You determine a concave mirror to have focal length f=10cm⁡f=10\operatorname{\mathrm{cm}}. If you submerge the mirror in water (n=1.33n=1.33), what would be the new focal length?

10cm⁡10\operatorname{\mathrm{cm}}

11.3cm⁡11.3\operatorname{\mathrm{cm}}

20cm⁡20\operatorname{\mathrm{cm}}

A concave makeup mirror has a focal length of 12.0cm⁡12.0\operatorname{\mathrm{cm}}. What is the mirror’s radius of curvature?

24.0cm⁡24.0\operatorname{\mathrm{cm}}

6.0cm⁡6.0\operatorname{\mathrm{cm}}

12.0cm⁡12.0\operatorname{\mathrm{cm}}

48.0cm⁡48.0\operatorname{\mathrm{cm}}

An object is placed 45.0cm⁡45.0\operatorname{\mathrm{cm}} from a concave mirror, producing a real image 15.0cm⁡15.0\operatorname{\mathrm{cm}} from the mirror. What is the focal length of the mirror?

11.25cm⁡11.25\operatorname{\mathrm{cm}}

7.5cm⁡7.5\operatorname{\mathrm{cm}}

30.0cm⁡30.0\operatorname{\mathrm{cm}}

60.0cm⁡60.0\operatorname{\mathrm{cm}}

What are the characteristics of images formed by concave mirrors depending on the object's position?

The characteristics of images formed by concave mirrors depend on the object's distance relative to the focal point (F) and center of curvature (C):When the object is beyond C, the image is real, inverted, and reduced, forming between F and C.If the object is between C and F, the image is real, inverted, and enlarged, forming beyond C.When the object is closer than F, the image is virtual, upright, and enlarged, appearing behind the mirror.At exactly the focal point, the image forms at infinity.

33. Geometric Optics

Concave Mirrors

33. Geometric Optics

Concave Mirrors: Videos & Practice Problems

Video Lessons Practice

Topic summary

Concave Mirrors are curved mirrors that reflect parallel light rays toward a common focal point, so they can form images that differ in size, orientation, and type. The key geometric points are the focal point and the center of curvature, with focal length and radius of curvature related by $f=\frac{R}{2}$ . For distant objects, incoming parallel rays focus near the focal point. Because this behavior depends on reflection, the focal length is set by mirror curvature and does not change with the surrounding medium.

Image location is found with the mirror equation, $ \frac{1}{d_O}+\frac{1}{d_I}=\frac{1}{f} $ , together with principal rays: the P-ray, F-ray, and C-ray. A real image forms where reflected rays actually meet; a virtual image forms where rays only appear to come from. For objects beyond C, the image is real, inverted, and reduced; between C and F, it is real, inverted, and enlarged; inside F, it is virtual, upright, and enlarged. At exactly F, the image is effectively at infinity.

Magnification connects image size and orientation through $m=\frac{h_I}{h_O}=-\frac{d_I}{d_O}$ . A negative magnification means inverted, a positive value means upright, and the magnitude shows whether the image is reduced or enlarged.

Concept

Intro to Spherical Mirrors

Video duration:

Intro to Spherical Mirrors Video Summary

Plane mirrors produce images that are always the same size as the object, which can limit their usefulness when magnification is needed. To create larger or smaller images, curved mirrors, also known as spherical mirrors, are used. Unlike plane mirrors that reflect light according to the law of reflection with images appearing the same distance behind the mirror, spherical mirrors focus light to a specific point called the focal point, labeled as F.

When parallel rays of light, often referred to as paraxial rays, strike a concave mirror (a mirror shaped like a cave, curving inward), they reflect and converge at the focal point. The focal length, denoted as f, is the distance from the mirror’s vertex (the center of the mirror surface) to this focal point. This focal length is related to the radius of curvature R of the mirror by the equation:

\[f = \frac{R}{2}\]

Here, R is the radius of curvature, which is the distance from the vertex to the center of curvature C, the center of the sphere from which the mirror segment is taken. This relationship is fundamental in understanding how concave mirrors focus light.

