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The **Snell’s law equation**, n_{1} sin θ_{1} = n_{2} sin θ_{2}, precisely describes the correlation between the angle of incidence (θ_{1}), the angle of refraction (θ_{2}), the refractive index of medium 1 (n_{1}), and the refractive index of medium 2 (n_{2}). 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, n
_{1}= 1 - Refractive index of medium 2, n
_{2}= 1.33 - Angle of incidence, θ
_{1}= 30° - Angle of refraction, θ
_{2}= ?

Applying the formula:

- 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 medium 1 has an angle of incidence of θ_{1} = 40° and a refractive index of n_{1} = 1.33. If the angle of refraction is θ_{2} = 45°, 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}= ?

Applying the formula:

- 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 medium 1 with a refractive index of n_{1} = 1.3 enters medium 2 with a refractive index of 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°

Applying the formula:

- 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 medium 1 with an unknown refractive index enters medium 2 with a refractive index of n_{2} = 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, n
_{2}= 1.2 - Angle of incidence, θ
_{1}= 20° - Angle of refraction, θ
_{2}= 35° - Refractive index of medium 1, n
_{1}= ?

Applying the formula:

- 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**.

## 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/

Deep

Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.