Snell’s law equation

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The Snell’s law equation, n₁ sin θ₁ = n₂ sin θ₂, precisely describes the correlation between the angle of incidence (θ₁), the angle of refraction (θ₂), the refractive index of medium 1 (n₁), and the refractive index of medium 2 (n₂). This equation shows that the product of the refractive index and the sine of the angle of incidence in medium 1 is equal to the product of the refractive index and the sine of the angle of refraction in medium 2.

Practice problems

Problem #1

A ray of light moving through medium 1 with a refractive index of 1 enters medium 2 with a refractive index of 1.33. If the angle of incidence is 30°, what is the angle of refraction?

Solution

Given data:

• Refractive index of medium 1, n₁ = 1
• Refractive index of medium 2, n₂ = 1.33
• Angle of incidence, θ₁ = 30°
• Angle of refraction, θ₂ = ?

Applying the formula:

• n₁ sin θ₁ = n₂ sin θ₂
• 1 × sin (30°) = 1.33 × sin θ₂
• 1 × 0.5 = 1.33 × sin θ₂
• 0.5 = 1.33 × sin θ₂
• sin θ₂ = 0.3759
• θ₂ = sin^-1 (0.3759)
• θ₂ = 22.07°

Therefore, the angle of refraction is 22.07°.

Problem #2

A laser beam traveling through medium 1 has an angle of incidence of θ₁ = 40° and a refractive index of n₁ = 1.33. If the angle of refraction is θ₂ = 45°, calculate the refractive index of medium 2.

Solution

Given data:

• Angle of incidence, θ₁ = 40°
• Refractive index of medium 1, n₁ = 1.33
• Angle of refraction, θ₂ = 45°
• Refractive index of medium 2, n₂ = ?

Applying the formula:

• n₁ sin θ₁ = n₂ sin θ₂
• 1.33 × sin (40°) = n₂ × sin (45°)
• 1.33 × 0.6427 = n₂ × 0.7071
• n₂ = 0.8547 ÷ 0.7071
• n₂ = 1.20

Therefore, the refractive index of medium 2 is 1.20.

Problem #3

A ray traveling through medium 1 with a refractive index of n₁ = 1.3 enters medium 2 with a refractive index of n₂ = 1.4. Calculate the angle of incidence if the angle of refraction is 25°.

Solution

Given data:

• Refractive index of medium 1, n₁ = 1.3
• Refractive index of medium 2, n₂ = 1.4
• Angle of incidence, θ₁ = ?
• Angle of refraction, θ₂ = 25°

Applying the formula:

• n₁ sin θ₁ = n₂ sin θ₂
• 1.3 × sin θ₁ = 1.4 × sin (25°)
• 1.3 × sin θ₁ = 1.4 × 0.4226
• 1.3 × sin θ₁ = 0.5916
• sin θ₁ = 0.5961 ÷ 1.3
• sin θ₁ = 0.4585
• θ₁ = sin^-1 (0.4585)
• θ₁ = 27.29°

Therefore, the angle of incidence is 27.29°.

Problem #4

A light traveling through medium 1 with an unknown refractive index enters medium 2 with a refractive index of n₂ = 1.2. If the angle of incidence is 20° and the angle of refraction is 35°, what is the refractive index of medium 1?

Solution

Given data:

• Refractive index of medium 2, n₂ = 1.2
• Angle of incidence, θ₁ = 20°
• Angle of refraction, θ₂ = 35°
• Refractive index of medium 1, n₁ = ?

Applying the formula:

• n₁ sin θ₁ = n₂ sin θ₂
• n₁ × sin (20°) = 1.2 × sin (35°)
• n₁ × 0.3420 = 1.2 × 0.5735
• n₁ × 0.3420 = 0.6882
• n₁ = 0.6882 ÷ 0.3420
• n₁ = 2.01

Therefore, the refractive index of medium 1 is 2.01.

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Related

• Newton’s second law equation
• Newton’s law of universal gravitation formula
• Newton’s law of cooling equation
• Coulomb’s law equation
• Snell’s law equation
• Ohm’s law equation
• Hooke’s law equation

External links

• https://www.britannica.com/science/Snells-law
• https://www.physicsclassroom.com/class/refrn/Lesson-2/Snell-s-Law
• https://www.omnicalculator.com/physics/snells-law
• https://study.com/learn/lesson/what-is-snells-law-equation-examples.html
• https://www.studysmarter.us/explanations/physics/waves-physics/snells-law/