

We use ray tracing to illustrate how images are formed by mirrors and to obtain numerical information about optical properties of the mirror. If we assume that a mirror is small compared with its radius of curvature, we can also use algebra and geometry to derive a mirror equation, which we do in the next section. Combining ray tracing with the mirror equation is a good way to analyze mirror systems.

## Image Formation by Reflection—The Mirror Equation

For a plane mirror, we showed that the image formed has the same height and orientation as the object, and it is located at the same distance behind the mirror as the object is in front of the mirror. Although the situation is a bit more complicated for curved mirrors, using geometry leads to simple formulas relating the object and image distances to the focal lengths of concave and convex mirrors.

Consider the object $$OP$$ shown in Figure $$\PageIndex{6}$$. The center of curvature of the mirror is labeled $$C$$ and is a distance $$R$$ from the vertex of the mirror, as marked in the figure. The object and image distances are labeled $$d_o$$ and $$d_i$$, and the object and image heights are labeled $$h_o$$ and $$h_i$$, respectively. Because the angles $$ϕ$$ and $$ϕ′$$ are alternate interior angles, we know that they have the same magnitude. However, they must differ in sign if we measure angles from the optical axis, so $$ϕ=−ϕ′$$. An analogous scenario holds for the angles $$θ$$ and $$θ′$$. The law of reflection tells us that they have the same magnitude, but their signs must differ if we measure angles from the optical axis. Thus, $$θ=−θ′$$. Taking the tangent of the angles $$θ$$ and $$θ′$$, and using the property that $$\tan(−θ)=−\tan θ$$, gives us

$\left. \begin{array}{rcl} \tanθ=\dfrac{h_o}{d_o} \\ \tanθ′=−\tanθ=\dfrac{h_i}{d_i} \end{array}\right\} =\dfrac{h_o}{d_o}=−\dfrac{h_i}{d_i} \label{eq51}$

$- \dfrac{h_o}{h_i}=\dfrac{d_o}{d_i}.\label{eq52}$

Similarly, taking the tangent of $$ϕ$$ and $$ϕ′$$ gives

$\left. \begin{array}{rcl} \tanϕ=\dfrac{h_o}{d_o-R} \\ \tanϕ′=−\tanϕ=\dfrac{h_i}{R-d_i} \end{array}\right\} =\dfrac{h_o}{d_o-R}=−\dfrac{h_i}{R-d_i} \nonumber$

$−\dfrac{h_o}{h_i}=\dfrac{d_o-R}{R-d_i}. \label{eq55}$

Combining Equation \ref{eq51} and \ref{eq55} gives

$\dfrac{d_o}{d_i}=\dfrac{d_o−R}{R−d_i}. \nonumber$

After a little algebra, this becomes

$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{2}{R}. \label{eq57}$

No approximation is required for this result, so it is exact. However, as discussed above, in the small-angle approximation, the focal length of a spherical mirror is one-half the radius of curvature of the mirror, or $$f=R/2$$. Inserting this into Equation \ref{eq57} gives the mirror equation :

$\underbrace{ \dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f}}_{\text{mirror equation}}. \label{mirror equation}$

The mirror equation relates the image and object distances to the focal distance and is valid only in the small-angle approximation (Equation \ref{sma}). Although it was derived for a concave mirror, it also holds for convex mirrors (proving this is left as an exercise). We can extend the mirror equation to the case of a plane mirror by noting that a plane mirror has an infinite radius of curvature. This means the focal point is at infinity, so the mirror equation simplifies to

$d_o=−d_i \nonumber$

which is the same equation obtained earlier .

Notice that we have been very careful with the signs in deriving the mirror equation. For a plane mirror, the image distance has the opposite sign of the object distance. Also, the real image formed by the concave mirror in Figure $$\PageIndex{6}$$ is on the opposite side of the optical axis with respect to the object. In this case, the image height should have the opposite sign of the object height. To keep track of the signs of the various quantities in the mirror equation, we now introduce a sign convention.

## Sign convention for spherical mirrors

Using a consistent sign convention is very important in geometric optics. It assigns positive or negative values for the quantities that characterize an optical system. Understanding the sign convention allows you to describe an image without constructing a ray diagram. This text uses the following sign convention:

• The focal length $$f$$ is positive for concave mirrors and negative for convex mirrors.
• The image distance $$d_i$$ is positive for real images and negative for virtual images.

Notice that rule 1 means that the radius of curvature of a spherical mirror can be positive or negative. What does it mean to have a negative radius of curvature? This means simply that the radius of curvature for a convex mirror is defined to be negative.

## Image Magnification

Let’s use the sign convention to further interpret the derivation of the mirror equation. In deriving this equation, we found that the object and image heights are related by

$−\dfrac{h_o}{h_i}=\dfrac{d_o}{d_i}. \label{eq61}$

See Equation \ref{eq52}. Both the object and the image formed by the mirror in Figure $$\PageIndex{6}$$ are real, so the object and image distances are both positive. The highest point of the object is above the optical axis, so the object height is positive. The image, however, is below the optical axis, so the image height is negative. Thus, this sign convention is consistent with our derivation of the mirror equation.

Equation \ref{eq61} in fact describes the linear magnification (often simply called “ magnification ”) of the image in terms of the object and image distances. We thus define the dimensionless magnification $$m$$ as follows:

$\underbrace{m=\dfrac{h_i}{h_o}}_{\text{linear magnification}}. \label{mag}$

If $$m$$ is positive, the image is upright, and if $$m$$ is negative, the image is inverted. If $$|m|>1$$, the image is larger than the object, and if $$|m|<1$$, the image is smaller than the object. With this definition of magnification, we get the following relation between the vertical and horizontal object and image distances:

$m=\dfrac{h_i}{h_o}=−\dfrac{d_i}{d_o}. \nonumber$

This is a very useful relation because it lets you obtain the magnification of the image from the object and image distances, which you can obtain from the mirror equation.

## Example $$\PageIndex{1}$$: Solar Electric Generating System

One of the solar technologies used today for generating electricity involves a device (called a parabolic trough or concentrating collector) that concentrates sunlight onto a blackened pipe that contains a fluid. This heated fluid is pumped to a heat exchanger, where the thermal energy is transferred to another system that is used to generate steam and eventually generates electricity through a conventional steam cycle. Figure $$\PageIndex{7}$$ shows such a working system in southern California. The real mirror is a parabolic cylinder with its focus located at the pipe; however, we can approximate the mirror as exactly one-quarter of a circular cylinder.

• If we want the rays from the sun to focus at 40.0 cm from the mirror, what is the radius of the mirror?
• What is the amount of sunlight concentrated onto the pipe, per meter of pipe length, assuming the insolation (incident solar radiation) is 900 W/m 2 ?
• If the fluid-carrying pipe has a 2.00-cm diameter, what is the temperature increase of the fluid per meter of pipe over a period of 1 minute? Assume that all solar radiation incident on the reflector is absorbed by the pipe, and that the fluid is mineral oil.