Unlike plane mirrors that produce virtual images (images that appear to be behind the mirror and cannot be projected onto a screen), concave mirrors can produce real images. Real images occur because the reflected rays actually converge at a point in front of the mirror. This means if a screen is placed at the focal point, the image can be projected and seen directly. This principle is essential in devices like telescopes, which gather light from distant objects and focus it to form clear images.

For example, consider a concave mirror with a radius of curvature of 3.4 meters pointed at the sun. Since sunlight arrives as parallel rays, these rays will focus at the mirror’s focal point. Using the formula for focal length, the distance from the mirror to the focal point is:

\[f = \frac{3.4\, \text{m}}{2} = 1.7\, \text{m}\]

This means all the sunlight will converge 1.7 meters from the mirror’s surface. This focusing effect is the basis for solar power plants, where concentrated sunlight heats water to generate energy.

Problem

You determine a concave mirror to have focal length $f equals 10$ . If you submerge the mirror in water ( $n equals 1.33$ ), what would be the new focal length?

$10 c m$

$11.3 c m$

$20 c m$

Cannot be determined

Problem

A concave makeup mirror has a focal length of $12.0 c m$ . What is the mirror’s radius of curvature?

$6.0 c m$

$12.0 c m$

$24.0 c m$

$48.0 c m$

Concept

Ray Diagrams for Concave Mirrors

Video duration:

Ray Diagrams for Concave Mirrors Video Summary

Understanding how images form with spherical concave mirrors involves analyzing the behavior of light rays reflecting off the mirror. When dealing with extended objects that are not far away, such as a pencil, the position and nature of the image can be determined using a principal ray diagram. This diagram is constructed by drawing at least two of three special rays from the tip of the object to the mirror and observing where the reflected rays intersect, as the image forms at this intersection point.

The three principal rays used in these diagrams are:

1. The Parallel Ray (P-ray): This ray travels parallel to the principal axis from the object’s tip toward the mirror. After reflection, it passes through the focal point (F) of the mirror.

2. The Focal Ray (F-ray): This ray passes through the focal point before striking the mirror. Upon reflection, it travels parallel to the principal axis.

3. The Center of Curvature Ray (C-ray): This ray heads directly toward the center of curvature (C) of the mirror. Since it strikes the mirror at a right angle, it reflects back along the same path.

By drawing any two of these rays from the object’s tip and extending their reflections, the point where they intersect indicates the location of the image. Notably, the image formed by a concave mirror is often inverted relative to the object, which is shown by drawing the image arrow upside down from the principal axis.

To precisely calculate the image position, the mirror equation is used, relating the object distance ($d_o$), image distance ($d_i$), and focal length (\(f\() as follows:

\[\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}\]

Here, \)d_o\) is the distance from the mirror to the object, $d_i$ is the distance from the mirror to the image, and $f$ is the focal length of the mirror, which is the distance from the mirror to the focal point.

When applying this equation, it is crucial to assign correct signs to each distance based on their positions relative to the mirror and the direction of light rays. The object distance $d_o$ is positive if the object is on the same side as the incoming (incident) rays, which is typically the case. The focal length \(f\( is positive if the focal point lies on the same side as the reflected (outgoing) rays, which is true for concave mirrors.

For example, if an object is placed 20 cm in front of a concave mirror with a focal length of 5 cm, the equation becomes:

\[\frac{1}{20} + \frac{1}{d_i} = \frac{1}{5}\]

Solving for \)d_i\) involves rearranging the equation:

\[\frac{1}{d_i} = \frac{1}{5} - \frac{1}{20} = 0.15\]

Taking the reciprocal gives:

\[d_i = \frac{1}{0.15} = 6.67 \text{ cm}\]

This positive value indicates that the image forms 6.67 cm from the mirror on the same side as the reflected rays, consistent with the ray diagram where the image appears beyond the focal point. This method of combining ray diagrams with the mirror equation and sign conventions provides a comprehensive understanding of image formation by concave mirrors, enabling accurate predictions of image location, size, and orientation.

