Exam-style Questions

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The angle of incidence is always measured from the normal to the ray - so that rules out diagrams B and D.

The critical angle is defined as the angle of incidence which produces an angle of refraction of 90 degrees. So the answer must be C.



  1. 1. Which of the following shows the correct path of a ray of light that is approaching the inside surface of a block of glass if the angle of incidence of the ray is the critical angle, c?

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    • A
    • B
    • C
    • D
  2. If a wave can be polarised it must be:

    • an electromagnetic wave
    • a longitudinal wave
    • a transverse wave
    • a seismic wave

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    It is only possible to polarise a transverse wave, not a longitudinal wave. Hence, although electromagnetic waves and some types of seismic waves are transverse, the best answer here is C - a transverse wave.

  3. a) Define the refractive index of a transparent medium.

    b) Fig 1.1 shows a ray of light X emitted by point light source embedded in a glass block of refractive index 1.49. The angle of incidence of X at the glass/air surface is 30°.

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    i) Calculate the angle of refraction of X.

    angle of refraction =________________________°

    ii) Complete Fig. 1.1 to show what happens to the ray X after it is incident at the glass/air interface.

    iii) Calculate the critical angle at the glass/air interface.

    critical angle = ____________________°

    iv) On Fig. 1.1 draw the complete path followed by another ray of light leaving the light source which reaches the glass/air interface at the critical angle (there is no need to measure the critical angle accurately but it should be labelled).

    c) i) Calculate the speed of light in glass of refractive index 1.49.

    speed =___________________ms-1

    ii) Calculate the minimum time taken for a light pulse to travel from end to end along a straight glass fibre of length 50.0km and refractive index 1.49.

    time = __________________s

    iii) Suggest a reason why the time taken might be slightly greater than that calculated in (ii).

    (Marks available: 15)

  4. A beam of light enters glass, of refractive index n = 1.5.

    Calculate the speed of light in glass.

    Speed of light in air = 3.0 x 108 m s -1

    speed = .................................. m s -1

    (2 Marks)

  5. a) Calculate the refractive index for light passing from medium 1 to medium 2.

    refractive index = .....................

    (2 Marks)

    b) Determine the value of i that corresponds to an angle of refraction, in medium 2, of 35°.

    i = ................................. °

    (2 Marks)

    (Marks available: 4)



Answer outline and marking scheme for question:


  1. a) Refractive Index = speed of light in air/speed of light in the medium

    {Allow sin i / sin r if i and r are correctly identified e.g. on a sketch}

    (1 mark)

    b) i) gna =1/1.49

    Correct substitution into n = sin i / sin r : sin r = 1.49 x sin 30

    → r = 48°

    {If n=1.49 is used, allow 2 marks for r = 19.6 (or 19.9° i.e. 2 ecf marks)

    (3 marks)

    ii) REFRACTED RAY correctly drawn on Fig 1.1 (i.e. r>30°)

    {allow ecf (error carried forward) from (i) for r = 19.6 °s i.e. for refracted ray bending towards the normal}

    (Partially) reflected ray drawn at 30° (roughly judged by eye)

    {NB allow this mark if partially refected ray is showm either here or in (iv)}

    (2 marks)

    iii) correct substitution into sin C = 1/n : sin C = 1/1.49 (1 mark)

    → C = 42°

    (2 marks)

    iv) Ray drawn to the right of X with C (or 42°) correctly labelled REFRACTED RAY along interface (1 mark)

    {ignore partially reflected ray unless mark in (b)(ii) was not gained}

    (2 marks)

    c) i) recall of n = ca/cgOR cg= (1/1.49) x 3.0 x 10 8

    cg= 2.01 x 108 (m/s)

    (2 marks)

    ii) time = dist./vel. OR t = (50 x 103)/(2.01 x 108) {allow ecf fro (i)}

    = 2.49 x 10-4 s

    (2 marks)

    iii) some light travels further because of Total Internal Reflection (WTTE)

    (1 mark)

    (Marks available: 15)

  2. a) speed in glass = speed in vacuo / n (1 Mark) or numerical m ;

    = 2 x 108 m s -1(1 Mark)

    (2 Marks)

    b) recall of R.I = ci /cr OR 3.0 x 108 / 1.9 x 108

    = 1.58 (R 1.6)

    (2 Marks)

    c) recall of R.I = sin i/sin r

    sin i = sin 35° x 1.58 implies hence i = 65° (64.9 to 67)

    (2 Marks)

    (Marks available: 4)