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Questions: 1 2 3 4 5 6 7 8


Refraction Solutions


  1. 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?

    We have light passing through media of different indices of refraction, so we need to apply the Law of Refraction at each interface. A sketch of a light ray's path is shown below.

    The angle is related to the 35.0° angle by the Law of Refraction, nairsin(35.0°) = nprismsin(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 θ 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.

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  2. 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 nr = 1.520 and for violet light is nv = 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,

    n1sin1) = n2sin2) .

    Assuming that n1 and θ1 refer to the air, the refracted angle in the glass is

    θ2 = arcsin(n1sin1)/n2) .

    For the each colour of light we get,

    θred = arcsin(1 × sin(45.00°) / 1.520) = 27.7233° ,

    θviolet = 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.

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  3. 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 γ is related to θ by the Law of Refraction, nairsin(θ) = nprismsin(γ). Solving for γ, we find γ = arcsin(1 × sin(40.0°) / 1.50) = 25.3740° .

    Next we need to do some geometry to find φ.

    α + ζ + η = π (1)
    ζ + γ = ½π (2)
    φ + η = ½π (3)

    Using equations (2) and (3) to eliminate ζ and η from equation (1), we find

    φ = α - θ = 45.0° - 25.3740° = 19.6260° .

    The angle φ is related to β by the Law of Refraction, nprismsin(φ) = nairsin(β). Solving for β, we find β = arcsin(1.50 × sin(19.6260°) / 1) = 30.2530 ° .

    The beam of light emerges at 30.3° to the normal.

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  4. To determine the refractive index of a transparent material, its critical angle is measured in air. If θC = 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, nsinC) = nairsin(90°). Solving for n, we find

    n = nair / sinC) = 1.00 / sin(40.5°) = 1.54 .

    The material has an index of refraction of 1.54 .

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  5. 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, nsinC) = nairsin(90°), to find θC, we get

    θC = arcsin(nair/n) = arcsin(1/1.33) = 48.7535° .

    Using geometry,

    R = dtanC) = (2.2 m)tan(48.7535°) = 2.51 m .

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  6. 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 = dactual(nair/nplastic) .

    In the second case the eye is in water and the apparent depth, d2, would be given by

    d2 = dactual(nwater/nplastic) .

    We have two equations in two unknowns. We eliminate the common factor dactual to get

    d2 = d1(nplastic/nair)(nwater/nplastic) = d1(nwater/nair) = (1.6)(1.33)/(1.00) = 2.1 cm .

    Under water, the medallion appears to be 2.1 cm under the plastic surface.

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  7. In the diagram below, parallel layers of air, water (nw = 1.33), and glass (ng = 1.20) are shown. A ray is incident from air at an angle θ, refracted in the water, and reflected off the glass. That reflected light is completely polarized. What is the angle of incidence θ?

    Reflected light is totally polarized when the angle of incidence is at Brewster's angle

    α = arctan(nglass/nwater) = arctan(1.20/1.33) = 42.059° .

    This angle is related to θ by the Law of Refraction,

    nairsin(θ) = nwatersin(α) .

    Thus we find the angle of incidence to be

    θ = arcsin[(nwater/nair)sin(α)] = arcsin[(1.33/1)sin(42.059°)] = 63.0° .

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  8. A beam of light of intensity Iinitial encounters two polarizers. The transmission axis of the first polarizer is tilted at 25° to the vertical. The transmission axis of the second polarizer is tilted at 55° to the vertical. Find the intensity of the light after it passes through both polarizers if the light is initially
    (a) unpolarized,
    (b) polarized vertically,
    (c) polarized horizontally.

    Polarized light passing through a polaroid is governed by Malus' Law, I = I0cos2(θ).

    (a) Since the light is initially unpolarized or random, only half the energy passes through. So after the first polarizer, I = ½Iinitial. This transmitted light is now polarized at 25° to the vertical. Passing through the second polaroid, the intensity is reduced to

    I = I0cos2(θ) = ½Iinitialcos2(55°-25°) = (3/8)Iinitial .

    (b) Since the light is initially polarized vertically, passing through the first polaroid reduces the intensity to

    I = I0cos2(θ) = Iinitialcos2(0°-25°) = 0.8214 Iinitial .

    This light is now polarized along the axis 25° to the vertical. Passing through the second polaroid, the intensity is further reduced to

    I = I0cos2(θ) = (0.8214 Iinitial)cos2(55°-25°) = 0.616 Iinitial .

    (c) Since the light is initially polarized horizontally, passing through the first polaroid reduces the intensity to

    I = I0cos2(θ) = Iinitialcos2(90°-25°) = 0.1786 Iinitial .

    This light is now polarized along the axis 25° to the vertical. Passing through the second polaroid, the intensity is further reduced to

    I = I0cos2(θ) = (0.1786 Iinitial)cos2(55°-25°) = 0.134 Iinitial .

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