# Snell’s Law Equation | Problems (With Solutions)

Snell’s law equation describes the relationship between the angle of incidence1), angle of refraction2), refractive index of medium 1 (n1) and refractive index of medium 2 (n2). Here’s the equation of snell’s law: n1 sin θ1 = n2 sin θ2

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

Using the equation of snell’s law,
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 the medium 1 has an angle of incidence, θ1 = 40° and refractive index, n1 = 1.33. If a laser beam has an angle of refraction, θ2 = 45° then 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 = ?

Using the equation of snell’s law,
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 the medium 1 of refractive index, n1 = 1.3 enters the medium 2 of refractive index, 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°

Using the equation of snell’s law,
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 the medium 1 of refractive index n1 enters the medium 2 of refractive index, n2 = 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, n2 = 1.2
Angle of incidence, θ1 = 20°
Angle of refraction, θ2 = 35°
Refractive index of medium 1, n1 = ?

Using the equation of snell’s law,
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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