Problem

For the following ray diagram, which ray is drawn incorrectly?

P-ray

F-ray

C-ray

None of the above.

Problem

An object is placed $45.0 c m$ from a concave mirror, producing a real image $15.0 c m$ from the mirror. What is the focal length of the mirror?

$7.5 c m$

$30.0 c m$

$60.0 c m$

$11.25 c m$

Example

Ray Diagrams for Concave Mirrors Example 1

Video duration:

Ray Diagrams for Concave Mirrors Example 1 Video Summary

When working with concave mirrors, understanding the relationship between the radius of curvature, focal length, and image formation is essential. The radius of curvature (R) is the distance from the mirror's surface to its center of curvature, and it directly relates to the focal length (f) by the formula $f = \frac{R}{2}$. For example, if a concave mirror has a radius of curvature of 24 centimeters, its focal length will be 12 centimeters.

To determine the object distance ($d_o$) when the image distance (\(d_i\() and focal length are known, the mirror equation is used:

\[\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}\]

In this context, the image distance is given as 20 centimeters, and since the image is real and formed by a concave mirror, both the image distance and focal length are positive values. By substituting the known values into the equation, we isolate \)d_o\):

\[\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i} = \frac{1}{12} - \frac{1}{20}\]

Calculating the right side yields approximately 0.0333, and taking the reciprocal gives the object distance:

\[d_o = 30 \text{ cm}\]

This means the object is located 30 centimeters from the mirror. Visualizing this with a ray diagram confirms that the object lies beyond the focal point but closer than the center of curvature, consistent with the formation of a real, inverted image between the focal point and center of curvature.

Understanding these relationships and sign conventions is crucial for solving mirror problems accurately. Positive values for focal length and image distance indicate real focal points and images on the same side as the reflected rays in concave mirrors. This approach not only aids in calculating distances but also helps in predicting image characteristics such as orientation and size.

Concept

Magnification in Geometric Optics

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Magnification in Geometric Optics Video Summary

In geometric optics, understanding how concave mirrors affect the size and orientation of an image is essential. The magnification of an image describes how much larger or smaller the image appears compared to the original object. This concept applies not only to mirrors but also to lenses. Magnification is defined as the ratio of the image height ($h_i$) to the object height (\(h_o\(), expressed as:

\[ M = \frac{h_i}{h_o} \]

Alternatively, magnification can be related to the distances of the image (\)d_i\)) and the object (\(d_o\() from the mirror or lens, using the formula:

\[ M = -\frac{d_i}{d_o} \]

The negative sign indicates the orientation of the image. A positive magnification means the image is upright, while a negative magnification means the image is inverted. Additionally, the absolute value of magnification reveals the size relationship: if \)|M| < 1\), the image is reduced; if $|M| > 1$, the image is enlarged.

For example, consider an object 4 cm tall placed 15 cm from a concave mirror with a focal length of 5 cm. If the image forms 7.5 cm in front of the mirror, the magnification can be calculated using the distances:

\[ M = -\frac{d_i}{d_o} = -\frac{7.5}{15} = -0.5 \]

This result indicates the image is inverted and half the size of the object. To find the height of the image, multiply the magnification by the object height:

\[ h_i = M \times h_o = -0.5 \times 4\, \text{cm} = -2\, \text{cm} \]

The negative height confirms the image is inverted, and the magnitude shows it is 2 cm tall, consistent with the calculated magnification. This approach allows for quick determination of image properties without needing to draw detailed ray diagrams, making it a powerful tool in optics problems.

Problem

A person stands $20.0 m$ in front of a large spherical mirror. A real image is formed $80.0 m$ in front of the mirror. What is the magnification of this image?

$4.0$

$0.25$

$negative 4.0$

$negative 0.25$

Example

Magnification in Geometric Optics Example 2

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Magnification in Geometric Optics Example 2 Video Summary

When observing distant objects like the planet Jupiter using a concave mirror telescope, understanding how the image size forms is essential. Jupiter, with a diameter of approximately $1.43 \times 10^{8}$ meters, is located about $5.88 \times 10^{11}$ meters away from Earth at its closest approach. Due to this vast distance, the light rays from Jupiter reaching the telescope are nearly parallel, allowing the concave mirror to focus these rays at its focal point.

