[ #1 | #2 | #3 | #4 | #5 | #6 | #7 | #8 | #9 | #10 | #11 | #12 | #13 | #14 | #15 | #16 | #17 | #18 | #19 | #20 | #21 | #22 | #23 | #24 | #25 | #26 | #27 | #28 | #29 | #30 | #31 | #32 | #33 | #34 | #35 | #36 | #37 | #38 | #39 | #40 | #41 | #42 ]

a. zero

b. 45 degrees

c. 90 degrees

d. 180 degrees

Answer: D

The angle of incidence is the angle between the incident ray and the normal. As this angle approaches 90 degrees, the reflected ray also approaches a 90 degree angle with the normal; thus, the angle between the incident and reflected ray approach 180 degrees.

a. 1.5 ft

b. 3.0 ft

c. 6.0 ft

d. 12.0 ft

Answer: C

If you stand 3 feet from the mirror, then your image is three feet on the other side of the mirror; this puts your image a total of six feet from you (3 feet to the mirror plus 3 more feet to the image).

a. 20.0 cm

b. 15.0 cm

c. 7.5 cm

d. 5.0 cm

Answer: B

Use the mirror equation:

where d_i = +30 cm and f = +10 cm.

Substitute and solve for d_o.

a. 1.4 cm

b. 16.7 cm

c. 12.4 cm

d. 7.1 cm

Answer: 7.1 cm

Use the mirror equation:

where d_i = +25 cm and d_o = +10 cm.

Substitute and solve for f.

a. 30.0 cm

b. 10.0 cm

c. 12.4 cm

d. 6.0 cm

Answer: D

Use the mirror equation:

where d_i = -10 cm and f = +15 cm. (Note that d_i is negative if the image is virtual.)

Substitute and solve for d_o.

a. 30.0 cm

b. 10.0 cm

c. 6.0 cm

d. 3.0 cm

Answer: A

Use the mirror equation:

where d_i = -10 cm and f = -15 cm. (Note that d_i is negative if the image is virtual.)

Substitute and solve for d_o.

a. 12.0 cm

b. 20.0 cm

c. 70.0 cm

d. 90.0 cm

Answer: D

First find d_i.

If the d_o = 30 cm and the M = +1.5 (Note: M is + since the image is upright), then use M = -d_i/d_o to find that d_i = -45 cm (the - d_i value is consistent with the image being virtual - located behind the mirror).

Now use the mirror equation to find the focal length:

where d_i = -45 cm and d_o = +30 cm.

Substitute and solve for f.

Answer: C

When you look at your image in a plane mirror, you see an upright image; it is located on the other side of the mirror (and thus is virtual); finally, it has the same dimensions (height, width) as yourself (the object).

a. virtual, upright and enlarged

b. real, inverted and reduced

c. virtual, upright and reduced

d. real, inverted and enlarged

Answer: D

For concave mirrors, when the object is located anywhere between F and C, the image is real, inverted, enlarged in size, and located beyond C. You should get this very result if you were to draw a ray diagram.

a. virtual, erect and enlarged

b. real, inverted and reduced

c. virtual, upright and reduced

d. real, inverted and enlarged

Answer: B

For concave mirrors, when the object is located anywhere beyond C, the image is real, inverted, reduced in size, and located between F and C. You should get this very result if you were to draw a ray diagram.

a. virtual, upright and enlarged

b. real, inverted and reduced

c. virtual, upright and reduced

Answer: A

For concave mirrors, when the object is located anywhere inside the F, the image is virtual, upright, enlarged in size, and located on the opposite side of the mirror. You should get this very result if you were to draw a ray diagram.

a. virtual, upright and enlarged

b. real, inverted and reduced

c. virtual, upright and reduced

d. real, inverted and enlarged

Answer: C

For convex mirrors, regardless of where the object is located, the image is virtual, upright, reduced in size, and located on the opposite side of the mirror. You should get this very result if you were to draw a ray diagram.

a. spherical aberration

b. mirages

c. chromatic aberration

d. light scattering

Answer: A

The problem of spherical aberration are caused by the inability of the outer portions of a concave spherical mirror to reflect and focus light to the same image locations as other portions of the mirror. A cure for the problem involves the use of a mirror with a different shape in the regions far from the principal axis; a parabolic mirror solves this problem.

a. real and upright.

b. real and inverted.

c. virtual and upright.

d. virtual and inverted.