First identify the physical principles involved. Part (a) is related to the optics of spherical mirrors. Part (b) involves a little math, primarily geometry. Part (c) requires an understanding of heat and density.

a. The sun is the object, so the object distance is essentially infinity: $$d_o=\infty$$. The desired image distance is $$d_i=40.0\,cm$$. We use the mirror equation (Equation \ref{mirror equation}) to find the focal length of the mirror:

\begin{align*} \dfrac{1}{d_o}+\dfrac{1}{d_i} &=\dfrac{1}{f} \nonumber \\[4pt] f &= \left(\dfrac{1}{d_o}+\dfrac{1}{d_i}\right)^{−1} \\[4pt] &= \left(\dfrac{1}{\infty}+\dfrac{1}{40.0\,cm}\right)^{−1} \\[4pt] &= 40.0 \,cm \end{align*} \nonumber

Thus, the radius of the mirror is

$R=2f=80.0\,cm. \nonumber$

b. The insolation is 900 W/m 2 . You must find the cross-sectional area $$A$$ of the concave mirror, since the power delivered is $$900\, W/m^2×A$$. The mirror in this case is a quarter-section of a cylinder, so the area for a length $$L$$ of the mirror is $$A=\frac{1}{4}(2πR)L$$. The area for a length of 1.00 m is then

\begin{align*} A&=\dfrac{\pi}{2}R(1.00m) \\[4pt] &=\dfrac{(3.14)}{2}(0.800\,m)(1.00\,m) \\[4pt] &=1.26\,m^2. \end{align*} \nonumber

The insolation on the 1.00-m length of pipe is then

$(9.00×10^2\dfrac{W}{m^2})(1.26\,m^2)=1130\,W. \nonumber$

c. The increase in temperature is given by $$Q=mcΔT$$. The mass $$m$$ of the mineral oil in the one-meter section of pipe is

\begin{align*} m &= ρV = ρπ\left(\dfrac{d}{2}\right)^2(1.00\,m) \nonumber \\[4pt] &=(8.00×10^2kg/m^3)(3.14)(0.0100\,m)^2(1.00\,m) \nonumber \\[4pt] &=0.251\,kg \end{align*} \nonumber

Therefore, the increase in temperature in one minute is

\begin{align*} \Delta T&= \dfrac{Q}{mc} \nonumber \\[4pt] &=\dfrac{(1130\,W)(60.0\,s)}{(0.251\,kg)(1670\,J⋅kg/°C)} \nonumber \\[4pt] &=162°\end{align*} \nonumber

Significance

An array of such pipes in the California desert can provide a thermal output of 250 MW on a sunny day, with fluids reaching temperatures as high as 400°C. We are considering only one meter of pipe here and ignoring heat losses along the pipe.

## Example $$\PageIndex{2}$$: Image in a Convex Mirror

A keratometer is a device used to measure the curvature of the cornea of the eye, particularly for fitting contact lenses. Light is reflected from the cornea, which acts like a convex mirror, and the keratometer measures the magnification of the image. The smaller the magnification, the smaller the radius of curvature of the cornea. If the light source is 12 cm from the cornea and the image magnification is 0.032, what is the radius of curvature of the cornea?

If you find the focal length of the convex mirror formed by the cornea, then you know its radius of curvature (it’s twice the focal length). The object distance is d o =12cm and the magnification is m=0.032. First find the image distance $$d_i$$ and then solve for the focal length $$f$$.

Start with the equation for magnification (Equation \ref{mag}) and solving for $$d_i$$ and inserting the given values yields

$d_i=−m d_o=−(0.032)(12\,cm)=−0.384\,cm \nonumber$

where we retained an extra significant figure because this is an intermediate step in the calculation. Solve the mirror equation for the focal length $$f$$ and insert the known values for the object and image distances. The result is

\begin{align*} \dfrac{1}{d_o}+\dfrac{1}{d_i} &= \dfrac{1}{f} \\[4pt] f &= \left(\dfrac{1}{d_o}+\dfrac{1}{d_i}\right)^{−1} \\[4pt] &= \left(\dfrac{1}{12cm}+\dfrac{1}{-0.384cm}\right)^{−1} \\[4pt] &=-40.0 \,cm \end{align*} \nonumber

The radius of curvature is twice the focal length, so

$R=2f=−0.80\,cm \nonumber$

The focal length is negative, so the focus is virtual, as expected for a concave mirror and a real object. The radius of curvature found here is reasonable for a cornea. The distance from cornea to retina in an adult eye is about 2.0 cm. In practice, corneas may not be spherical, which complicates the job of fitting contact lenses. Note that the image distance here is negative, consistent with the fact that the image is behind the mirror. Thus, the image is virtual because no rays actually pass through it. In the problems and exercises, you will show that, for a fixed object distance, a smaller radius of curvature corresponds to a smaller the magnification.

## PROBLEM-SOLVING STRATEGY: SPHERICAL MIRRORS

• Step 1. First make sure that image formation by a spherical mirror is involved.
• Step 2. Determine whether ray tracing, the mirror equation, or both are required. A sketch is very useful even if ray tracing is not specifically required by the problem. Write symbols and known values on the sketch.
• Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns).
• Step 4. Make a list of what is given or can be inferred from the problem as stated (identify the knowns).
• Step 5. If ray tracing is required, use the ray-tracing rules listed near the beginning of this section.
• Step 6. Most quantitative problems require using the mirror equation. Use the examples as guides for using the mirror equation.
• Step 7. Check to see whether the answer makes sense. Do the signs of object distance, image distance, and focal length correspond with what is expected from ray tracing? Is the sign of the magnification correct? Are the object and image distances reasonable?

## Departure from the Small-Angle Approximation

The small-angle approximation (Equation \ref{smallangle}) is a cornerstone of the above discussion of image formation by a spherical mirror. When this approximation is violated, then the image created by a spherical mirror becomes distorted. Such distortion is called aberration. Here we briefly discuss two specific types of aberrations: spherical aberration and coma.

## Spherical aberration

Consider a broad beam of parallel rays impinging on a spherical mirror, as shown in Figure $$\PageIndex{8}$$. The farther from the optical axis the rays strike, the worse the spherical mirror approximates a parabolic mirror. Thus, these rays are not focused at the same point as rays that are near the optical axis, as shown in the figure. Because of spherical aberration, the image of an extended object in a spherical mirror will be blurred. Spherical aberrations are characteristic of the mirrors and lenses that we consider in the following section of this chapter (more sophisticated mirrors and lenses are needed to eliminate spherical aberrations).

## Coma or Comatic Aberration

Coma is similar to spherical aberration, but arises when the incoming rays are not parallel to the optical axis, as shown in Figure $$\PageIndex{8b}$$. Recall that the small-angle approximation holds for spherical mirrors that are small compared to their radius. In this case, spherical mirrors are good approximations of parabolic mirrors. Parabolic mirrors focus all rays that are parallel to the optical axis at the focal point. However, parallel rays that are not parallel to the optical axis are focused at different heights and at different focal lengths, as show in Figure $$\PageIndex{8b}$$. Because a spherical mirror is symmetric about the optical axis, the various colored rays in this figure create circles of the corresponding color on the focal plane.

Although a spherical mirror is shown in Figure $$\PageIndex{8b}$$, comatic aberration occurs also for parabolic mirrors—it does not result from a breakdown in the small-angle approximation (Equation \ref{smallangle}). Spherical aberration, however, occurs only for spherical mirrors and is a result of a breakdown in the small-angle approximation. We will discuss both coma and spherical aberration later in this chapter, in connection with telescopes.

• Physics Concept Questions And Answers

## Mirror Equation Questions

A mirror is a polished shiny object usually made of glass which reflects most of the rays of light falling upon it.

Mirror equation is $$\begin{array}{l}\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\end{array}$$ .

• u is the Object distance
• v is the Image distance
• f is the Focal Length given by f=R/2
• R is the radius of curvature of the spherical mirror

## Important Mirror Equation Questions with Answers

1. What are the various types of mirrors?

The various types of mirrors are the plane mirror, convex mirror, and concave mirror.

2. ______ is a ray of light that hits the surface.

• Reflected ray
• Incident ray
• Refracted ray

Explanation: A ray of light that hits the surface is known as an incident ray.

3. The distance between the pole of the mirror and the object is known as ______.

• Image distance
• Object distance
• Focal Length
• Focal distance

Explanation: Object distance is the distance between the pole of the mirror and the object.

4. Image distance is represented by which letter?

Explanation: Image distance is represented by the letter ‘v’.

