Refraction of light
- The refractive index of paraffin is 1.47 and that of glass is 1.55. Determine the critical
angle of a ray of light travelling from glass to paraffin
- The diagram figure 1 below shows a ray of light incident on glass air boundary:
fig. 1
A second ray strikes the boundary at the same point C at an angle of incident greater than ao.
(i) On the diagram, draw the second ray before and after striking the boundary
- a) State Snell’s law
- b) When does total internal reflection occur?
- c) The figure below represents a ray of light falling normally on the curved surface of a
semi- circular glass block A at an angle of 32° at O and emerging into air at an angle of 48°
Calculate the absolute refractive index of the glass of which the block is made.
(Assume air is a vacuum)
- Figure 2 below shows a ray of light traveling from glass to water
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Calculate the angle q if the refractive index of glass and water are 3/2 and 4/3 respectively (3mks)
- Figure 3 shows light rays moving from medium 1 to medium 2. If the refractive index of medium
1 is 4/3 and that of medium 2 is 3/2. Calculate angle r
- (a) The diagram below shows a glass prism and an incident ray striking the face marked AB.
The critical angle of the glass is 42o. Use it to answer the questions that follow:-
(i) Complete the diagram showing the path of the emergent ray
(ii) Calculate the angle of refraction of the resultant emergent ray
- (a) (i) What is a critical angle as used in refraction of light?
(ii) State one condition under which total internal reflection occurs
(b) Calculate the value of the critical angle c in the figure below
(c) (i) Show that m = v + 1
f
where m = linear magnification , V= Image distance and f is the focal length of lens
(ii) In the table below shows readings obtained out of an experiment to determine focal length
of a converging lens
| Image distance V (cm) | 17.1 | 18.3 | 20 | 23 | 30 |
| Object distance (u) | 40 | 35 | 30 | 25 | 20 |
Plot a graph of 1 against 1 and determine the focal length of the lens from the graph.
V u (Use the graph paper provided).
- a) The Fig.9 shows a ray of sunlight incident to face AB of a glass prism. -•
Fig. 9
- i) Complete the diagram showing the observation on the screen.
- ii) Explain the observation on the screen.
iii) State why the spectrum formed above is not pure.
- b) i) You are provided with four equilateral prisms and four convex lenses. Sketch a diagram
showing how all the eight can be arranged to make a simple prism binoculars.
- ii) State one reason why prisms produce better optical instruments than plane mirrors………………………………………………………………………………………………………..
Refraction of light
- ahp = 1.47 and ahg = 1.55
ghp= gha x ahp
|
= ahp
ahg
= 1.47 =0.9484
|
1.5
Sin C = 1 = 0.9484
h
C = sin-1(0.9484)
C = 71.5o
|
- (i) for incident and reflected ray
(ii) The ray undergoes total internal reflection. Since angle of incident is greater than ao the
critical angle.
- a) The ratio the sin Ø of the angle of incidence to the sin e of the angle of refraction
is a constant for a pair of media
- b) When a ray is moving from an optically dense medium to a less optically dense
medium or when the angle of incidence in the optically dense medium is greater than the critical angle
- c) ang = sin i/sin r
= sin 48°/sin 32°
= 1.40, Accept 1.402
- d) Separation of colours of light from white light
|
- gnw = gna x anw
= 2/3 x 4/3
|
= 8/9
8/9 = sinq
sin 40
|
sin q = 8/9sin 40 = 0.5713
|
= 34.84o
- If the refractive index of medium 1 is 4/3 and that of medium 2 is 3/2. Calculate angle r n1sinq1 = n2 sinq2
4/3 sin 35 = 3/2 sinq2
sinq2 = 4/3 X 2/3 sin 35 = 0.5098
q2 = 30.654
- a) i
- ii) n = 1
sin 42
Sin 25 = I√
Sin r R
Sin 25 = sin 42√
Sin r
Sin r = Sin 25
Sin 42
= 0.631593
r = Sin -1 (0.631593)
= 39.17° (accept 39.2°)√
- (a) (i)-When a ray is moving from an optically denser medium to a less optically dense medium.
– When the angle of incidence in the optically denser medium is greater than the
critical angle (any 1)
(b) Sin C = n2 = 1.3 = 0.866
n1 1.5
ÐC = sin-10.866 \ÐC = 60.1o
(c) (i) From the len’s formula 1 = 1/V+ 1/u and dividing both sides by V,
V = 1 + V/u , but V/u = M
V/f = 1 + M and making M the subject ;
M = V/f -1
(ii) Graph: – scale used (1mk)
– Labeling axis
– Straight line
– Points
– Gradient/slope
1/V = 1/u – 1/f
1/f = 1/u + 1/V or 1/V = 1/f – 1/u
Gradient = Negative
1/V Intercept = 1/f
Refraction of light
- ahp = 1.47 and ahg = 1.55
ghp= gha x ahp
|
= ahp
ahg
= 1.47 =0.9484
|
1.5
Sin C = 1 = 0.9484
h
C = sin-1(0.9484)
C = 71.5o
|
- (i) for incident and reflected ray
(ii) The ray undergoes total internal reflection. Since angle of incident is greater than ao the
critical angle.
- a) The ratio the sin Ø of the angle of incidence to the sin e of the angle of refraction
is a constant for a pair of media
- b) When a ray is moving from an optically dense medium to a less optically dense
medium or when the angle of incidence in the optically dense medium is greater than the critical angle
- c) ang = sin i/sin r
= sin 48°/sin 32°
= 1.40, Accept 1.402
- d) Separation of colours of light from white light
|
- gnw = gna x anw
= 2/3 x 4/3
|
= 8/9
8/9 = sinq
sin 40
|
sin q = 8/9sin 40 = 0.5713
|
= 34.84o
- If the refractive index of medium 1 is 4/3 and that of medium 2 is 3/2. Calculate angle r n1sinq1 = n2 sinq2
4/3 sin 35 = 3/2 sinq2
sinq2 = 4/3 X 2/3 sin 35 = 0.5098
q2 = 30.654
- a) i
- ii) n = 1
sin 42
Sin 25 = I√
Sin r R
Sin 25 = sin 42√
Sin r
Sin r = Sin 25
Sin 42
= 0.631593
r = Sin -1 (0.631593)
= 39.17° (accept 39.2°)√
- (a) (i)-When a ray is moving from an optically denser medium to a less optically dense medium.
– When the angle of incidence in the optically denser medium is greater than the
critical angle (any 1)
(b) Sin C = n2 = 1.3 = 0.866
n1 1.5
ÐC = sin-10.866 \ÐC = 60.1o
(c) (i) From the len’s formula 1 = 1/V+ 1/u and dividing both sides by V,
V = 1 + V/u , but V/u = M
V/f = 1 + M and making M the subject ;
M = V/f -1
(ii) Graph: – scale used (1mk)
– Labeling axis
– Straight line
– Points
– Gradient/slope
1/V = 1/u – 1/f
1/f = 1/u + 1/V or 1/V = 1/f – 1/u
Gradient = Negative
1/V Intercept = 1/f