Answer: C

Look at yourself in a plane mirror and you see your image - it is upright. The image is located on the other side of the mirror since reflected rays diverge upon reflection; when mirrors produce images on the the opposite side of the mirror, the images are said to be virtual.

Answer: B

For plane mirrors, the image distance is the same as the object distance (d_i =-d_o). The only way to modify the image distance is to modify the object distance.

Answer: B

The portion of mirror required to view the full image of an object is always one-half the height of the object (for plane mirrors only). For the skeptics who believe C is the more believable answer, see the proof given on the Physics Classroom tutorial page.

a. always be real.

b. always be virtual.

c. be either real or virtual.

d. will always be magnified.

Answer: C

Concave mirrors can produce real images (if the object is beyond the focal point) and virtual images (if the object is located in front of the focal point). These images can be either magnified in size (if object is in front of F or between C and F), reduced in size (if object is located beyond C),or the same size as the object (if object is located at C).

a. always be real.

b. always be virtual.

c. may be either real or virtual.

d. will always be magnified.

Answer: B

Convex mirrors always (always) produce images which are virtual, upright, and reduced in size. Since reflected light rays always diverge from each other, these convex mirrors can never produce real images. Draw a ray diagram to convince yourself that this is true.

a. at the center of curvature.

b. at the focal point.

c. at infinity.

d. at a point half way to the focal point.

Answer: B

The focal point is the location where light rays traveling parallel to the principal axis converge. (Perhaps you remember the demo with the concave (converging) mirror and the pencil or dollar bill.)

Answer: ABEG

A is true; like any plane mirror image, the image formed by two mirrors oriented at right angles is the same size as the object.

B is true; the middle image is the same distance from the crease of the dual mirror system as the image is from the crease. However, image distances are usually measured as the smallest distance measured perpendicularly to the mirror. For middle images, this distance would be measured to an extended mirror line and it would be the same distance as that from the object to that same mirror.

C is false; the middle image, unlike the primary images, does not exhibit this feature of left-right reversal.

D is false; the middle image is a virtual image; it is upright and located behind the mirror (like any virtual image).

E is true; just look at the middle image and you will see an upright image.

F is false; a magnification of -1 means that the image would be inverted (negative M) and the same size as the object (1); the middle image is the same size as the object but it is not inverted.

G is true; light will reflect twice - once off each mirror - prior to traveling to the viewer's eye.

a. increase

b. decreases

c. remains the same

Answer: B

Consider two mirrors arranged at a 0-degree angle - parallel to each other. There would be an infinite number of images, one located directly behind the other forming a line which seems to extend forever. Suppose the mirrors are adjusted to a 30-degree angle (see question #29 below). The mirrors would create 5 images of objects placed between the mirrors. Now suppose that the angle is increased to a 90-degree angle (see question #29 below). Finally, suppose the mirrors are adjusted to a 180-degree angle to form a single plane. At such an angle, only one image would be formed. As the angle is increased, the number of images decreases.

Answer: ACDF

A is true; virtual images are always located behind the mirror.

B is false; virtual images are always upright.

C is true; a virtual image is reduced in size if formed by a convex mirror; it is magnified in size when created by concave mirrors; it is the same size as the object when created by plane mirrors.

D is true; convex and plane mirrors always form virtual images; concave mirrors will form virtual images of objects placed between the focal point and the mirror.

E is false; while virtual images are not real, they can be seen if sighted at in the mirror - just consider your plane mirror image which you probably view everyday.