5. What is the formula for finding the focal length?

Explanation: The formula to find the focal length is f=R/2.

6. What is the mirror formula?

• $$\begin{array}{l}\frac{1}{v}-\frac{1}{u}={f}\end{array}$$
• $$\begin{array}{l}\frac{1}{v}+\frac{1}{u}={f}\end{array}$$
• $$\begin{array}{l}\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\end{array}$$
• $$\begin{array}{l}\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\end{array}$$

Answer: c) $$\begin{array}{l}\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\end{array}$$

Explanation: The mirror formula is $$\begin{array}{l}\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\end{array}$$ .

7. Mirror equation holds good for _____.

• Concave mirror
• Convex mirror
• Concave and convex mirror
• Plane mirror

Answer: c) Concave and convex mirror.

Explanation: Mirror equation holds good for a concave mirror and convex mirror.

8. What are the types of spherical mirrors?

The types of spherical mirrors are concave mirrors and convex mirrors.

9. The angle formed between the normal and the incident ray at the point of incidence is known as ______.

• Angle of reflection
• Angle of incidence
• Angle of refraction
• Angle of interference

Explanation: The angle of incidence is the angle formed between the normal and the incident ray at the point of incidence.

10. The ray that is 90° to the surface or the ray which is perpendicular to the reflecting surface is called ______.

Explanation: Normal ray is the ray that is 90° to the surface or the ray which is perpendicular to the reflecting surface.

## Practice Questions

• What is a mirror equation?
• Explain the mirror equation.
• What is a convex mirror?
• Define refracted ray.
• What is a spherical mirror?

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• 16.1 Reflection
• Introduction
• 1.1 Physics: Definitions and Applications
• 1.2 The Scientific Methods
• 1.3 The Language of Physics: Physical Quantities and Units
• Section Summary
• Key Equations
• Concept Items
• Critical Thinking Items
• Multiple Choice
• Extended Response
• 2.1 Relative Motion, Distance, and Displacement
• 2.2 Speed and Velocity
• 2.3 Position vs. Time Graphs
• 2.4 Velocity vs. Time Graphs
• 3.1 Acceleration
• 3.2 Representing Acceleration with Equations and Graphs
• 4.2 Newton's First Law of Motion: Inertia
• 4.3 Newton's Second Law of Motion
• 4.4 Newton's Third Law of Motion
• 5.1 Vector Addition and Subtraction: Graphical Methods
• 5.2 Vector Addition and Subtraction: Analytical Methods
• 5.3 Projectile Motion
• 5.4 Inclined Planes
• 5.5 Simple Harmonic Motion
• 6.1 Angle of Rotation and Angular Velocity
• 6.2 Uniform Circular Motion
• 6.3 Rotational Motion
• 7.1 Kepler's Laws of Planetary Motion
• 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity
• 8.1 Linear Momentum, Force, and Impulse
• 8.2 Conservation of Momentum
• 8.3 Elastic and Inelastic Collisions
• 9.1 Work, Power, and the Work–Energy Theorem
• 9.2 Mechanical Energy and Conservation of Energy
• 9.3 Simple Machines
• 10.1 Postulates of Special Relativity
• 10.2 Consequences of Special Relativity
• 11.1 Temperature and Thermal Energy
• 11.2 Heat, Specific Heat, and Heat Transfer
• 11.3 Phase Change and Latent Heat
• 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium
• 12.2 First law of Thermodynamics: Thermal Energy and Work
• 12.3 Second Law of Thermodynamics: Entropy
• 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators
• 13.1 Types of Waves
• 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period
• 13.3 Wave Interaction: Superposition and Interference
• 14.1 Speed of Sound, Frequency, and Wavelength
• 14.2 Sound Intensity and Sound Level
• 14.3 Doppler Effect and Sonic Booms
• 14.4 Sound Interference and Resonance
• 15.1 The Electromagnetic Spectrum
• 15.2 The Behavior of Electromagnetic Radiation
• 16.2 Refraction
• 16.3 Lenses
• 17.1 Understanding Diffraction and Interference
• 17.2 Applications of Diffraction, Interference, and Coherence
• 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge
• 18.2 Coulomb's law
• 18.3 Electric Field
• 18.4 Electric Potential
• 18.5 Capacitors and Dielectrics
• 19.1 Ohm's law
• 19.2 Series Circuits
• 19.3 Parallel Circuits
• 19.4 Electric Power
• 20.1 Magnetic Fields, Field Lines, and Force
• 20.2 Motors, Generators, and Transformers
• 20.3 Electromagnetic Induction
• 21.1 Planck and Quantum Nature of Light
• 21.2 Einstein and the Photoelectric Effect
• 21.3 The Dual Nature of Light
• 22.1 The Structure of the Atom
• 22.2 Nuclear Forces and Radioactivity
• 22.3 Half Life and Radiometric Dating
• 22.4 Nuclear Fission and Fusion
• 23.1 The Four Fundamental Forces
• 23.2 Quarks
• 23.3 The Unification of Forces
• A | Reference Tables

## Section Learning Objectives

By the end of this section, you will be able to do the following:

• Explain reflection from mirrors, describe image formation as a consequence of reflection from mirrors, apply ray diagrams to predict and interpret image and object locations, and describe applications of mirrors
• Perform calculations based on the law of reflection and the equations for curved mirrors

## Teacher Support

• (D) investigate behaviors of waves, including reflection, refraction, diffraction, interference, resonance, and the Doppler effect;
• (E) describe and predict image formation as a consequence of reflection from a plane mirror and refraction through a thin convex lens; and
• (F) describe the role of wave characteristics and behaviors in medical and industrial applications.

In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Mirrors and Lenses, as well as the following standards:

• (D) investigate behaviors of waves, including reflection, refraction, diffraction, interference, resonance, and the Doppler effect.

## Section Key Terms

Characteristics of mirrors.

[BL] Recall that, in geometry, angles are numbers that tell how far two straight lines are spread apart. The lines must be straight lines for the number to have meaning.

[OL] Geometry is the study of relationships involving points, lines, angles, and shapes. In this chapter, we are focused on the first three ideas.

[AL] In this chapter, we apply equations that use trigonometric functions that describe the properties of angles. Trigonometric functions are ratios of the lengths of two sides of a right triangle. There are six possible ratios; therefore, there are six such functions.

There are three ways, as shown in Figure 16.2 , in which light can travel from a source to another location. It can come directly from the source through empty space, such as from the Sun to Earth. Light can travel to an object through various media, such as air and glass. Light can also arrive at an object after being reflected, such as by a mirror. In all these cases, light is modeled as traveling in a straight line, called a ray . Light may change direction when it encounters the surface of a different material (such as a mirror) or when it passes from one material to another (such as when passing from air into glass). It then continues in a straight line—that is, as a ray. The word ray comes from mathematics. Here it means a straight line that originates from some point. It is acceptable to visualize light rays as laser rays (or even science fiction depictions of ray guns ).

Because light moves in straight lines, that is, as rays, and changes directions when it interacts with matter, it can be described through geometry and trigonometry. This part of optics, described by straight lines and angles, is therefore called geometric optics . There are two laws that govern how light changes direction when it interacts with matter: the law of reflection , for situations in which light bounces off matter; and the law of refraction , for situations in which light passes through matter. In this section, we consider the geometric optics of reflection .

[BL] Explain that light bounces is a simplification. The geometry of the path of a bouncing ball is similar to that of light, but what happens at the point of impact is different at the molecular level.

[OL] Indicate that the terms right angle , perpendicular , and normal line all mean the same thing: a vertical line at a 90° angle to a flat surface.

