Tung Wah Group of Hospitals Kap Yan Directors’ College
F3 Physics Extra Exercise 2
Refraction
Answers and explanations are provided at the end of this file.
Multiple Choice
1. The diagram shows a beam of light passing through a block of
transparent plastic. There is a spherical air bubble trapped in the
block as shown. Which of the rays, A, B, C or D correctly shows the
direction of the light beam in the bubble?
A. Ray A
B. Ray B
C. Ray C
D. Ray D
2. The figure shows the refraction of a ray of light by a rectangular glass
block of refractive index 1.5. What is the angle of emergence (angle a in
the figure)?
A. 20o
B. 25o
C. 30o
D. 60o
3. The refractive index of glass is 1.5. Which of the following statements
is/are correct?
(1)
Speed of light in glass
 1.5
Speed of light in air
(2) For a ray of light passing from air to glass, the angle of incidence is greater than the angle of refraction.
(3) The critical angle of glass is 41.8o.
A. (1) only
B. (2) only
C. (1) and (2) only
D. (2) and (3) only
4. A ray of light enters a transparent rectangular block and travels along the path as shown in the figure. What
is angle ?
A. 15o
B. 33o
C. 57o
D. 59o
5. A long chain of length l is hung from the centre of a circular disc of
diameter d floating on water as shown. If the critical angle for water is c,
what is the shortest value of l in order that the whole chain can be seen by
an observer above the water surface?
A. (d /2) sin c
C. (d /2) tan c
B. (d /2) cos c
D. d /(2 tan c)
6. The figure shows a light ray passing from a liquid to air. What is the
critical angle of the liquid?
A. 35.4o
C. 41.8o
B. 40.7o
D. 45o
P.1
Structured Questions
1. The figure shows a small coin C placed at the centre of the bottom of a can. The
coin cannot be seen by the eye. After pouring in some liquid until the can is full,
the coin can just be seen. (You may regard the coin as a point.)
*(a) Explain why the eye cannot see the coin if the can is empty.
(4 marks)
(b) (i) Draw a ray diagram to show why the coin can be seen after the liquid is
poured in the can.
(2 marks)
(ii) Mark on your diagram the apparent position of the coin using the letter
C'.
(2 marks)
(c) (i) Calculate angle ABD.
(2 marks)
(ii) Calculate angle CBD
(2 marks)
(iii) Hence, or otherwise, find the refractive index of the liquid.
(2 marks)
2. The figure on the answer sheet shows a section of an optical fibre
made of perspex. A light ray is incident at an angle θ as shown.
(a) Calculate the critical angle of perspex. (Refractive index of
perspex = 1.6)
(2 marks)
(b) On your answer sheet, draw a ray diagram in each of the following cases to show the path of ray if
(i) θ = 60o,
(ii) θ = 30o.
Calculate the angles that the ray makes with the normal at the point where the ray is turned.
(5 marks)
(c) Find the greatest value of θfor the ray to be trapped inside the optical fibre.
(2 marks)
(d) Name one application of optical fibre.
(1 mark)
3. Two light rays, 1 and 2, are incident normally onto a semi-circular glass block of radius 10 cm and
refractive index 1.50 as shown in the figure. O is the center of the semi-circular glass.
(a) Calculate the critical angle of glass.
(2 marks)
(b) On the answer sheet, show the paths of light rays 1 and 2 until they emerge from the block at the curved
surface.
(2 marks)
* (c) Light ray 2 is displaced gradually towards the left, i.e. towards the edge. It is found that at a certain
position, the emergent ray (ray passed out of the block) suddenly disappears. Explain why this happens
and calculate the distance of the light ray from O when such a change occurs.
(5 marks)
--End of Extra Exercise 2--
Answers to F3 physics Extra Exercise 2
Multiple Choice
1-6 A D D B D B
Explanations to MC
1. The light ray is traveling from plastic to air, so it is from denser to less dense media and thus the refracted
ray bends away from normal. C is along the radius and is in normal direction. B is the original direction, so
bending away from normal should be A.
2. According to notes, the incident ray and the emergent ray are parallel after passing through the rectangular
glass block, so the required angle is the angle of incidence, i.e. 60o.
3. Statement (1) is wrong. The correct statement should be
Speed of light in air / speed of light in glass = 1.5
4. Refer to the figure.
sin c = 1/n => sin 50o = 1/n
=> n = 1.305
b = 40o (alternate angles)
sin a / sin b = n
sin a / sin 40o = 1.305 => a = 57.05o
 = 90o – a = 90o – 57.05o = 32.95o
5. Refer to notes
6. sin 30o / sin 50o = 1/n
sin c = 1/n = 1/ 1.532
=> n = 1.532
=> c = 40.7o
Structured Questions
1. (a) Light travels in straight lines.
Light from C to the eye
is blocked by the cup. Hence the coin cannot be seen.
(b) (i)
(ii)
(c) (i) Angle ABD = tan-1 10/10 = 45o
(ii) Angle CBD = tan-1 5/10 = 26.6o
sin 45 0
 1.58
(iii) Refractive index =
sin 26.6 0
1A
1A
1A+1C
1A for incident ray
1A for refracted ray to the eye
C’ above C 1A
C’ on the straight line 1A
1A+1A
1A+1A
1A+1A
2. (a) sin c = 1/n
1M
sin c = 1/1.6
c = 38.7o
1A
(b) (i) If  = 60o, angle of incidence = 30o < 38.7o,
refraction occurs.
sin 30o / sin r = 1/1.6
1M
r = 53.1o
1A
(ii) If  = 30o, angle of incidence = 60o > critical angle,
so total internal reflection occurs.
Correct refraction 1A
1A for total internal reflection
Correct reflection 1A
(c) For trapped inside the optical fibre, the light ray in Perspex should be incident at critical angle. So
 + c = 90o
 = 90o- 38.7o = 51.3o
(d) Endoscope / telecommunication
3. (a) sin c = 1/n = 1/1.50 = 0.667
c = 41.8o
(b) 1A for ray 1. 1A for ray 2
(1M)
(1A)
(c) As the light ray moves to the left, when the angle of incidence at the glass air boundary is greater than
critical angle, total internal reflection occurs. The emergent ray (or refracted ray) no longer exists.
This occurs when
sin 41.8o = OP / 10
OP = 8.94 cm