PHYS 1100 Reflection and Refraction Solutions

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Physics 1100: Reflection and Refraction 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 0.9 m.
Two plane mirrors are inclined to one another at an angle α. 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. Express your answer in terms of α.

Using geometry

α + β + η = π, (1)

γ + β = ½π, (2)

φ + η = 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
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, S = +25. Using the lens formula
1/S + 1/S´ = 1/f .
Isolating 1/S´,
1/S´ = 1/f - 1/S = (S-f) / fS .
Inverting, we find
S´ = fS / (S-f) = (30)(25)/(25 - 30) = -150 cm .
Since S´ < 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 = -S´/S = -f / (S-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, S = +25. Using the lens formula
1/S + 1/S´ = 1/f .
Isolating 1/S´,
1/S´ = 1/f - 1/S = (S-f) / fS .
Inverting, we find
S´ = fS / (S-f) = (-30)(25)/[25 - (-30)] = -150/11 cm = -13.64 cm .
Since S´ < 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 = -S´/S = -f / (S-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.
A beam of sunlight encounters a plate of crown glass at an angle of 45.00°. If the index of refraction for red light in crown glass is n_r = 1.520 and for violet light is n_v = 1.538, find the angle between the violet ray and the red ray in the glass.
The angle that the light is refracted or bent is given by the Law of Refraction,
n₁sin(θ₁) = n₂sin(θ₂) .
Assuming that n₁ and θ₁ refer to the air, the refracted angle in the glass is
θ₂ = arcsin(n₁sin(θ₁)/n₂) .
For the each colour of light we get,
θ_red = arcsin(1 × sin(45.00°) / 1.520) = 27.7233° ,
θ_violet = arcsin(1 × sin(45.00°) / 1.538) = 27.3714° .
The difference is Δθ = 0.35°.
This phenomenon where different colours of light refract at different angles is called Dispersion and leads to the rainbow of light formed by prisms.
A beam of light is incident at an angle θ = 40.0° on a triangular prism of angle α = 45.0° and index n = 1.50 as shown below. Find the angle β at which it emerges.

The angle is related to by the Law of Refraction, n_airsin(θ) = n_prismsin(γ). Solving for , we find γ = arcsin(1 × sin(40.0°) / 1.50) = 25.3740° .
Next we need to do some geometry to find .

α + ζ + η = π , (1)

ζ + γ = ½π, (2)

φ + η = ½π, (3)

Using equations (2) and (3) to eliminate and from equation (1), we find
φ = α - θ = 45.0° - 25.3740° = 19.6260°.
The angle is related to by the Law of Refraction, n_prismsin(φ) = n_airsin(β). Solving for , we find β = arcsin(1.50 × sin(19.6260°) / 1) = 30.2530° .
The beam of light emerges at 30.3° to the normal.
A silver medallion is sealed within a transparent block of plastic. An observer in air , viewing the medallion from directly above, sees the medallion at an apparent depth of 1.6 cm beneath the top surface of the medallion. How far below the top surface would the medallion appear if the observer (not wearing goggles) and the block were under water?
In the first case the eye is in air and the apparent depth, d₁, would be given by
d₁ = d_actual(n_air/n_plastic) .
In the second case the eye is in water and the apparent depth, d₂, would be given by
d₂ = d_actual(n_water/n_plastic) .
We have two equations in two unknowns. We eliminate the common factor d_actual to get
d₂ = d₁(n_plastic/n_air)(n_water/n_plastic) = d₁(n_water/n_air) = (1.6)(1.33)/(1.00) = 2.1 cm .
Under water, the medallion appears to be 2.1 cm under the plastic surface.
A ray of light is incident at 35.0° on a 5.00 mm thick plane of glass with refractive index n = 1.70. What is the displacement of the ray?