[AL] Recount and explain all the possible interactions of light with matter. Light can be absorbed at the surface of an opaque object. Some colors of light may be absorbed and others reflected. Light often is partially absorbed and partially reflected. It may also be transmitted through a transparent material, such as water or glass. Typically, if the surface of a transparent material is smooth, such as that of a window pane, light is transmitted partially and reflected partially.

Whenever we look into a mirror or squint at sunlight glinting from a lake, we are seeing a reflection. How does the reflected light travel from the object to your eyes? The law of reflection states: The angle of reflection , θ r θ r , equals the angle of incidence , θ i θ i . This law governs the behavior of all waves when they interact with a smooth surface, and therefore describe the behavior of light waves as well. The reflection of light is simplified when light is treated as a ray. This concept is illustrated in Figure 16.3 , which also shows how the angles are measured relative to the line perpendicular to the surface at the point where the light ray strikes it. This perpendicular line is also called the normal line , or just the normal . Light reflected in this way is referred to as specular (from the Latin word for mirror : speculum ).

We expect to see reflections from smooth surfaces, but Figure 16.4 , illustrates how a rough surface reflects light. Because the light is reflected from different parts of the surface at different angles, the rays go in many different directions, so the reflected light is diffused . Diffused light allows you to read a printed page from almost any angle because some of the rays go in different directions. Many objects, such as people, clothing, leaves, and walls, have rough surfaces and can be seen from many angles. A mirror, on the other hand, has a smooth surface and reflects light at specific angles.

When we see ourselves in a mirror, it appears that our image is actually behind the mirror. We see the light coming from a direction determined by the law of reflection. The angles are such that our image is exactly the same distance behind the mirror, d i , as the distance we stand away from the mirror, d o . Although these mirror images make objects appear to be where they cannot be (such as behind a solid wall), the images are not figments of our imagination. Mirror images can be photographed and videotaped by instruments and look just as they do to our eyes, which are themselves optical instruments. An image in a mirror is said to be a virtual image , as opposed to a real image . A virtual image is formed when light rays appear to diverge from a point without actually doing so.

Figure 16.5 helps illustrate how a flat mirror forms an image. Two rays are shown emerging from the same point, striking the mirror, and reflecting into the observer’s eye. The rays can diverge slightly, and both still enter the eye. If the rays are extrapolated backward, they seem to originate from a common point behind the mirror, allowing us to locate the image. The paths of the reflected rays into the eye are the same as if they had come directly from that point behind the mirror. Using the law of reflection—the angle of reflection equals the angle of incidence—we can see that the image and object are the same distance from the mirror. This is a virtual image, as defined earlier.

## Fun In Physics

Mirror mazes.

Figure 16.6 is a chase scene from an old silent film called The Circus , starring Charlie Chaplin. The chase scene takes place in a mirror maze. You may have seen such a maze at an amusement park or carnival. Finding your way through the maze can be very difficult. Keep in mind that only one image in the picture is real—the others are virtual.

One of the earliest uses of mirrors for creating the illusion of space is seen in the Palace of Versailles, the former home of French royalty. Construction of the Hall of Mirrors ( Figure 16.7 ) began in 1678. It is still one of the most popular tourist attractions at Versailles.

## Grasp Check

Only one Charlie in this image ( Figure 16.8 ) is real. The others are all virtual images of him. Can you tell which is real? Hint—His hat is tilted to one side.

• The virtual images have their hats tilted to the right.
• The virtual images have their hats tilted to the left.
• The real images have their hats tilted to the right.
• The real images have their hats tilted to the left.

## Watch Physics

Virtual image.

This video explains the creation of virtual images in a mirror. It shows the location and orientation of the images using ray diagrams, and relates the perception to the human eye.

• The distances of the image and the object from the mirror are the same.
• The distances of the image and the object from the mirror are always different.
• The image is formed at infinity if the object is placed near the mirror.
• The image is formed near the mirror if the object is placed at infinity.

Have students construct a ray diagram for an object reflected in a plane mirror. Point out to them that all information can be represented in the diagram by using just paper, a pencil, a ruler, and a protractor. Students may use the preceding video and Figure 16.5 to help them to draw the necessary rays for the diagram. Have them compare the position and orientation of the virtual image with that of the object, paying particular attention to the identical distances that the object and image have with respect to the mirror surface.

[BL] Ask students to define virtual and dispel any misconceptions. Explain the term in relation to geometric optics.

[OL] Explain that a real focal point is a point at which there is a concentration of light energy that can be transformed into other useful forms. At a virtual focal point , on the other hand, light energy cannot be concentrated because no light actually goes to that point.

[AL] Explain the difference between a parabolic shape and a spherical shape. Use drawings of a cross-section of each. Point out that, for a short section of a curved mirror with very little curvature, a spherical mirror approximates a parabolic one.

Some mirrors are curved instead of flat. A mirror that curves inward is called a concave mirror , whereas one that curves outward is called a convex mirror . Pick up a well-polished metal spoon and you can see an example of each type of curvature. The side of the spoon that holds the food is a concave mirror; the back of the spoon is a convex mirror. Observe your image on both sides of the spoon.

## Tips For Success

You can remember the difference between concave and convex by thinking, Concave means caved in .

Ray diagrams can be used to find the point where reflected rays converge or appear to converge, or the point from which rays appear to diverge. This is called the focal point , F. The distance from F to the mirror along the central axis (the line perpendicular to the center of the mirror’s surface) is called the focal length , f . Figure 16.9 shows the focal points of concave and convex mirrors.

Images formed by a concave mirror vary, depending on which side of the focal point the object is placed. For any object placed on the far side of the focal point with respect to the mirror, the rays converge in front of the mirror to form a real image, which can be projected onto a surface, such as a screen or sheet of paper However, for an object located inside the focal point with respect to the concave mirror, the image is virtual. For a convex mirror the image is always virtual—that is, it appears to be behind the mirror. The ray diagrams in Figure 16.10 show how to determine the nature of the image formed by concave and convex mirrors.

The information in Figure 16.10 is summarized in Table 16.1 .

## Concave and Convex Mirrors

• Silver spoon and silver polish, or a new spoon made of any shiny metal

Instructions

• Choose any small object with a top and a bottom, such as a short nail or tack, or a coin, such as a quarter. Observe the object’s reflection on the back of the spoon.
• Observe the reflection of the object on the front (bowl side) of the spoon when held away from the spoon at a distance of several inches.
• Observe the image while slowly moving the small object toward the bowl of the spoon. Continue until the object is all the way inside the bowl of the spoon.
• You should see one point where the object disappears and then reappears. This is the focal point.

Describe the differences in the image of the object on the two sides of the focal point. Explain the change. Identify which of the images you saw were real and which were virtual.

• The height of the image became infinite.
• The height of the object became zero.
• The intensity of intersecting light rays became zero.
• The intensity of intersecting light rays increased.

[BL] [OL] Ask students to identify as many examples as they can of curved mirrors that are used in everyday applications. Supply any they miss: security mirrors, mirrors for entering and exiting a driveway with poor visibility, rear-view mirrors, mirrors for application of cosmetics, and so on.

## Parabolic Mirrors and Real Images

This video uses ray diagrams to show the special feature of parabolic mirrors that makes them ideal for either projecting light energy in parallel rays, with the source being at the focal point of the parabola, or for collecting at the focal point light energy from a distant source.

• The rays do not polarize after reflection.
• The rays are dispersed after reflection.
• The rays are polarized after reflection.
• The rays become parallel after reflection.

## Teacher Demonstration

Have students use the demonstration in the video to construct a ray diagram that shows that rays from an object (upright arrow) placed at the focal point of a concave mirror emerge parallel to the central axis.

You should be able to notice everyday applications of curved mirrors. One common example is the use of security mirrors in stores, as shown in Figure 16.11 .