In geometric optics, the image formed by a concave mirror of an object at a very large distance (effectively at infinity) appears near the mirror’s focal length, $f$. For this telescope, the focal length is given as 1.5 meters. Since the object distance $d_o$ (distance from Jupiter to the mirror) is extremely large compared to the focal length, the image distance \(d_i\( can be approximated as equal to the focal length:

\[d_i \approx f = 1.5 \text{ m}\]

The magnification \)M\) of the mirror relates the image height $h_i$ to the object height \(h_o\( and is also connected to the object and image distances by the formula:

\[M = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\]

Given the diameter of Jupiter as the object height \)h_o = 1.43 \times 10^{8}\) m and the object distance $d_o = 5.88 \times 10^{11}$ m, the image height $h_i$ can be calculated by substituting $d_i$ with the focal length:

\[h_i = M \times h_o = \left(-\frac{d_i}{d_o}\right) \times h_o = -\frac{1.5}{5.88 \times 10^{11}} \times 1.43 \times 10^{8}\]

Performing this calculation yields an image diameter of approximately:

\[h_i \approx 3.65 \times 10^{-4} \text{ m} = 0.365 \text{ mm}\]

This result shows that despite Jupiter’s massive size, its image formed by a simple concave mirror telescope is extremely small—less than half a millimeter in diameter—due to the immense distance between Earth and Jupiter. This highlights the importance of additional optical components, such as lenses, in telescopes to magnify and enlarge the image for better observation.

In summary, the size of the image formed by a concave mirror telescope depends on the focal length of the mirror and the ratio of the object’s size to its distance. When the object is very far away, the image distance approximates the focal length, simplifying calculations and allowing us to determine the tiny scale of the image produced by the telescope.

Example

Magnification in Geometric Optics Example 3

Video duration:

Magnification in Geometric Optics Example 3 Video Summary

When working with concave mirrors, understanding the relationship between object distance, image distance, magnification, and focal length is essential. In this example, a concave mirror forms a real, inverted image that is twice as tall as the object, which is positioned 24 centimeters from the mirror. To determine the focal length of the mirror, we start by analyzing the magnification and mirror equations.

The magnification (M) of a mirror is defined as the ratio of the image height to the object height, and it also relates to the distances by the formula:

\[ M = -\frac{d_i}{d_o} \]

Here, $d_o$ is the object distance, and $d_i$ is the image distance. Since the image is real and inverted, the magnification is negative, so $M = -2$. Given that the object distance $d_o$ is 24 cm, we can solve for the image distance \(d_i\(:

\[ -2 = -\frac{d_i}{24} \implies d_i = 48 \text{ cm} \]

With both distances known, the mirror equation allows us to find the focal length \)f\):

\[ \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f} \]

Substituting the values:

\[ \frac{1}{24} + \frac{1}{48} = \frac{1}{f} \]

Calculating the left side:

\[ \frac{2}{48} + \frac{1}{48} = \frac{3}{48} = \frac{1}{16} \]

Therefore, the focal length is:

\[ f = 16 \text{ cm} \]

This example highlights how the magnification and object distance can be used together to find the focal length of a concave mirror. The negative sign in magnification indicates the image inversion typical of real images formed by concave mirrors. Understanding these relationships is crucial for solving problems involving spherical mirrors and their image formation properties.

Concept

Ray Diagrams for Various Distances in Concave Mirrors

Video duration:

Ray Diagrams for Various Distances in Concave Mirrors Video Summary

When analyzing concave mirrors, the position of the object relative to key points such as the center of curvature (C) and the focal point (F) significantly influences the characteristics of the image formed. The center of curvature is located at a distance of twice the focal length, expressed as $C = 2f$, where $f$ is the focal length. Understanding how the image properties change with object placement is essential for mastering ray diagrams and predicting image behavior.

