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Refraction and total internal reflection questions
Light changes speed when it crosses into a new medium, and Snell's law relates the change of direction to the two refractive indices. Beyond the critical angle it does not cross at all but reflects back inside, which is how an optical fibre carries light along its length.
18 original questions · 52 marks · the refraction and total internal reflection notes · Waves
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Define the refractive index of a material in terms of the speed of light.
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Refractive index n = speed of light in a vacuum / speed of light in the material = c/v (1). It is always greater than or equal to 1 (1).Light travels at 2.0 × 108 m s−1 in a glass block. Calculate the refractive index of the glass.
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n = c/v = (3.0 × 108)/(2.0 × 108) (1)
n = 1.5 (1)State the two conditions required for total internal reflection to occur.
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The light must be travelling in the more optically dense medium (higher n) towards the boundary with a less dense one (1), and the angle of incidence must exceed the critical angle (1).A ray of light passes from air into a glass block. State what happens, if anything, to the speed, the frequency and the wavelength of the light.
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The speed decreases and the wavelength decreases in the same ratio (1). The frequency is unchanged (1).Explain why light travelling from air into glass can never be totally internally reflected at the boundary, however large the angle of incidence.
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Total internal reflection requires the light to be travelling towards a medium of lower refractive index (1). Air has a lower refractive index than glass, so this light is heading into the denser medium and no critical angle exists: sin θc = n2/n1 would be greater than 1 (1).A ray of light passes from air into glass of refractive index 1.5 at an angle of incidence of 40°. Calculate the angle of refraction.
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n1 sinθ1 = n2 sinθ2 (1)
sinθ2 = sin 40°/1.5 (1)
θ2 = 25.4° (1)Calculate the critical angle for a glass–air boundary where the glass has a refractive index of 1.5.
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sinθc = 1/n = 1/1.5 (1)
θc = 41.8° (1)Light of wavelength 600 nm in air enters glass of refractive index 1.5. Calculate its wavelength and speed in the glass.
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λglass = λ/n = 600/1.5 (1)
= 400 nm (1)
v = c/n = (3.0 × 108)/1.5 = 2.0 × 108 m s−1 (1)A ray of light travels from water (refractive index 1.33) into air, meeting the surface at an angle of incidence of 30° in the water. Calculate the angle of refraction in the air.
A ray of sunlight strikes the flat surface of a pond at 25° to the water surface. The refractive index of the water is 1.33. Calculate the angle of refraction in the water.
A ray of light passes from crown glass of refractive index 1.52 into flint glass of refractive index 1.66. The angle of incidence at the boundary is 50°. Calculate the angle of refraction, and state which way the ray bends.
State what is meant by modal dispersion in a step-index optical fibre, and give one design feature that reduces it.
An optical fibre has a core of refractive index 1.50 and a cladding of refractive index 1.40. Calculate the critical angle at the core–cladding boundary.
A ray inside a glass block (refractive index 1.5) strikes a face at an angle of incidence of 45°. Deduce whether the ray is totally internally reflected or refracted out of the block.
Describe the structure of a step-index optical fibre and state the purpose of the cladding.
A manufacturer is designing a step-index optical fibre with a core of refractive index 1.52. To limit modal dispersion, the critical angle at the core–cladding boundary must be at least 80°. Three cladding glasses are available: P (refractive index 1.38), Q (refractive index 1.45) and R (refractive index 1.50). Deduce which cladding the manufacturer should choose.
A straight optical fibre is 15 km long and its core has a refractive index of 1.48. The speed of light in a vacuum is 3.0 × 108 m s−1. Show that light travels along the core at about 2 × 108 m s−1. Go on to calculate the minimum time for a light pulse to travel the length of the fibre.
A step-index optical fibre carries digital data as short pulses of light. Explain the two causes of pulse broadening in the fibre, state how the design of the fibre and the choice of light source reduce each one, and explain why pulse broadening limits the rate at which data can be sent.
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