Some telescopes also use curved mirrors and no lenses (except in the eyepieces) both to magnify images and to change the path of light. Figure 16.12 shows a Schmidt-Cassegrain telescope. This design uses a spherical primary concave mirror and a convex secondary mirror. The image is projected onto the focal plane by light passing through the perforated primary mirror. The effective focal length of such a telescope is the focal length of the primary mirror multiplied by the magnification of the secondary mirror. The result is a telescope with a focal length much greater than the length of the telescope itself.

A parabolic concave mirror has the very useful property that all light from a distant source, on reflection by the mirror surface, is directed to the focal point. Likewise, a light source placed at the focal point directs all the light it emits in parallel lines away from the mirror. This case is illustrated by the ray diagram in Figure 16.13 . The light source in a car headlight, for example, is located at the focal point of a parabolic mirror.

Parabolic mirrors are also used to collect sunlight and direct it to a focal point, where it is transformed into heat, which in turn can be used to generate electricity. This application is shown in Figure 16.14 .

## The Application of the Curved Mirror Equations

[BL] [OL] Review operations for manipulating fractions and for rearranging equations involving fractional values of variables.

[AL] Demonstrate how to solve equations of the type 1 a = 1 b + 1 c 1 a = 1 b + 1 c for any of the variables in terms of the other two. Rearrange so that the variable solved for is not a reciprocal.

Curved mirrors and the images they create involve a fairly small number of variables: the mirror’s radius of curvature, R ; the focal length, f ; the distances of the object and image from the mirror, d o and d i , respectively; and the heights of the object and image, h o and h i , respectively. The signs of these values indicate whether the image is inverted, erect (upright), real, or virtual. We now look at the equations that relate these variables and apply them to everyday problems.

Figure 16.15 shows the meanings of most of the variables we will use for calculations involving curved mirrors.

The basic equation that describes both lenses and mirrors is the lens/mirror equation

This equation can be rearranged several ways. For example, it may be written to solve for focal length.

Magnification, m , is the ratio of the size of the image, h i , to the size of the object, h o . The value of m can be calculated in two ways.

This relationship can be written to solve for any of the variables involved. For example, the height of the image is given by

We saved the simplest equation for last. The radius of curvature of a curved mirror, R , is simply twice the focal length.

We can learn important information from the algebraic sign of the result of a calculation using the previous equations:

• A negative d i indicates a virtual image; a positive value indicates a real image
• A negative h i indicates an inverted image; a positive value indicates an erect image
• For concave mirrors, f is positive; for convex mirrors, f is negative

Now let’s apply these equations to solve some problems.

## Worked Example

Calculating focal length.

A person standing 6.0 m from a convex security mirror forms a virtual image that appears to be 1.0 m behind the mirror. What is the focal length of the mirror?

The person is the object, so d o = 6.0 m. We know that, for this situation, d o is positive. The image is virtual, so the value for the image distance is negative, so d i = –1.0 m.

Now, use the appropriate version of the lens/mirror equation to solve for focal length by substituting the known values.

f = d i d o d o + d i = ( − 1.0 ) ( 6.0 ) 6.0 + ( − 1.0 ) = − 6.0 5.0 = − 1.2  m f = d i d o d o + d i = ( − 1.0 ) ( 6.0 ) 6.0 + ( − 1.0 ) = − 6.0 5.0 = − 1.2  m

The negative result is expected for a convex mirror. This indicates the focal point is behind the mirror.

## Calculating Object Distance

Electric room heaters use a concave mirror to reflect infrared (IR) radiation from hot coils. Note that IR radiation follows the same law of reflection as visible light. Given that the mirror has a radius of curvature of 50.0 cm and produces an image of the coils 3.00 m in front of the mirror, where are the coils with respect to the mirror?

We are told that the concave mirror projects a real image of the coils at an image distance d i = 3.00 m. The coils are the object, and we are asked to find their location—that is, to find the object distance d o . We are also given the radius of curvature of the mirror, so that its focal length is f = R /2 = 25.0 cm (a positive value, because the mirror is concave, or converging). We can use the lens/mirror equation to solve this problem.

Because d i and f are known, the lens/mirror equation can be used to find d o .

Rearranging to solve for d o , we have

Entering the known quantities gives us

Note that the object (the coil filament) is farther from the mirror than the mirror’s focal length. This is a case 1 image ( d o > f and f positive), consistent with the fact that a real image is formed. You get the most concentrated thermal energy directly in front of the mirror and 3.00 m away from it. In general, this is not desirable because it could cause burns. Usually, you want the rays to emerge parallel, and this is accomplished by having the filament at the focal point of the mirror.

Note that the filament here is not much farther from the mirror than the focal length, and that the image produced is considerably farther away.

## Practice Problems

What is the focal length of a makeup mirror that produces a magnification of 1.50 when a person’s face is 12.0 cm away? Construct a ray diagram using paper, a pencil and a ruler to confirm your calculation.

Use these questions to assess student achievement of the section’s learning objectives. If students are struggling with a specific objective, these questions will help identify which one, and then you can direct students to the relevant content.

How does the object distance, d o , compare with the focal length, f, for a concave mirror that produces an image that is real and inverted?

• d o > f, where d o and f are object distance and focal length, respectively.
• d o < f, where d o and f are object distance and focal length, respectively.
• d o = f, where do and f are object distance and focal length, respectively.
• d o = 0, where do is the object distance.

Use the law of reflection to explain why it is not a good idea to polish a mirror with coarse sandpaper.

• The surface becomes smooth. A smooth surface produces a sharp image.
• The surface becomes irregular. An irregular surface produces a sharp image.
• The surface becomes smooth. A smooth surface transmits but does not reflect light.
• The surface becomes irregular. An irregular surface produces a blurred image.
• It is real and erect.
• It is real and inverted.
• It is virtual and inverted.
• It is virtual and erect.

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## Concave Mirror: Problems with Answers for AP Physics 2

Here, using the mirror equation and magnification formula for curved mirrors a number of problems on the concave mirror are solved which is helpful for the AP Physics 2 exam.

We tried to illustrate all properties of the image formed by a concave mirror using equations and ray diagrams for a deeper understanding.

## Summary of concave spherical mirrors

Mirror equation is $\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}$ Where $f, d_i, d_o$ are the focal length, image distance and the object distance, respectively.

The magnification formula for spherical mirrors is also written as $M=-\frac{d_i}{d_o}=\frac{h_i}{h_o}$ where $h_i, h_o$ are the image and object heights, respectively.

## Sign Rules for Concave mirror:

• The focal length is positive $f>0$.
• The image distance must be inserted positive $d_i>0$ if the image is formed in front of the mirror.
• The image distance must be inserted negative $d_i<0$ if the image is formed back of the mirror.

Note : In front of the mirror is where the object is placed.

## Concave Mirror Solved Problems:

Problem (1): A pencil is placed 6 cm in front of a concave mirror having a radius of curvature of 40 cm.  (a) What is its focal length? (b) Determine the position of the image formed by this mirror.  (c) Draw a ray diagram and verify your previous results. (d) What do we conclude from this problem?

Solution: (a) Recall that the focal length $f$ of a spherical mirror is related to the radius of curvature $R$ by the following formula $f=\frac R2$ Substituting the known value into the above, we get $f=\frac R2=\frac {40}{2}=20\quad {\rm cm}$  (b) Using the mirror equation below and solving for unknowns such as image distance $d_i$ or object distance $d_o$, we complete our calculation. $\frac{1}{d_i}+\frac{1}{d_o}=\frac{1}{f}$ To use mirror equation, we must note the sign rules involved in this equation:

• $f<0$ for convex mirror and $f>0$ for concave mirror.
• The image distance is always formed behind the convex mirror so $d_i<0$.
• In a concave mirror, if the image is formed on the same side of the object so $d_i>0$ otherwise $d_i<0$ must be inserted.

