# Snell’s law equation

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.

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## 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.