For an object placed beyond the center of curvature (object distance $d_o > 2f$), the reflected rays converge between the focal point and the center of curvature ($f < d_i < 2f$), producing a real, inverted, and reduced image. This means the image forms on the same side as the object but is smaller in size and upside down relative to the object.

When the object is positioned between the center of curvature and the focal point ($f < d_o < 2f$), the image forms beyond the center of curvature ($d_i > 2f$). In this case, the image remains real and inverted but becomes enlarged compared to the object. This relationship highlights a useful shortcut: if the object is closer to the mirror than the center of curvature but beyond the focal point, the image will be larger and located farther away than the center of curvature.

In contrast, if the object is placed closer to the mirror than the focal point (\(d_o < f\(), the reflected rays diverge, and the image cannot be formed by actual ray convergence. Instead, the rays appear to originate from a point behind the mirror, creating a virtual image. This virtual image is upright, enlarged, and located on the opposite side of the mirror from the object. This scenario is unique because the image is not formed by real ray intersections but by the apparent backward extension of diverging rays.

These principles can be summarized and applied using the mirror equation:

\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]

where \)d_o\) is the object distance, $d_i$ is the image distance, and $f$ is the focal length. The sign conventions for $d_i$ help distinguish between real and virtual images: a positive $d_i$ indicates a real image formed on the same side as the reflected rays converge, while a negative $d_i$ corresponds to a virtual image formed behind the mirror.

By mastering the use of principal rays—specifically the parallel ray (P), focal ray (F), and center of curvature ray (C)—one can accurately construct ray diagrams to visualize image formation. Typically, two of these three rays are sufficient to determine the image location and characteristics. The parallel ray travels parallel to the principal axis and reflects through the focal point; the focal ray passes through the focal point and reflects parallel to the principal axis; and the center of curvature ray passes through the center of curvature and reflects back on itself.

Understanding these relationships enhances problem-solving efficiency in optics, allowing for quick predictions of image type (real or virtual), orientation (inverted or upright), and size (enlarged or reduced) based on object placement relative to $f$ and $C$. This foundational knowledge is crucial for applications involving concave mirrors in fields such as physics, engineering, and optical design.

Problem

Draw the principal-ray diagram for an object placed exactly at the focal point of a concave mirror. Where does the image form?

Diagram of a concave mirror showing principal rays from an object at the focal point, with rays reflected parallel. ; at infinity

Diagram showing principal rays from an object at the focal point of a concave mirror, with reflected rays parallel and image at infinity. ; between the mirror and the focal point

Ray diagram showing an object at the focal point of a concave mirror with reflected rays and image formation. ; behind the mirror

Concave mirror with object at focal point, showing principal rays reflecting parallel and image forming at infinity. ;at infinity

Problem

An object is placed $10 c m$ in front of a concave mirror. If a real, enlarged image is formed even farther from the mirror than the object, which of the following is the most likely value for the focal length of the mirror?

$f equals 4$

$f equals 8$

$f equals 10$

$f equals 20$

Example

Ray Diagrams for Various Distances in Concave Mirrors Example 4

Video duration:

Ray Diagrams for Various Distances in Concave Mirrors Example 4 Video Summary

When working with concave mirrors, understanding how to determine the image location, size, and characteristics is essential. Consider a concave mirror with a focal length of 15 centimeters and an object placed 20 centimeters in front of it. To analyze this setup, begin by drawing a principal ray diagram. This diagram helps visualize the positions of the focal point (F) and the center of curvature (C), which are crucial for accurately locating the image.

Since the focal length (f) is 15 cm, the center of curvature (C) is twice that distance, at 30 cm. The object is positioned between F and C, at 20 cm from the mirror. By drawing two principal rays—one parallel to the optic axis that reflects through the focal point, and another passing through the focal point that reflects parallel to the optic axis—you can find the image where these rays intersect. This intersection occurs beyond C, indicating the image is real, inverted, and larger than the object.