In this problem, the mirror is concave so focal length must be put with a plus sign $f>0$ in the mirror equation. \begin{align*} \frac{1}{f}&=\frac{1}{d_i}+\frac{1}{d_o}\\ \\ \frac{1}{20}&=\frac{1}{d_i}+\frac{1}{6}\\ \\ \Rightarrow \frac{1}{d_i}&=\frac{1}{20}-\frac{1}{6}\\\\&=\frac{6-20}{20\times 6}\\ \\ &=\frac{-14}{20\times 6}\end{align*} Flipping both sides of above relation, we get $d_i=-\frac{120}{14}=8.57\quad {\rm cm}$ As you can see, we obtained a negative value for image distance for an object in front of a concave mirror. Thus, we conclude that the image must be formed on the opposite side of the mirror, or in other words, behind the mirror.

This image which is formed on the opposite side of the object is called a virtual image.

Therefore, the image of the above pencil is located about 9 cm behind the mirror.

(c) In the following figure, using two rays we obtained the image of the pencil.

(d) In this problem, the object sits between the focal point and the concave mirror. As the ray diagram above showed, once there was such a case, we obtain a virtual, upright, and magnified image. But how much bigger? see next problem.

Problem (2): A candle 6 cm tall is placed at a distance of 8 cm in front of a concave mirror whose radius of curvature is 20 cm. Determine the position, size, orientation, and nature (real or virtual) of the image.

Solution : Known data is object' height $h_o=6\,{\rm cm}$, object distance $d_o=10\,{\rm cm}$, and the radius of curvature of the mirror $R=20\,{\rm cm}$. The focal length is obtained as below $f=\frac R2=10\quad {\rm cm}$ As you can see, $d_o<f$ that is the object is located between the focal point and the concave mirror. In such cases, we have a virtual, upright, and bigger image (see the previous problem).

Applying the mirror equation get the position of the image $d_i$ as below \begin{align*} \frac{1}{f}&=\frac{1}{d_i}+\frac{1}{d_o}\\ \\ \frac{1}{10}&=\frac{1}{d_i}+\frac{1}{8}\\ \\ \Rightarrow \frac{1}{d_i}&=\frac{1}{10}-\frac{1}{8}\\\\&=\frac{8-10}{10\times 8}\\ \\ &=\frac{-2}{80}\end{align*} Reversing both sides of the above, we have $d_i=-40\,{\rm cm}$. Because image distance $d_i$ is negative, the image is on the opposite side of the mirror and hence is virtual.

To quantify how much bigger the image is in the curved mirrors, we use the magnification formula for curved mirrors as below $M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$ Where $h_i$, $h_o$ are the image and object heights. $M=-\frac{d_i}{d_o}=-\frac{-40}{8}=5$ Thus, the object is magnified as big as 5 times.

Problem (3): A certain concave spherical mirror has a focal length of 10 cm. An object is located 10 cm in front of it. Find the properties of the image.

Solution : Here, the object is at the focal point. Use the mirror equation to find the image position as below \begin{align*} \frac{1}{f}&=\frac{1}{d_i}+\frac{1}{d_o}\\ \\ \frac{1}{10}&=\frac{1}{d_i}+\frac{1}{10}\\ \\ \frac{1}{d_i}&=0 \\ \\ \Rightarrow d_i&=\infty \end{align*} Thus, the image is formed at infinity!. Using the magnification formula, we can see that it is also infinite.

From this problem, we conclude that when an object is placed at the focal point, its image is formed at infinity with infinite in size.

Problem (4): In front of a concave mirror whose radius of curvature is 20 cm we place a candle, 5 cm tall, at a distance of 25 cm away from it. Determine (a) the position,(b) the size, and (c) the nature (real or virtual) and orientation of the image.

Solution : All we need to find the image properties is to apply the mirror equation and magnification formula. In this problem, the radius of curvature of the concave curved mirror is given which is related to the focal length by the following formula $f=\frac R2=\frac {20}{2}=10\,{\rm cm}$ The object is placed outside the center of the curvature of the concave mirror because $d_o>2f$. Now, we want to find the properties of the image formed by this object in this position.

(a) First of all, using the mirror equation we find the image distance from the concave mirror but recall that, according to mirror sign rules, for the concave mirror, the focal length must be inserted as $f>0$. \begin{align*} \frac{1}{f}&=\frac{1}{d_i}+\frac{1}{d_o}\\ \\ \frac{1}{10}&=\frac{1}{d_i}+\frac{1}{25}\\ \\ \frac{1}{d_i}&=\frac{1}{10}-\frac{1}{25}\\\\&=\frac{25-10}{10\times 25}\\\\&=\frac{15}{25} \\ \\ \Rightarrow d_i&=\frac{25}{15}=16.7\,{\rm cm} \end{align*} Because $d_i>0$, the image is in front of the mirror and is real (on the same side of the object). In other words, the image is formed between the focal point and the center of the curvature.

(b) Magnification in curved mirrors is the ratio of the image size to the object size or the ratio of the image distance to the object distance with a negative. $M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

From the first expression, we can find the magnification of the mirror as $M=-\frac{d_i}{d_o}=\frac{16.7}{25}=-0.6$ The image is 0.6 the size of the object.

The image height is also obtained by the second expression as below \begin{align*} M&=\frac{h_i}{h_o}\\\\ -0.6&=\frac{h_i}{5}\\\\ \Rightarrow  h_i&=(-0.6)(5)=-3\,{\rm cm}\end{align*} The minus sign indicates that the image is inverted.

(c) Because $d_i>0$, the image is real. Since the magnification is negative $M<0$, the image is inverted.

All of the above findings can also be obtained by a concave mirror ray diagram as shown in the figure below.

The notes we learn from this problem are that when an object is placed outside the center of curvature of a concave mirror, then the image is real, inverted, reduced in size, and is formed between focal point $f$ and center of curvature $C$.

For more practicing, you can also check out the following pages Converging Lens Problems with Solution Diverging Lens Problems with Solution

Author : Dr. Ali Nemati Date Published: 5/14/2021

## Mirror and Lens Problems for Class 10

This article present the general Glossary of Mirror and lens and step by step method to solve Mirror and Lens problems

## Glossary of Mirror and Lens

Plane Mirror A plane mirror is a mirror with a planar reflective surface. For light rays striking a plane mirror, the angle of reflection equals the angle of incidence.It always form virtual image

Concave or Converging Mirrors A curved mirror that has its center farther away from you than its edges is called a concave mirror. A concave mirror tends to reflect parallel rays through a point called its focus – in other words a concave mirror converges light. Therefore concave mirrors are often referred to as converging mirrors. This converging effect makes sense if you imagine a concave mirror to be an array of small plane mirrors.You can demonstrate the converging of parallel light rays by a concave mirror quite dramatically for yourself. A relatively small concave mirror placed in the sun will gather enough light to ignite a small piece of paper quite quickly

Convex or Diverging Mirrors It is  a curved mirror that has its center closer to you than its edges is called a convex mirror.A convex mirror tends to reflect parallel rays so that they spread out, or diverge. For this reason, convex mirrors are often referred to as diverging mirrors. This diverging effect makes sense if you think of a convex mirror as an array of small plane mirrors.

Convex lens It is also called converging lens, real or virtual rays come together at a point

Concave lens

It is also called diverging lens because spreads light out

Real Image when light really comes together-will show up on a screen

Virtual Image light rays not really coming together-never on a screen

## How to solve the Mirror and Lens Problems

Step 1) The concept of flats mirror,concave mirror/lens and convex mirror/lens should be clear. The formula used are Mirror equation/lens maker equation and Magnification equation. Step 2) The sign convention and ray diagram should also be clear Step 3) Read the Problem statement clearly and find out the clues in the problem. Find the known and unknown variable. This step is crucial and you need to apply the sign convention also. Step 4) Draw a ray diagram. The more careful you are in constructing this, the better idea you’ll have of where the image is. Step 5) Apply the appropriate  equation to determine the unknown. You may have solve two simultaneous equation in some case. Step 6) The result of the steps 4 and 5 are consistent with each other.

