Grade 11 Physical Sciences · Waves, Sound and Light

Apply Snell's law

Use n₁ sin θ₁ = n₂ sin θ₂ to find an angle, refractive index or critical angle when light passes from air into glass.

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Grade 11 - Waves, Sound, and Light | Geometric Optics (Snell’s Law) | November 2017 | Mlungisi NkosiMlungisi Nkosi Maths & Science

Snell's Law & Index of Refraction Practice Problems - PhysicsThe Organic Chemistry Tutor

Calculating Critical Angle using Snell's LawScience Chomp

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Question WE2 · Siyavula Grade 11 Physical Sciences, 5.6 Snell's Law, Worked example 2 · 3 marks

WE2 Light is refracted at the boundary between water and an unknown medium. If the angle of incidence is

25°

, and the angle of refraction is

20, 6°

, calculate the refractive index of the unknown medium and identify the material. (The refractive index of water is

1, 333

; typical glass has a refractive index between

1, 5

and

1, 9

.)(3)

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More past paper questions: apply snell's law

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3. Question WE1 · Siyavula Grade 11 Physical Sciences, 5.5 Refraction, Worked example 1 · 2 marks
WE1 Calculate the speed of light through glycerine which has a refractive index of $1, 4729$ .(2)
Show the memo (2 ticks)Memo
1. ✓
  Formula and substitution $n = \frac{c}{v}$ , so $v = \frac{c}{n}$
  $v = \frac{3 \times 1 0 ^{8}}{1 , 4729}$
  1 mark · Apply Snell's law
2. ✓
  Answer with unit $v = 2, 04 \times 1 0^{8}$ m·s $^{- 1}$
  1 mark · Apply Snell's law
4. Question WE3 · Siyavula Grade 11 Physical Sciences, 5.6 Snell's Law, Worked example 3 · 3 marks
WE3 A light ray with an angle of incidence of $35°$ passes from water to air. Find the angle of refraction using Snell's Law (refractive index of water $1, 333$ , of air about $1$ ). Discuss the meaning of your answer.(3)
Show the memo (3 ticks)Memo
1. ✓
  Snell's law substitution $n_{1} sin θ_{1} = n_{2} sin θ_{2}$
  $1, 33 sin 35° = 1 sin θ_{2}$
  $sin θ_{2} = 0, 763$
  1 mark · Apply Snell's law
2. ✓
  Angle of refraction $θ_{2} = 49, 7°$ or $130, 3°$
  Since $130, 3°$ is larger than $90°$ , $θ_{2} = 49, 7°$ .
  1 mark · Apply Snell's law
3. ✓
  MeaningThe light ray passes from a medium of high refractive index to one of low refractive index. Therefore, the light ray is bent away from the normal.
  1 mark · Explain refraction and total internal reflection
5. Question WE4 · Siyavula Grade 11 Physical Sciences, 5.6 Snell's Law, Worked example 4 · 3 marks
WE4 A light ray passes from water to diamond with an angle of incidence of $75°$ . Calculate the angle of refraction (refractive index of water $1, 333$ , of diamond $2, 42$ ). Discuss the meaning of your answer.(3)
Show the memo (3 ticks)Memo
1. ✓
  Snell's law substitution $n_{1} sin θ_{1} = n_{2} sin θ_{2}$
  $1, 33 sin 75° = 2, 42 sin θ_{2}$
  $sin θ_{2} = 0, 531$
  1 mark · Apply Snell's law
2. ✓
  Angle of refraction $θ_{2} = 32, 1°$
  1 mark · Apply Snell's law
3. ✓
  MeaningThe light ray passes from a medium of low refractive index to one of high refractive index. Therefore, the light ray is bent towards the normal.
  1 mark · Explain refraction and total internal reflection

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