To calculate the exact image distance (d_i), use the mirror equation:

\[\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}\]

Here, d_o is the object distance (20 cm), and f is the focal length (15 cm). Since the object is in front of the mirror, d_o is positive. The focal length for a concave mirror is also positive. Rearranging to solve for d_i:

\[\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{15} - \frac{1}{20} = \frac{4}{60} - \frac{3}{60} = \frac{1}{60}\]

Taking the reciprocal gives:

\[d_i = 60 \text{ cm}\]

This positive image distance confirms the image forms on the same side as the reflected rays, consistent with a real image. The image is located 60 cm from the mirror, which is three times farther than the object distance, indicating significant magnification.

To find the magnification (M), use the formula relating image and object distances:

\[M = -\frac{d_i}{d_o} = -\frac{60}{20} = -3\]

The negative sign indicates the image is inverted relative to the object, and the magnitude greater than 1 shows the image is enlarged. This aligns with the ray diagram observation that the image is upside down and larger than the object.

In summary, when an object is placed between the focal point and the center of curvature of a concave mirror, the resulting image is real, inverted, and magnified. The mirror equation and magnification formula provide precise quantitative descriptions of the image’s position and size, reinforcing the importance of combining graphical and mathematical methods in geometric optics.

Example

Ray Diagrams for Various Distances in Concave Mirrors Example 5

Video duration:

Ray Diagrams for Various Distances in Concave Mirrors Example 5 Video Summary

When analyzing object distances for concave mirrors, understanding the relationship between object distance, image distance, and focal length is crucial. Consider an object placed 8 centimeters in front of a concave mirror, which forms an image 24 centimeters away on the opposite side of the mirror. This detail indicates the image is virtual and upright, as concave mirrors produce virtual images on the opposite side only when the object is located between the focal point and the mirror.

To determine the focal length of the mirror, the mirror equation is applied:

\[\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}\]

Here, $d_o$ is the object distance, $d_i$ is the image distance, and $f$ is the focal length. The sign conventions are essential: distances measured in the direction of the incoming light (usually in front of the mirror) are positive, while those on the opposite side are negative. Since the object is in front of the mirror, $d_o = +8\, \text{cm}$. The image forms on the opposite side, so $d_i = -24\, \text{cm}$.

Substituting these values into the mirror equation gives:

\[\frac{1}{8} + \frac{1}{-24} = \frac{1}{f}\]

Calculating the left side:

\[\frac{1}{8} - \frac{1}{24} = \frac{3}{24} - \frac{1}{24} = \frac{2}{24} = \frac{1}{12}\]

Therefore, the focal length is:

\[f = 12\, \text{cm}\]

This positive focal length confirms the mirror is concave, with the focal point located 12 centimeters from the mirror on the same side as the object. Visualizing this setup through a ray diagram helps reinforce the concept: rays originating from the object reflect off the mirror and appear to diverge from a point behind the mirror, forming a virtual, upright, and magnified image.

Understanding these principles allows for accurate predictions of image characteristics based on object placement relative to the focal point, emphasizing the importance of sign conventions and the mirror equation in geometric optics.

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33. Geometric Optics - Part 1 of 2

5 topics 9 problems

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33. Geometric Optics - Part 2 of 2

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Here's what students ask on this topic:

In a concave mirror, the focal length ( $f$ ) is directly related to the radius of curvature ( $R$ ) by the equation $f = \frac{R}{2}$ . This means the focal length is exactly half the radius of curvature. The radius of curvature is the distance from the mirror's vertex to its center of curvature, while the focal length is the distance from the vertex to the focal point where parallel rays converge after reflection. This relationship is fundamental in understanding how concave mirrors focus light and form images.

To locate the image formed by a concave mirror, you draw a principal ray diagram using any two of the three key rays: the P-ray (parallel to the axis, then reflects through the focal point), the F-ray (passes through the focal point, then reflects parallel to the axis), and the C-ray (passes through the center of curvature and reflects back on itself). The image forms at the point where the reflected rays intersect. If the rays actually converge, the image is real; if they only appear to diverge from a point behind the mirror, the image is virtual. This method visually shows the image's position, size, and orientation relative to the object.