F is true. Virtual images result when incident light from the object strikes the mirror, reflects and diverges. The virtual image is located behind the mirror at the spot where the reflected rays would converge if extended behind the mirror.

G is false; only real images which are formed in front of mirrors can be projected onto the screen. Virtual images are located behind the mirror where the light does not travel.

Answer: CG

A is false; real images are always located in front of the mirror.

B is false; real images are always inverted and never upright.

C is true; a real image can be magnified if placed between C and F of a concave mirror; it can be reduced in size if placed beyond C in front of a concave mirror; and it can be the same size as the object if placed at C in front of a concave mirror.

D is false; convex and plane mirrors always form virtual images and can never form real images; only a concave mirror can form both types of images.

E is false; real images can be seen when sighting at the image location (just like virtual images can).

F is false; real images result when reflected light rays converge to a point.

G is true; real images (unlike virtual images) can be projected onto a sheet of paper. Because real images form in front of the mirror at the locations where reflected rays converge, the image can be projected onto a sheet of paper if it is placed at this image location.

Answer: AEF

A results in a magnified image. If the image distance is greater than 0, then the image is a real image. If the image distance is greater than the object distance, then according to the magnification ratio, the image height must also be equal to the object height.

In B and C, a convex and a plane mirror never creates a magnified image. Convex mirrors always create images which are reduced in size. Plane mirrors always create images which are the same size as the object.

In D, the object is placed beyond the center of curvature of a concave mirror. In such an instance, the image is reduced in size and located between C and F.

In E, the object is placed between the focal point and a concave mirror. At such an object location, the image is virtual, magnified, and upright.

In F, the object is placed in between points C and F of a concave mirror. For such an object location, the image is real, magnified, and located beyond the center of curvature.

Answer: BCE

In A, the image is real and its distance is further from the mirror than the object. This could only occur when the object is placed between C and F of a concave mirror. For such an object location, the image is real, inverted, and magnified in size.

In B, a convex mirror is used. Convex mirrors always produce upright, virtual images.

In C, a plane mirror is used. Plane mirrors always produce upright, virtual images.

In D, the object is located beyond the center of curvature of a concave mirror. At such an object location, the image would be real, inverted and reduced in size.

In E, the object is located in front of the focal point of a concave mirror. At such an object location, the image would be virtual, upright, and magnified.

In F, the object is located between the center of curvature and the focal point of a concave mirror; the image would be real, inverted, and magnified.

a. TRUE

b. FALSE

Answer: A

This is a true statement. The image characteristics of being upright and being virtual go hand in hand. If the image is virtual, then it is upright; and vice versa. Only real images are inverted and only inverted images are real.

Answer and Explanation:

The path of the reflected ray at each boundary can be determined by sketching a normal at the point where the incident ray strikes the mirror surface. Once the normal is constructed the angle of incidence can be measured. For instance, the black dashed line at the location where ray 1 strikes the mirror is the normal line. The angle between ray 1 and the normal is approximately 62 degrees. The reflected ray should be drawn at the same angle from the normal - 62 degrees - but on the opposite side of the normal line. Ray 2 is thus drawn at a 62 degrees from the dashed normal line. Ray 2 is extended towards the second mirror. The process of measuring the angle of incidence and drawing the reflected ray (ray 3) is repeated.

Answer and Explanation:

The images of the arrow objects can be found by using the principle that the image distance equals the object distance. The two extremities of the arrows can be used to find the two extremities of the image. The entire image will stretch from the image extremities.

The eye will be able to see any ray of light which bounces off the edges of the mirror or between the edges. The region shaded in yellow represents images within the visual field of this particular eye. When the yellow region is extended behind the mirror, it becomes obvious that the eye is unable to see the leftmost image and can see only a portion of the image located second from the left.

Answer and Explanation:

The images have been shown in pink above. The images and mirrors have been labeled to facilitate the discussion below.

