Physical Science 20 - Unit 6: Light Review (math solutions)
4. A photon of ultra violet light has 15.3 eV of energy. What is the frequency and wavelength of the light?
𝐸 =ℎ·𝑓
𝑐 =𝑓·λ
15.3 𝑒𝑉(1.60 × 10−19 𝐽/𝑒𝑉) = (6.63 × 10−34 𝐽 · 𝑠) × 𝑓
2.448×10−18 𝐽
=
6.63×10−34 𝐽·𝑠
15

𝑐
λ=𝑓
3.00×108 𝑚/𝑠
λ = 3.692×1015 𝐻𝑧
λ = 8.13 × 10−8 𝑚
𝑓
3.69 × 10 𝐻𝑧 = 𝑓
λ = 81.3 𝑛𝑚
5. Red light has a wavelength of 750 nm, how much energy is in one photon of the light.
If
𝑐
𝐸 = ℎ · 𝑓 and
ℎ·𝑐
𝑓 = λ then
𝐸=
ℎ·𝑐
λ
and since we want it in electronvolts, we get:
𝐸 = λ(1.60×10−19 𝐽/𝑒𝑉)
𝐸=
(6.63×10−34 𝐽·𝑠)×(3.00×108 𝑚/𝑠)
=
(750×10−9 𝑚)×(1.60×10−19 𝐽/𝑒𝑉)
1.989×10−25 𝑒𝑉
1.20×10−25
𝐸 = 1.66 𝑒𝑉
6. How far away is 17 light years in meters?
𝑑 = 𝑐 · Δ𝑡
365𝑑𝑎𝑦
24ℎ
3600𝑠
)×(
)×(
)
1𝑦𝑟
1𝑑𝑎𝑦
1ℎ
𝑑 = 3.00 × 108 𝑚/𝑠 × 17 𝑦𝑟 × (
𝑑 = 1.61 × 1017 𝑚
7. Calculate the absolute refractive index for a clear plastic material, if the velocity of light in the plastic is
2.5 x 108 m/s.
3.00×108 𝑚/𝑠
2.50×108 𝑚/𝑠
𝑐
𝑣
𝑛= =
𝑛 = 1.20
8. A ray of light in air has an incident angle of 40.8° on the surface of the same plastic used in the last
question. Determine the angle of refraction in the plastic.
𝑛1 = 1.0003
𝑛2 = 1.20
𝑛1 · sin 𝑖 = 𝑛2 · sin 𝑅
𝑛
sin 𝑅 = 𝑛1 · sin 𝑖 =
2
𝑖 = 40.8°
𝑅 =?
1.003
· sin 40.8°
1.20
sin 𝑅 = 0.5447
𝑅 = 33°
Light is slowing down and bending towards the normal.
9. As a ray of light passes from air into ruby, it enters with a refracted angle of 22°. What is the angle of
incidence?
𝑛1 = 1.0003
𝑛2 = 1.54
𝑛1 · sin 𝑖 = 𝑛2 · sin 𝑅
𝑛
𝑖 =?
𝑅 = 22°
1.54
sin 𝑖 = 𝑛2 · sin 𝑅 = 1.0003 · sin 22°
1
sin 𝑖 = 0.5767
𝑖 = 35°
Light is slowing down and bending towards the normal.
10. Calculate the relative index of refraction between:
a. Sodium chloride and zircon
1𝑛2
=
𝑛2(𝑧𝑖𝑟𝑐𝑜𝑛)
𝑛1(𝑁𝑎𝐶𝑙)
=
1.92
1.53
= 1.25
Light slows down and bends towards the normal.
b. Flint glass and Quartz
1𝑛2
=
𝑛2(𝑞𝑢𝑎𝑟𝑡𝑧)
𝑛1(𝑓𝑙𝑖𝑛𝑡 𝑔𝑙𝑎𝑠𝑠)
=
1.46
1.65
= 0.885
Light speeds up and bends away from the normal.
c. Having an incident angle of 15° in one medium and refracted angle 36° in the other.
1𝑛2
=
sin 𝑖
sin 𝑅
sin 15°
0.2588
= sin 36° = 0.5878 = 0.440
Light speeds up and bends away from the normal.
d. Having an incident angle of 63° in one medium and refracted angle 51° in the other.
1𝑛2
=
sin 𝑖
sin 𝑅
sin 63°
0.8910
= sin 51° = 0.7771 = 1.15
Light slows down and bends towards the normal.
11. Find the critical angle for light going from: Remember sin 90° = 1
a. Zircon to crystal glass
𝑛1 = 1.92 𝑛2 = 1.54 𝑖𝑐 =?
𝑛1 · sin 𝑖𝑐 = 𝑛2 · sin 𝑅𝑐
sin 𝑖𝑐 =
𝑛2
𝑛1
=
𝑅𝑐 = 90°
1.54
1.92
sin 𝑖𝑐 = 0.8020
𝑖𝑐 = 53°
b. Crown glass to glycerine
𝑛1 = 1.52 𝑛2 = 1.47 𝑖𝑐 =?
𝑛1 · sin 𝑖𝑐 = 𝑛2 · sin 𝑅𝑐
sin 𝑖𝑐 =
𝑛2
𝑛1
=
𝑅𝑐 = 90°
1.47
1.52
sin 𝑖𝑐 = 0.9671
𝑖𝑐 = 75.4°
c. If the angle of incidence for each of the above is 60°, does the light ray reflect or refract?
Part a. Light will refract into the crystal glass
Part b. Light will undergo TIR
d. Air to water
It is not possible to have total internal reflection when going into a denser medium. Light will
always refract when n2>n1. For this reason, there is no critical angle.