Refraction of light
angle of a ray of light travelling from glass to paraffin
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
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)
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Calculate the angle q if the refractive index of glass and water are 3/2 and 4/3 respectively (3mks)
1 is 4/3 and that of medium 2 is 3/2. Calculate angle r
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
(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).
Fig. 9
iii) State why the spectrum formed above is not pure.
showing how all the eight can be arranged to make a simple prism binoculars.
Refraction of light
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
|
(ii) The ray undergoes total internal reflection. Since angle of incident is greater than ao the
critical angle.
is a constant for a pair of media
medium or when the angle of incidence in the optically dense medium is greater than the critical angle
= sin 48°/sin 32°
= 1.40, Accept 1.402
|
= 2/3 x 4/3
|
= 8/9
8/9 = sinq
sin 40
|
sin q = 8/9sin 40 = 0.5713
|
= 34.84o
4/3 sin 35 = 3/2 sinq2
sinq2 = 4/3 X 2/3 sin 35 = 0.5098
q2 = 30.654
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°)√
– 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
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
|
(ii) The ray undergoes total internal reflection. Since angle of incident is greater than ao the
critical angle.
is a constant for a pair of media
medium or when the angle of incidence in the optically dense medium is greater than the critical angle
= sin 48°/sin 32°
= 1.40, Accept 1.402
|
= 2/3 x 4/3
|
= 8/9
8/9 = sinq
sin 40
|
sin q = 8/9sin 40 = 0.5713
|
= 34.84o
4/3 sin 35 = 3/2 sinq2
sinq2 = 4/3 X 2/3 sin 35 = 0.5098
q2 = 30.654
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°)√
– 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