## How to solve Multiple Lenses Problem

Step 1)Do the first lens as if the others weren’t there. Step 2)Use the image formed by this lens as the object of the next lens step 3)Repeat this process for all the lenses in the system Step 4)The total magnification is just the product of the individual magnifications of each lens.

## Some Mirror and Lens Problems

Question (1)

An object is placed 25 cm in front of a concave mirror of focal length 60 cm. Calculate the image distance and the magnification. Characterize the image.

Question (2)

An object is placed 25 cm in front of a convex mirror of focal length 30 cm. Calculate the image distance and the magnification. Characterize the image.

Question (3)

Two plane mirrors are inclined to one another at an angle 30. A ray traveling in the plane as shown below is incident on one of the mirrors. Applying the Law of Reflection, show that the path of the ray after the two reflections is deviated by an angle which is independent of the angle of incidence.

Question (4)

Ram obtains a blurred image of an object on the screen by using the concave mirror, In order to obtain sharp sharp image on the screen, he will have to shift the mirror (a)to a position very far from screen (b)towards the screen (c) away from the screen (d) either away or towards depending on the position of the object

Question (5)

Refractive index of media A, B, C and D are A 1.31 B 1.52 C 1.45 D 1.64

In which of the four media is the speed of light (i) maximum (ii) minimum?

Question (6)

An object of size 5 cm is kept at a distance of 25 cm from the optical center of a converging lens of focal length 10cm. Calculate the distance of the image from the lens and size of the image.

Question (7)

Explain reasons for each of the following: (i) The sky appears to be blue during day time to a person on earth. (ii) The sky near the horizon appears to have a reddish heu at the time of sunset and sunrise. (iii) The sky appears dark instead of blue to an astronaut. (iv) The stars appears to twinkle. (v) The planets do not twinkle.

Question (8)

How does the speed of light change as the index of refraction increases?

Question (9)

How does the size of the critical angle change as the index of refraction increases?

Question (10)

Could an index of refraction ever be less than 1? What would this imply about the speed of light in that medium?

Question (11)

A 2.25-cm-tall object is 8.5 cm to the left of a convex lens of 5.5-cm focal length. Find the image position and height

Question (12)

Does your bathroom mirror show you older or younger than your actual age?

Question (13)

A concave makeup mirror is designed so that a person 25 cm in front of it sees an upright image magnified by a factor of 2. What is the radius of curvature of the mirror?

## 2 thoughts on “Mirror and Lens Problems for Class 10”

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## Solve Problems on Mirror and Magnification Formula

Binoculars are usually used to observe distant objects. While looking through this device, it has been noticed that the size of the objects visible via the binoculars differs from what we observe with our naked eyes. This is due to the fact that binoculars concentrate light from distant objects, allowing us to see them in more detail by making them look larger. This was a demonstration of the mirror formula and magnification.

## Mirror Formula

The mirror formula shows the relationship between the object distance, image distance, and the focal length of the spherical mirror. Hence, the formula is given as,

1/u + 1/v  = 1/f

where,

• u is the object distance,
• v is the image distance, and
• f is the focal length of the mirror
While solving numerical problems using mirror formula one must remember two important things according to the sign convention for spherical mirrors, that are: If object is at left side of the mirror the object distance is taken as negative, if it is in right it is positive. For focal length the sign is depend on type of mirror we are using if mirror is concave then focal length is taken negative and if mirror is convex then focal length is positive .

Mirror formula in terms of Radius of Curvature:

Since, the radius of curvature (R) is two times its focal length (f), that is

Hence, the Mirror Formula can be written as:

1/u + 1/v = 1/f = 2/R

## Magnification

One must have noticed that there is simply an increase or decrease in the size of image by concave or convex mirror, this is a magnification of the object.

• Magnification is defined as the ratio of the height of the image (h i ) to the height of the object (h o ).

Magnification = Height of image / Height of object

m = H i  / H o

Magnification can also be defined as the ratio of the image distance (v) and the object distance from the mirror (v).

Magnification = -Image distance / Object distance

Determination of the Nature of the Image formed by a spherical mirror: The positive magnitude of magnification shows us that a virtual and erect image is formed. The negative  magnitude of magnification shows us that a real and inverted image is formed.

## Problems on Mirror Formula and Magnification Formula

Problem 1: An object is placed at a distance of 2 times of focal length from the pole of the convex mirror, Calculate the linear magnification.

Let the Focal length of mirror = f So, the object distance, u = -2f  The formula to calculate image distance we use mirror formula as, 1 / v + 1 / u = 1 / f Therefore, 1 / v + 1 / -2f = 1 / f               1 / v  = 1 / f + 1 / 2f                         = 3 / 2f or v = 2f / 3 Magnification is given as,  m = – v / u = -(2f/3) / (-2f)  =   1/3

Problem 2. If the image is a distance of 6 cm and the object is at 12 cm in the front of the concave mirror, Calculate the magnification formed.

Given that, The distance of object, u = – 12 cm The distance of image, v = – 6 cm Since, Magnification is given by,  m = – v / u Therefore, m = – (-6 / -12)  = -0.5 Hence, the image will be diminished by nearly half as size of object.

Problem 3: In the experiment height of the image is 12 cm whereas the height of the object is 3 cm, would you determine the magnification formed.

Given that,  Height of image = 12 cm  Height of object = 3 cm Magnification in terms of height is given by, m = height of image / height of object = 12 / 3  = 4 Therefore magnification is 4 .

Problem 4: In the case of a concave mirror if the object is placed at the distance of 12 cm. Determine the image distance from the mirror if the height of the object to image ratio is 1:2.

Given that, The object distance, u = -12 cm Ratio of object to image height = 1/2 Magnification = height of image / height of object   = 1/ (1/2)  = 2 Now, magnification in terms of distance of object and image from the mirror, m = – v / u = – v / -12  2 = v / 12   or  v = 12 × 2  = 24 Therefore the distance of image from the mirror is equal to 24 cm .

Problem 5: Calculate the change in the size of the image formed, if the object distance is 18 cm and the distance of the image is 6 cm from the concave mirror.

Given that, The object distance, u = -12 cm Image distance, v = – 6cm Magnification,  m = – v/u  = – (-6 / -18)  = -1/3 which means that size of image is 1/3 rd of the size of object.

Problem 6: The radius of curvature of the rear view convex mirror of the truck is 6 m. If the car is 8 m from the mirror of the truck. Calculate the distance at which the image is formed.

Given that, Radius of curvature, R = 6 m  Object distance, u = -8 m Focal length is half of Radius of curvature, f = R/2  =  6/2  = 3 m Using mirror formula  1 / v + 1 / u = 1 / f 1 / v + 1 / -8 = 1 / 3 1 / v  = 1 / 3 + 1 / 8 = 11 / 24 v = 24 / 11 m The image is formed at distance of 24 / 11 behind the mirror.

Problem 7: A concave mirror produces an image of size n times that of the object and of focal length f. If the image is real then find the distance of the object from the mirror.

Given that Size of image = n × size of object n = Size of image /  size of object  = magnification Since the image is real, it must be inverted hence magnification will be negative,  m = -n Let d is the distance of object then, m = -v/u   -n = -v / d  or v = nd Therefore, the mirror formula: 1 / f = 1/v + 1/u becomes, 1/f = 1/nd + 1/d or 1/f = 1/d(1/n + 1) or 1/d = n/ f(n + 1) Therefore, d = f (n + 1)/ n

Problem 8: Where should the object be placed to obtain a magnification of 1/3? If an object is placed at a distance of 60 cm from a convex mirror, then the magnification produced is 1/2.

