**Snell’s law equation** describes the relationship between the **angle of incidence** (θ_{1}), **angle of refraction** (θ_{2}), **refractive index of medium 1** (n_{1}) and **refractive index of medium 2** (n_{2}). Here’s the equation of snell’s law: **n**_{1}** sin θ**_{1}** = n**_{2}** sin θ**_{2}

Let’s solve some problems based on this equation, so you’ll get a clear idea.

## Snell’s Law Practice Problems

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

Solution:

Given data:

Refractive index of medium 1, n_{1} = 1

Refractive index of medium 2, n_{2} = 1.33

Angle of incidence, θ_{1} = 30°

Angle of refraction, θ_{2} = ?

Using the equation of snell’s law,

n_{1} sin θ_{1} = n_{2} sin θ_{2}

1 × sin (30°) = 1.33 × sin θ_{2}

1 × 0.5 = 1.33 × sin θ_{2}

0.5 = 1.33 × sin θ_{2}

sin θ_{2} = 0.3759

θ_{2} = sin^{-1} (0.3759)

θ_{2} = 22.07°

Therefore, the angle of refraction is **22.07°**.

**Problem 2:** A laser beam traveling through the medium 1 has an angle of incidence, θ_{1} = 40° and refractive index, n_{1} = 1.33. If a laser beam has an angle of refraction, θ_{2} = 45° then calculate the refractive index of medium 2.

Solution:

Given data:

Angle of incidence, θ_{1} = 40°

Refractive index of medium 1, n_{1} = 1.33

Angle of refraction, θ_{2} = 45°

Refractive index of medium 2, n_{2} = ?

Using the equation of snell’s law,

n_{1} sin θ_{1} = n_{2} sin θ_{2}

1.33 × sin (40°) = n_{2} × sin (45°)

1.33 × 0.6427 = n_{2} × 0.7071

n_{2} = 0.8547 ÷ 0.7071

n_{2} = 1.20

Therefore, the refractive index of medium 2 is **1.20**.

**Problem 3:** A ray traveling through the medium 1 of refractive index, n_{1} = 1.3 enters the medium 2 of refractive index, n_{2} = 1.4. Calculate the angle of incidence, if the angle of refraction is 25°.

Solution:

Given data:

Refractive index of medium 1, n_{1} = 1.3

Refractive index of medium 2, n_{2} = 1.4

Angle of incidence, θ_{1} = ?

Angle of refraction, θ_{2} = 25°

Using the equation of snell’s law,

n_{1} sin θ_{1} = n_{2} sin θ_{2}

1.3 × sin θ_{1} = 1.4 × sin (25°)

1.3 × sin θ_{1} = 1.4 × 0.4226

1.3 × sin θ_{1} = 0.5916

sin θ_{1} = 0.5961 ÷ 1.3

sin θ_{1} = 0.4585

θ_{1} = sin^{-1} (0.4585)

θ_{1} = 27.29°

Therefore, the angle of incidence is **27.29°**.

**Problem 4:** A light traveling through the medium 1 of refractive index n_{1} enters the medium 2 of refractive index, n_{2} = 1.2. If the angle of incidence is 20° and the angle of refraction is 35°, then what is the refractive index of medium 1?

Solution:

Given data:

Refractive index of medium 2, n_{2} = 1.2

Angle of incidence, θ_{1} = 20°

Angle of refraction, θ_{2} = 35°

Refractive index of medium 1, n_{1} = ?

Using the equation of snell’s law,

n_{1} sin θ_{1} = n_{2} sin θ_{2}

n_{1} × sin (20°) = 1.2 × sin (35°)

n_{1} × 0.3420 = 1.2 × 0.5735

n_{1} × 0.3420 = 0.6882

n_{1} = 0.6882 ÷ 0.3420

n_{1} = 2.01

Therefore, the refractive index of medium 1 is **2.01**.

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