PHYS 1220 Images formed by Mirrors Solutions

Questions: 1 2 3 4 5 6 7 8

Images formed by Mirrors Solutions

Determine the minimum height of a wall mirror that will permit a 1.8-metre tall person to view his or her entire height. Sketch rays from the top and bottom of the person, and determine the proper placement of the mirror such that the full image is seen, regardless of the person's distance from the mirror.
To see the top of your head, a light ray must leave your head and be reflected in the mirror. Similarly to see your foot, a light ray must leave your foot and be reflected in the mirror. The reflected rays obey the Law of Reflection, θ_incident = θ_reflected.

Examining the diagram, we see that the required portion of the mirror is about one-half the person's height or .9 m .
Two plane mirrors are inclined to one another at an angle α. A ray travelling 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. Express your answer in terms of α.
Light rays obey the Law of Reflection when they encounter plane mirrors. The path of the ray is sketched below.

Using geometry

α + β + η = π (1)

γ + β = ½π (2)

η + φ = ½π (3)

θ = π - 2γ - 2φ (4)

Using the first three equations to eliminate γ and φ from equation (4), we find

θ = π - 2α
In the diagram below are two plane mirrors set at an angle of 60° to one another. A point P is on the bisector of the angle between the two mirrors. Find all the possible images.

Images in plane mirrors are always the same distance behind a mirror as the object is in front of the mirror.

For the object at P there will be a primary image formed in each mirror A and B. These images are labelled P_1A and P_1B in the diagram. The subscript indicate that these are the primary images and which mirror you have to be looking at to see the image.

Next an image in one mirror will act as an object for the second mirror as long as the image is in front of the plane of the second mirror. So we will get a set of secondary images. The image of P_1A is labelled P_2B in the diagram. The image of P_1B is labelled P_2A in the diagram.

The images labelled P_2A and P_2B are still in front of the plane of the other mirror. So we will get a set of tertiary images. The image of P_2A is labelled P_3B in the diagram. The image of P_2B is labelled P_3A in the diagram.

There are no images P_3A and P_3B as P_3A is behind the plane of mirror B and P_3B is behind the plane of mirror A.

We thus have a total of 6 images.
Three mirrors are arranged as in the diagram below. Find all the images of point P.

Images in plane mirrors are always the same distance behind a mirror as the object is in front of the mirror.

For the object at P there will be a primary image formed in each mirror A, B, and C. These images are labelled P_1A, P_1B, and P_1C in the diagram. The subscript indicate that these are the primary images and which mirror you have to be looking at to see the image.

Next an image in one mirror will act as an object for the second mirror as long as the image is in front of the plane of the second mirror. P_1B is on the plane of mirrors A and C, so it will not cause any further reflections. Image P_1A is in front of the plane of mirror C, we get a secondary image P_2C. Image P_1C is in front of the plane of mirror A, we get a secondary image P_2A.
Image P_2A is in front of the plane of mirror C, we get a tertiary image P_3C. Image P_2C is in front of the plane of mirror A, we get a tertiary image P_3A. The placement of the mirrors is such that the image locations of P_3A and P_3C coincide.
The tertiary images are behind the plane of the mirrors, so there will be no further reflections.

There is a total of five images.
Use the Principle Ray Technique to find the image, created by a concave spherical mirror, of an object placed (a) between C and F, (b) at C, and (c) between F and the mirror. In each case, characterize the image, if possible.

Image is real, inverted, approximately twice as large, past C.

Image is real, inverted, same size, at C.

Image is virtual, erect, approximately 30% larger, behind mirror.
When people stand in front of a type of mirror found in amusement parks, they see themselves with small heads and large lower torsos. Explain how this is accomplished.

Funhouse mirrors are two mirrors, of different focal lengths, joined together. In each case the image must be erect or upright which means that the image must be virtual. The top mirror must shrink an object, i.e. M_up < 1. The bottom mirror must enlarge the object, M_bottom < 1.

This can be achieved with an S-shaped mirror.
An object is placed 25 cm in front of a concave mirror of focal length 30 cm. Calculate the image distance and the magnification. Characterize the image.

Since the mirror is concave, f = +30. Since the object is in front of the object, o = +25. Using the lens formula
1/o + 1/i = 1/f .
Isolating 1/i,

1/i = 1/f - 1/o = (o-f) / fo .
Inverting, we find
i = fo / (o-f) = (30)(25)/(25 - 30) = -150 cm .
Since i < 0, the image is behind the mirror. As a result it must be virtual, since no rays can come from behind the mirror.

The magnification is given by

M = -i/o = -f / (o-f) = -30 / (25 - 30) = +6 .
The image is six times larger than the object. The image is erect since M is positive.
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.

Since the mirror is convex, f = -30. Since the object is in front of the object, o = +25. Using the lens formula
1/o + 1/i = 1/f .
Isolating 1/i,

1/i = 1/f - 1/o = (o-f) / fo .
Inverting, we find

i = fo / (o-f) = (-30)(25)/[25 - (-30)] = -150/11 cm = -13.64 cm .
Since i < 0, the image is behind the mirror. As a result it must be virtual, since no rays can come from behind the mirror.

The magnification is given by

M = -i/o = -f / (o-f) = -(-30) / [25 - (-30)] = +30/55 = +6/11 = +0.55 .
The image is 55% as large as the object. The image is erect since M is positive.

Questions? mike.coombes@kpu.ca

α + β + η	= π	(1)
γ + β	= ½π	(2)
η + φ	= ½π	(3)
θ	= π - 2γ - 2φ	(4)