Given that, u = -60 cm m  = 1/2  So, -v/u = 1/2  and  v/60 = 1/2 or v = 30 cm Since, the mirror formula is: 1 / v + 1 / u = 1 / f Therefore, 1 / 30 + 1 / (-60) = 1/f 1/f = ( 2-1 ) / 60 = 1 / 60 f = 60 cm Now for magnification = 1 / 3, – v / u = 1 / 3 or  v = – u / 3  using mirror formula  1 / v + 1 / u = 1 / f 1 / (-u/3) + 1/ u = 1/ 60 -3/ u + 1/u = 1/60 -2/ u = 1/60 or  u = -120 cm object should be placed at 120 cm in front of mirror to get magnification of 1/3.

Problem 9: In the case of a concave mirror, if the object distance is 11 cm, its focal length is 11 cm then, Calculate the image distance.

Given that, Distance of object, u = -11 cm Focal length, f = -11cm Using mirror formula, 1 / v + 1 / u = 1 / f Therefore, 1 / v + 1 / -11 = 1/ -11 So, 1/v = 0 or  v = infinity This means that image will be at infinity if object is present at the focal length.

Problem 10: If the object distance is 32 cm in front of the concave mirror, the focal length of the mirror is 16 cm. State the nature and the size of the image formed.

Given that, Object distance, u = -32 cm Focal length , f = -16 cm For image distance use mirror formula, 1 / v + 1 / u = 1 / f Therefore, 1/ v + 1/ -32 = 1/ -16 or  1/ v = 1/ -16 + 1/ 32 or 1/ v = -1 / 16 So, v = -16 cm Hence the image is located 16 cm in front of the mirror. and the image formed is real and inverted.  Size of image will be same as that of object, as it is located at center of curvature.

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## Ray Optics: Reflection and Mirrors

#### IMAGES

1. Mirror Formula

2. Derivation of mirror equation||Mirror Equation useful for solving

3. Mirror Formula

4. 15.3 Mirror Equation

5. Spherical Mirrors & The Mirror Equation

6. How to Calculate the Focal Point of a Concave Mirror Using the Mirror

#### VIDEO

1. Solving mirror cube

2. solving mirror cube from r u r, u, #shorts #cubing

3. Solving Mirror cube in 10 sec 🤯 #rubikscube

4. Solving mirror block in 0.1 sec

5. Solving The Mirror Cube

6. solving mirror cube 🤯 #viral #shorts

1. Physics Tutorial: The Mirror Equation

As a demonstration of the effectiveness of the mirror equation and magnification equation, consider the following example problem and its solution. Example Problem #1 A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and the image size.

2. Mirror equation example problems (video)

Mirror equation example problems Google Classroom About Transcript In this video David solves a few exmaple problems involving concave and convex mirrors using the mirror equation and magnification equation. Questions Tips & Thanks Want to join the conversation? Sort by: Top Voted Coco 7 years ago Now I'm confused...

3. Mirror Equation

Mirror equation is an equation relating object distance and image distance with focal length. It is also known as the mirror formula. In a spherical mirror: The distance between the object and the pole of the mirror is called Object distance (u). The distance between the image and the pole of the mirror is called Image distance (v).

4. 25.7 Image Formation by Mirrors

Learning Objectives By the end of this section, you will be able to: Illustrate image formation in a flat mirror. Explain with ray diagrams the formation of an image using spherical mirrors. Determine focal length and magnification given radius of curvature, distance of object and image.

5. Mirror equation example problems

6. Mirror formula (video)

more. As far as I know, there are majorly 4 formulae in light: the mirror formula, lens formula and magnification. All three are sign sensitive as in a mirror and lens, each side has a different denotation. Even the power of lens formula is sign sensitive, as it is calculated using focal length of a lens.

7. 2.3: Spherical Mirrors

Solve the mirror equation for the focal length $$f$$ and insert the known values for the object and image distances. ... PROBLEM-SOLVING STRATEGY: SPHERICAL MIRRORS. Step 1. First make sure that image formation by a spherical mirror is involved. Step 2. Determine whether ray tracing, the mirror equation, or both are required. A sketch is very ...

8. 2.2 Spherical Mirrors

Solve the mirror equation for the focal length f and insert the known values for the object and image distances. The result is. 1 d o + 1 d i = 1 f f = (1 d o + 1 d i) ... Most quantitative problems require using the mirror equation. Use the examples as guides for using the mirror equation. Step 7. Check to see whether the answer makes sense.

9. Mirror Equation Questions

1. What are the various types of mirrors? The various types of mirrors are the plane mirror, convex mirror, and concave mirror. 2. ______ is a ray of light that hits the surface. Normal ray Reflected ray Incident ray Refracted ray Answer: c) Incident ray. Explanation: A ray of light that hits the surface is known as an incident ray. 3.

10. The Mirror Equation (Concave Mirrors)

0:00 / 5:24 The Mirror Equation (Concave Mirrors) The Science Classroom 55K subscribers Subscribe 1.6K Share Save 171K views 8 years ago In this video we will learn the mirror equation...

11. 16.1 Reflection

Now let's apply these equations to solve some problems. Worked Example. ... /2 = 25.0 cm (a positive value, because the mirror is concave, or converging). We can use the lens/mirror equation to solve this problem. Solution. Because d i and f are known, the lens/mirror equation can be used to find d o. 1 f = 1 d i + 1 d o 1 f = 1 d i + 1 d o.

12. PDF Physics

The mirror equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f). The equation is stated as follows: Magnification Equation

13. How to Solve Mirror or Lens Equation Problems

Step 1: We will first determine all of the known quantities. We have been given the object height, the object distance, and the focal length (which we know to be positive because the mirror is...

14. Concave Mirror: Problems with Answers for AP Physics 2

Concave Mirror Solved Problems: Problem (1): A pencil is placed 6 cm in front of a concave mirror having a radius of curvature of 40 cm. (a) What is its focal length? (b) Determine the position of the image formed by this mirror. (c) Draw a ray diagram and verify your previous results. (d) What do we conclude from this problem?

15. Mirror and Lens Problems for Class 10

How to solve the Mirror and Lens Problems. Step 1) The concept of flats mirror,concave mirror/lens and convex mirror/lens should be clear. The formula used are Mirror equation/lens maker equation and Magnification equation. Step 2) The sign convention and ray diagram should also be clear. Step 3) Read the Problem statement clearly and find out ...

16. How to Solve Concave Mirror Problems

Steps for Solving Concave Mirror Problems. Step 1: Make a list of the known quantities given in the problem. Step 2: Determine if the unknown quantities require you to use the mirror equation, the ...

17. Physics Video Tutorial

The Mirror Equation Video Tutorial The Mirror Equation Video Tutorial explains how to use the mirror equation and the magnification ratio to solve Physics word problems. Four examples are discussed. The video lesson answers the following questions:

18. Solve Problems on Mirror and Magnification Formula

1/u + 1/v = 1/f = 2/R Magnification One must have noticed that there is simply an increase or decrease in the size of image by concave or convex mirror, this is a magnification of the object. Magnification is defined as the ratio of the height of the image (h i) to the height of the object (h o ). Magnification = Height of image / Height of object

19. Reflection and Mirrors Problem Sets

Includes 7 problems. Problem Set RM4 - Concave Mirrors 1. Use the mirror and magnification equations to determine the image distance, image height, and magnification for objects placed at several strategic locations in front of a concave mirror. Includes 3 problems. Problem Set RM5 - Concave Mirrors 2.