For the right angle mirror, there are three images; two are primary images and the third is a secondary image. The primary images I₁ and I₂ are simple an image resulting from the reflection of light off of mirrors M₁ and M₂. When viewing these images, light from the object will reflect off one of these mirrors and travel towards the viewer's eye. The secondary image I₃ is an image of an image. When viewing the secondary image, light must first reflect off one of the mirrors and then the second mirror before traveling towards the viewer's eye. Elaborate step-by-step directions for locating these images are found elsewhere; see the Physics Classroom tutorial page.

As the angle between mirrors is decreased as shown on the right, more images are formed. Two of these images are primary images - I₁ and I₂; the other images are images of images (I₃ and I₄) or images of images of images (I₅). Image I₁ is the image created by mirror M₁ and image I₂ is the image created by mirror M₂. When viewing these images, light from the object will reflect off the respective mirror and travel towards the viewer's eye. I₃ is the image of an image - the image of I₂ as reflected across mirror M₁. A quick measurement will reveal that the distance from I₂ to mirror M₁ is equal to the distance from image I₃ to mirror M₁. I₄ is the image of an image - the image of I₁ as reflected across mirror M₂. A quick measurement will reveal that the distance from I₁ to mirror M₂ is equal to the distance from image I₃ to mirror M₂. When viewing the secondary images I₃ and I₄, light will reflect off both mirrors before traveling to the viewer's eye. Image I₅ is the image of an image of an image. A quick measurement will review that the distance from I₃ to mirror M₂ is equal to the distance from I₅ to mirror M₂. Similarly, the distance from I₄ to mirror M₁ is equal to the distance from I₅ to mirror M₁. When viewing this image, light will reflect off one of the mirrors twice and the other mirror once before traveling to the viewer's eye.

30. Consider the following object in front of a plane mirror. Construct a ray diagram to show how light travels from the object (arrow) to the mirror and ultimately to the eye; then indicate the portion of the mirror needed in order for the eye to view the image.

Answer and Explanation:

The image of the arrow object can be found by using the principle that the image distance equals the object distance. The two extremities of the arrow can be used to find the two extremities of the image. The entire image will stretch from the image extremities. This is shown in blue.

For the eye to see this image, it must sight along a line at the top of the image and the bottom of the image. As the eye sights along this line, a ray of light comes to the eye along this line of sight. This ray is the reflected ray. The light originates at the object and strikes the mirror at the location where the line of sight intersects the mirror. The light will automatically follow the law of reflection at the mirror (if the image is accurately located and the lines of sight are accurately drawn). The path of light is shown in red; note that dashed lines are used as extensions of the original line of sight extending behind the mirror.

Answer and Explanation:

The primary and secondary images are drawn in a shade of pink. Once located, the ray diagram can be constructed. The principle that "to see an image, the eye must sight along a line at the image" is utilized. The set of green and blue lines represent the path of the light from the object to the eye. As the eye sights at the top of the middle image, light travels along the line of sight marked by the blue line and extended backwards to the top of the middle image. The light bouncing off the vertical mirror must first come from the horizontal mirror. The ray coming from the horizontal mirror to the vertical mirror is determined using the law of reflection. A normal line is drawn (in black) and the angle of reflection is measured; this angle is then used to determine the angle of incidence at the vertical mirror's surface. A ray is then traced backwards at this angle to the horizontal mirror. The light originated from the top of the object; thus, a ray is drawn from the top of the object to the horizontal mirror. These light rays for viewing the top of the image are drawn in blue in the diagram above. The process is repeated for viewing the bottom of the image; these rays are drawn in green.

Complete step-by-step instructions for right angle mirror ray diagrams are described elsewhere. See the Physics Classroom tutorial page.

Answer:

The images are drawn in green. The light rays are drawn in red or orange. The image descriptions are stated in green below each ray diagram. Elaborate step-by-step directions for constructing ray diagrams are provided on a separate page - see the links below. The plane mirror image is found by drawing any two (or more) rays and reflecting them according to the law of reflection. The reflected rays will always intersect at an image point.