Questions on refraction and Snell's law

Question 1

Define the refractive index of a material in terms of the speed of light.

Accepted Answer

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).

Question 2

Light travels at 2.0 × 10⁸ m s⁻¹ in a glass block. Calculate the refractive index of the glass.

Accepted Answer

n = c/v = (3.0 × 10⁸)/(2.0 × 10⁸) (1) n = 1.5 (1)

Question 3

State the two conditions required for total internal reflection to occur.

Accepted Answer

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).

Question 4

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.

Accepted Answer

The speed decreases and the wavelength decreases in the same ratio (1). The frequency is unchanged (1).

Question 5

Explain why light travelling from air into glass can never be totally internally reflected at the boundary, however large the angle of incidence.

Accepted Answer

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 θ_c = n₂/n₁ would be greater than 1 (1).

Question 6

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.

Accepted Answer

n₁ sinθ₁ = n₂ sinθ₂ (1) sinθ₂ = sin 40°/1.5 (1) θ₂ = 25.4° (1)

Question 7

Calculate the critical angle for a glass–air boundary where the glass has a refractive index of 1.5.

Accepted Answer

sinθ_c = 1/n = 1/1.5 (1) θ_c = 41.8° (1)

Question 8

Light of wavelength 600 nm in air enters glass of refractive index 1.5. Calculate its wavelength and speed in the glass.

Accepted Answer

λ_glass = λ/n = 600/1.5 (1) = 400 nm (1) v = c/n = (3.0 × 10⁸)/1.5 = 2.0 × 10⁸ m s⁻¹ (1)

Question 9

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.

Accepted Answer

n₁ sinθ₁ = n₂ sinθ₂ (1) 1.33 × sin 30° = 1 × sinθ₂ (1) θ₂ = 41.7° (1)

Question 10

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.

Accepted Answer

The angle of incidence is measured from the normal: θ₁ = 90° − 25° = 65° (1). sinθ₂ = sin 65°/1.33 (1), so θ₂ = 43.0° (1).

Question 11

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.

Accepted Answer

n₁ sinθ₁ = n₂ sinθ₂: sinθ₂ = 1.52 × sin 50°/1.66 (1), so θ₂ = 44.5° (1). The ray bends towards the normal because it enters the optically denser medium, where light travels more slowly (1).

Question 12

State what is meant by modal dispersion in a step-index optical fibre, and give one design feature that reduces it.

Accepted Answer

Rays reflecting along the fibre at different angles travel different total distances, so parts of a pulse arrive at different times and the pulse broadens (1). A very narrow core reduces it by forcing all rays onto nearly the same path (1).

Question 13

A student measures the refractive index of a glass block by tracing rays through it. For six angles of incidence she plots sin θ₁, the sine of the angle in the air, against sin θ₂, the sine of the angle in the glass. Her line of best fit passes through the origin and through the point at which sin θ₂ = 0.41 and sin θ₁ = 0.62. Determine the refractive index of the glass, and state what the line passing through the origin shows.

Accepted Answer

Air has n = 1.00, so for a ray going from air into the solid Snell's law reduces to n = sin θ₁/sin θ₂ (1). The gradient of the graph is therefore n, and it must be read from the line of best fit rather than from any single plotted point (1). n = 0.62/0.41 = 1.5 (1). A straight line through the origin shows that sin θ₁ is proportional to sin θ₂, which is Snell's law itself (1).

Question 14

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.

Accepted Answer

sinθ_c = n_cladding/n_core (1) = 1.40/1.50 (1) θ_c = 69.0° (1)

Question 15

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.

Accepted Answer

sinθ_c = 1/n = 1/1.5 (1) θ_c = 41.8° (1) 45° is greater than the critical angle, so the ray undergoes total internal reflection and does not leave the block (1)

Question 16

Describe the structure of a step-index optical fibre and state the purpose of the cladding.

Accepted Answer

It has a central core of high refractive index surrounded by a cladding of slightly lower refractive index (1). The cladding gives a well-defined boundary for total internal reflection (1), protects the core, and reduces signal loss and crossover between fibres (1).

Question 17

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.

Accepted Answer

sinθ_c = n_cladding/n_core (1). For P: θ_c = 65.2°; for Q: θ_c = 72.5° (1); for R: θ_c = 80.7° (1). Only R gives a critical angle of at least 80°, so the manufacturer should choose cladding R (1).

Question 18

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.

Accepted Answer

v = c/n = (3.0 × 10⁸)/1.48 (1) = 2.03 × 10⁸ m s⁻¹, which is about 2 × 10⁸ m s⁻¹ (1). t = d/v = (15 × 10³)/(2.03 × 10⁸) (1) = 7.4 × 10⁻⁵ s (1).

Question 19

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.

Accepted Answer

Modal dispersion: rays reflecting at different angles travel different total path lengths along the fibre (1), so parts of the same pulse arrive at different times and the pulse spreads out (1). A very narrow core reduces this by forcing every ray onto nearly the same path (1). Material dispersion: different wavelengths travel at slightly different speeds in the glass, so a pulse of non-monochromatic light spreads in time (1); using monochromatic light (a single wavelength, for example from a laser) removes this spread (1). Broadened pulses can overlap their neighbours so that individual pulses can no longer be distinguished; pulses must therefore be sent less often, limiting the data rate (1).

Question 20

A student is given a rectangular glass block, a sheet of paper, four optical pins and a protractor. Describe how she obtains one pair of values of θ₁ and θ₂, the angles the ray makes with the normal in the air and in the glass, and explain why she repeats the measurement at several different angles of incidence.

Accepted Answer

Lay the block on the paper, draw round its outline, mark a point on one long face and rule the normal at that point (1). Push two pins into the paper in line with a ray entering at that point, mark the incident ray, and measure θ₁ between it and the normal (1). Look through the block from the far side and push in two more pins so that all four appear in a line through the glass, then mark the emergent ray (1). Lift the block off and join the entry point to the exit point: that line is the path the light took inside the block, which could not be watched directly, and θ₂ is measured between it and the normal (1). One tracing gives one value of n, and a poor one; taking six angles spread between about 20° and 70° and plotting them lets a line of best fit average the readings, and the graph tests Snell's law at the same time (1).

Question 21

A step-index optical fibre has a core of refractive index 1.50 and cladding of refractive index 1.47. A ray travelling inside the core meets the core-cladding boundary while travelling at 8.0° to the axis of the fibre. Deduce whether this ray is totally internally reflected at the boundary.

Accepted Answer

The normal to the boundary is at right angles to the axis, so the angle of incidence at the wall is 90° − 8.0° = 82.0° (1). sinθ_c = n_cladding/n_core = 1.47/1.50 (1), so θ_c = 78.5° (1). The angle of incidence 82.0° is greater than the critical angle, so the ray is totally internally reflected and stays in the core (1).