The characteristics of images formed by concave mirrors depend on the object's distance relative to the focal point ( $F$ ) and center of curvature ( $C$ ):

When the object is beyond $C$ , the image is real, inverted, and reduced, forming between $F$ and $C$ .
If the object is between $C$ and $F$ , the image is real, inverted, and enlarged, forming beyond $C$ .
When the object is closer than $F$ , the image is virtual, upright, and enlarged, appearing behind the mirror.

At exactly the focal point, the image forms at infinity.

The image distance ( $d_I$ ) in a concave mirror is calculated using the mirror equation: $\frac{1}{d_O} + \frac{1}{d_I} = \frac{1}{f}$ , where $d_O$ is the object distance and $f$ is the focal length. To find $d_I$ , rearrange the equation: $\frac{1}{d_I} = \frac{1}{f} - \frac{1}{d_O}$ . Then take the reciprocal to get $d_I$ . Sign conventions are important: distances are positive if measured in the direction of incoming light (usually in front of the mirror) and negative otherwise. This equation helps precisely determine where the image forms.

Magnification ( $m$ ) describes how the size and orientation of the image compare to the object. It is given by $m = \frac{h_I}{h_O} = - \frac{d_I}{d_O}$ , where $h_I$ and $h_O$ are the heights of the image and object, and $d_I$ and $d_O$ are their distances from the mirror. A negative magnification means the image is inverted, while a positive magnification means it is upright. The absolute value indicates size: if | $m$ | < 1, the image is reduced; if | $m$ | > 1, it is enlarged. This helps predict image properties without drawing diagrams.

Real images are formed when reflected rays actually converge at a point; they can be projected onto a screen and are typically inverted. In concave mirrors, real images form when the object is beyond the focal point. Virtual images occur when reflected rays diverge but appear to come from a point behind the mirror; these images cannot be projected and are upright. Virtual images form when the object is placed between the mirror and the focal point. Understanding this distinction is crucial for applications like telescopes and solar concentrators.

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Concave Mirrors: Videos & Practice Problems

Intro to Spherical Mirrors

Intro to Spherical Mirrors Video Summary

You determine a concave mirror to have focal length $f equals 10$ . If you submerge the mirror in water ( $n equals 1.33$ ), what would be the new focal length?

A concave makeup mirror has a focal length of $12.0 c m$ . What is the mirror’s radius of curvature?

Ray Diagrams for Concave Mirrors

Ray Diagrams for Concave Mirrors Video Summary

For the following ray diagram, which ray is drawn incorrectly?

An object is placed $45.0 c m$ from a concave mirror, producing a real image $15.0 c m$ from the mirror. What is the focal length of the mirror?

Ray Diagrams for Concave Mirrors Example 1

Ray Diagrams for Concave Mirrors Example 1 Video Summary

Magnification in Geometric Optics

Magnification in Geometric Optics Video Summary

A person stands $20.0 m$ in front of a large spherical mirror. A real image is formed $80.0 m$ in front of the mirror. What is the magnification of this image?

Magnification in Geometric Optics Example 2

Magnification in Geometric Optics Example 2 Video Summary

Magnification in Geometric Optics Example 3

Magnification in Geometric Optics Example 3 Video Summary

Ray Diagrams for Various Distances in Concave Mirrors

Ray Diagrams for Various Distances in Concave Mirrors Video Summary

Draw the principal-ray diagram for an object placed exactly at the focal point of a concave mirror. Where does the image form?

An object is placed $10 c m$ in front of a concave mirror. If a real, enlarged image is formed even farther from the mirror than the object, which of the following is the most likely value for the focal length of the mirror?

Ray Diagrams for Various Distances in Concave Mirrors Example 4

Ray Diagrams for Various Distances in Concave Mirrors Example 4 Video Summary

Ray Diagrams for Various Distances in Concave Mirrors Example 5

Ray Diagrams for Various Distances in Concave Mirrors Example 5 Video Summary

Do you want more practice?

Here's what students ask on this topic:

What is the relationship between the focal length and the radius of curvature in a concave mirror?

How do you use principal ray diagrams to locate the image formed by a concave mirror?