The angle θ is related to the 35.0° angle by the Law of Refraction, n_airsin(35.0°) = n_prismsin(θ). Solving for θ, we find θ = arcsin(1 × sin(35.0°) / 1.70) = 19.7184° .
We will need to do some geometry to relate and t to d. First let x be the length of the ray in the glass. The separation of the ray is related to x by d = xsin(35°-θ). Similarly, the thickness of the glass t is related to x by t = xcos(θ). Eliminating the common factor x,
d = t sin(35° - θ) / cos(θ) = (5.00 mm)sin(35°-19.7184°)/cos(19.71484°) = 1.40 mm .
The ray has shifted 1.40 mm to the side.
To determine the refractive index of a transparent material, its critical angle is measured in air. If θ_c = 40.5°, what is the index of refraction of the solid?
The critical angle is the angle in the material for which the ray emerging into air is refracted at 90°. Using the Law of Refraction, nsin(θ_C) = n_airsin(90°). Solving for n, we find
n = n_air / sin(θ_C) = 1.00 / sin(40.5°) = 1.54 .
The material has an index of refraction of 1.54 .
A point source of light is submerged 2.2 m below the surface of a lake and emits rays in all directions. On the surface of the lake, directly above the source, the area illuminated is a circle. What is the maximum radius that this circle could have?

The circle is defined by the light that escapes. Light at angles greater than the critical angle is reflected back into the water. At the critical angle, the light emerging into the air is refracted at 90°. Using the Law of Refraction, nsin(θ_C) = n_airsin(90°), to find θ_C, we get
θ_C = arcsin(n_air/n) = arcsin(1/1.33) = 48.7535° .
Using geometry,
R = dtan(θ_C) = (2.2 m)tan(48.7535°) = 2.51 m .
Use the Principle Ray Technique to find the image, created by a converging lens, of an object placed between the first focal point F and the lens. Characterize the image.

The image, I, is erect, larger, and virtual. It is located between the left focal point and the object at O. We assume that the observer in on the right side.
Use the Principle Ray Technique to find the image, created by a diverging lens, of an object placed between F and the lens. Characterize the image.

The image, I, is erect, smaller, and virtual. It is located between the object, O, and the lens.
The convex lens of a magnifying glass has a focal length of 20 cm. At what distance from the postage stamp must you hold this lens if the image of the stamp is to be twice as large as the stamp and
(a) the image is inverted, or
(b) the image is erect?
For a convex lens, f = +20 cm.
(a) Given that the image is inverted and twice as large, we have M = -2. Now our definition of magnification is M = -S/S, therefore S = 2S. Using the lens formula
1/S + 1/S = 1/f .
Substituting S = 2S, this becomes
1/S +1/2S = 1/f ,
or
S = 3f/2 = 3(20 cm)/2 = 30 cm.
Thus S = 2(30 cm) = 60 cm.
(b) Since the image is erect and twice as large, then M = +2. Therefore S = -2S. We find
S = ½f = ½(20 cm) = 10 cm .
Thus S = -2(10 cm) = -20 cm.
A lens forms an image of an object. The object is 20.0 cm from the lens. The image is 5.00 cm from the lens on the same side as the object.
(a) What is the focal length of the lens? Is the lens converging or diverging?
(b) If the object is 2.00 cm tall, how tall is the image? Is it erect or inverted?
(a) We given that S = 20.0 cm and that S = -5.00 cm. The minus sign for S indicates that the image is on the same side of the lens as the object. Using the lens equation
1/f = 1/S + 1/S = 1/20 + 1/(-5) = -3/20 .
The focal length of the lens is thus -20/3 = -6.67 cm. The minus sign indicates that the lens is diverging or concave.
(b) The magnification is given by
M = -S/S = -(-5)/20 = ¼ .
As well, the size of the image is given by
H_image = Mh_object = ¼(2.00 cm) = 0.50 cm .
The magnification is positive so the image is erect. As well the image is on the same side as the object, so it is virtual and virtual images are erect.

Questions? mike.coombes@kwantlen.ca

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