32º
Figure 1
θ
r
n
1
= 1
n
2
= 1.37
Assignment 2 Solutions
Question 1
(8 Marks)
3
2
2
O
2Al
3O
4Al
→
+
O
4H
+
3CO
5O
+
H
C
2
2
2
8
3
→
2NH
N
+
3H
3
2
2
→
6
4
2
3
2
2
2
O
S
Na
+
2NaI
O
S
2Na
+
I
→
Question 2
(8 Marks)
(i)
Calculating the angle of refraction
We can calculate the angle
θ
r
using Snell’s law;
sin
sin
i
1
2
θ
θ
n
n
r
=
Remember that the angles are always taken with
respect to the normal to the surface, so the angle of
incidence is;
º
8
5
32

90
=
=
i
θ
We are given that
n
1
= 1, and
n
2
= 1.37, then using Snell’s law to calculate the
refracted angle gives;
sin58
1
sin
37
.
1
=
r
θ
º
24
.
38
1.37
58
sin
arcsin
=
=
r
θ
Therefore, the angle of refraction is 38º, or (this answer cannot be accurate to more
than 2 significant figures).
(ii)
Calculating the wavelength of the light in the liquid
We know that for the light beam in vacuum;
Wavelength
nm
470
0
=
λ
Velocity
m/s
10
3
8
×
=
c
And we know that the speed of light is different in different media depending on the
refractive index;
v
c
n
=
,
or
n
c
v
=
(1)
where
n
is the absolute refractive index of the medium,
v
is the velocity of light in the
medium.
The wavelength and frequency of a wave are related to its velocity by
f
λ
=
v
, which
means that the frequency is given by f=v/
λ
.
The frequency will be the same in both
media – this is because the frequency represents the number of wave cycles arriving
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 Spring '08
 Beavan
 Light, Sin, refractive index, Geometrical optics