What are the characteristics of images formed by concave mirrors depending on the object's position?

How do you calculate the image distance using the mirror equation for a concave mirror?

What does the magnification of a concave mirror tell us about the image?

What is the difference between real and virtual images formed by concave mirrors?

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Concave Mirrors: Videos & Practice Problems

Intro to Spherical Mirrors

Intro to Spherical Mirrors Video Summary

You determine a concave mirror to have focal length f=10cmf=10\(\operatorname{\mathrm{cm}\)}. If you submerge the mirror in water (n=1.33n=1.33), what would be the new focal length?

A concave makeup mirror has a focal length of 12.0cm12.0\(\operatorname{\mathrm{cm}\)}. What is the mirror’s radius of curvature?

Ray Diagrams for Concave Mirrors

Ray Diagrams for Concave Mirrors Video Summary

For the following ray diagram, which ray is drawn incorrectly?

An object is placed 45.0cm45.0\(\operatorname{\mathrm{cm}\)} from a concave mirror, producing a real image 15.0cm15.0\(\operatorname{\mathrm{cm}\)} from the mirror. What is the focal length of the mirror?

Ray Diagrams for Concave Mirrors Example 1

Ray Diagrams for Concave Mirrors Example 1 Video Summary

Magnification in Geometric Optics

Magnification in Geometric Optics Video Summary

A person stands 20.0m20.0\(\operatorname{\mathrm{m}\)} in front of a large spherical mirror. A real image is formed 80.0m80.0\(\operatorname{\mathrm{m}\)} in front of the mirror. What is the magnification of this image?

Magnification in Geometric Optics Example 2

Magnification in Geometric Optics Example 2 Video Summary

Magnification in Geometric Optics Example 3

Magnification in Geometric Optics Example 3 Video Summary

Ray Diagrams for Various Distances in Concave Mirrors

Ray Diagrams for Various Distances in Concave Mirrors Video Summary

Draw the principal-ray diagram for an object placed exactly at the focal point of a concave mirror. Where does the image form?

An object is placed 10cm10\(\operatorname{\mathrm{cm}\)} in front of a concave mirror. If a real, enlarged image is formed even farther from the mirror than the object, which of the following is the most likely value for the focal length of the mirror?

Ray Diagrams for Various Distances in Concave Mirrors Example 4

Ray Diagrams for Various Distances in Concave Mirrors Example 4 Video Summary

Ray Diagrams for Various Distances in Concave Mirrors Example 5

Ray Diagrams for Various Distances in Concave Mirrors Example 5 Video Summary

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Here's what students ask on this topic:

What is the relationship between the focal length and the radius of curvature in a concave mirror?

What is the relationship between the focal length and the radius of curvature in a concave mirror?

How do you use principal ray diagrams to locate the image formed by a concave mirror?

How do you use principal ray diagrams to locate the image formed by a concave mirror?

What are the characteristics of images formed by concave mirrors depending on the object's position?

What are the characteristics of images formed by concave mirrors depending on the object's position?

How do you calculate the image distance using the mirror equation for a concave mirror?

How do you calculate the image distance using the mirror equation for a concave mirror?

What does the magnification of a concave mirror tell us about the image?

What does the magnification of a concave mirror tell us about the image?

What is the difference between real and virtual images formed by concave mirrors?

What is the difference between real and virtual images formed by concave mirrors?

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You determine a concave mirror to have focal length $f equals 10$ . If you submerge the mirror in water ( $n equals 1.33$ ), what would be the new focal length?

A concave makeup mirror has a focal length of $12.0 c m$ . What is the mirror’s radius of curvature?

An object is placed $45.0 c m$ from a concave mirror, producing a real image $15.0 c m$ from the mirror. What is the focal length of the mirror?

A person stands $20.0 m$ in front of a large spherical mirror. A real image is formed $80.0 m$ in front of the mirror. What is the magnification of this image?

An object is placed $10 c m$ in front of a concave mirror. If a real, enlarged image is formed even farther from the mirror than the object, which of the following is the most likely value for the focal length of the mirror?