## What is a Critical Angle?

**Definition:** The critical angle is the angle of incidence in the denser medium when the angle of refraction in the less dense medium is ninety degrees (90^{0}). The critical angle is a concept in optics that describes the angle of incidence at which light transitions from one medium to another with a higher refractive index, typically from a denser medium (e.g., glass or water) to a less dense one (e.g., air). When the angle of incidence surpasses the critical angle, light no longer refracts into the second medium but instead undergoes total internal reflection, staying within the denser medium.

The formula to calculate the critical angle (**θ _{c}**) is given by:

**θ _{c} = sin^{-1}(n_{2} / n_{1})**

Where:

**θ**is the critical angle._{c}**n**is the refractive index of the denser medium._{2}**n**is the refractive index of the less dense medium._{1}

This phenomenon has numerous practical applications, such as in fibre optics, where it allows for the transmission of light signals over long distances within the core of optical fibres, preventing signal loss.

At values above the critical angle, the total internal reflection occurs, that is, the refracted ray disappears and a strong reflected ray appears.

A critical angle is the angle of incidence for which the refracted ray emerges tangent to the surface of the angle.

## How to Find Critical Angle

### Critical Angle Formula

**The critical angle formula is **

**θ _{c} = sin^{-1} (1 / n) ** [Where θ

_{c}= critical angle, n = refractive index]

We can also use

**θ _{c} = sin^{-1} (n_{a} / n_{g})**

Therefore, the critical angle formula that can help us to find critical angle based on the question available to us are: **θ _{c} = sin^{-1} (1 / n)** or

**θ**. Where: θ

_{c}= sin^{-1}(n_{a}/ n_{g})_{c}= critical angle, n = refractive index.

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### Explanation of Critical Angle

The critical angle is a term in physics and optics to describe the angle of incidence of a ray of light at which it refracts at an angle of 90 degrees to the normal (the line perpendicular to the surface of the medium it is entering).

Let’s assume we allow a ray of light to move from one medium to another. For example, say **air to water** or from **water to glass**. This ray of light will change direction because of the change in the speed of light in any of the mediums.

Therefore, we refer to the change in the direction of the ray of light as **refraction**. The amount of refraction depends on the angle of incidence of the light ray and the refractive index of the two media.

Therefore, to explain the **critical angle** in the simplest term. We can say that **it’s the angle of incidence at which the refracted ray of light is parallel to the boundary between the two media** **at 90 degrees**. In other words, the angle of incidence results in the refracted ray being bent at a 90-degree angle from the normal.

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## Snell’s Law

We can calculate critical angle by applying **Snell’s law. The law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two media.**

**sin i / sin r = sin θ _{c} / sin 90^{0} = _{g}n_{a} = 1 / _{a}n_{g} = 1 / n =**

**n**/

_{a}**n**

_{g}i = Angle of incidence

r = Angle of refraction = 90^{0}

θ_{c} = Critical Angle

** _{g}n_{a}** = Refractive index of the ray of light from glass to air

** _{a}n_{g}** = Refractive index of the ray of light from air to glass

n = Refractive index of glass by definition

## How to Derive Critical Angle Formula

When the angle of refraction is 90 degrees, the sine of the angle of refraction is equal to one (1). Therefore, the sine of the critical angle is similar to the reciprocal of the refractive index of the two media.

sin r = sin 90^{0} = 1

and sin i = sin θ_{c}

**Thus, sin i / sin r = sin θ _{c} / sin 90^{0} = sin θ_{c} / 1 = sin θ_{c} = _{g}n_{a} = 1 / _{a}n_{g} = 1 / n = n_{a} / n_{g}**

As stated earlier, the critical angle formula can be written as: **θ _{c} = sin^{-1} (1 / n)** or

**θ**. Where: θ

_{c}= sin^{-1}(n_{a}/ n_{g})_{c}= critical angle, n = refractive index.

The critical angle is an important concept in optics. This is because it determines whether a light ray will be refracted or reflected when it encounters a boundary between two media.

If the angle of incidence is greater than the critical angle. The ray of light will be reflected back into the medium from which it came. This phenomenon is called **total internal reflection**. We use it in various optical devices, such as fiber optic cables, and prism-based cameras.

## Relationship Between Refractive Index and Critical Angle

If the refractive angle is 90^{0}

By applying Snell’s law, the refractive index for glass – air boundary is: ** _{g}n_{a}** = (sin θ

_{c}) / (sin 90

^{0})

Now, the refractive index from air to glass is: _{a}**n**_{g} = (sin 90^{0}) / (sin θ_{c})

But sin 90^{0} = 1

Therefore, the refractive index from air to glass is: _{a}**n**_{g} = 1 / (sin θ_{c})

## Total Internal Reflection

When a ray of light passes from an optically denser medium to an optically less dense medium, there exists a weak internal reflection and a strong refraction. But as the angle of incidence increases, the angle of refraction also increases at the same time.

The intensity of the reflected ray gets stronger and that of the refractive ray becomes weaker. When the angle of incidence exceeds the critical angle, there exists no refraction and a total reflection occurs. We refer to this phenomenon in physics as **total internal reflection.**

The conditions under which total internal reflection occurs are:

- Light rays must travel through a denser medium to a less dense medium
- The angle of incidence must exceed the critical angle.

## Critical Angle Practice Problems

Here are practice problems to help you understand how to calculate critical angle problems:

### Problem 1

What is the critical angle for light traveling from water to air? The refractive index of water = 4/3

**Data**

The refractive index of water, n = 4/3 = 1.33

**Unknown **

Critical angle = ?

**Formula**

Critical angle, θ_{c} = sin^{-1} (1 / n)

#### Solution

We will insert our data into the formula

θ_{c} = sin^{-1} (1 / n) = sin^{-1} (1 / 1.33)

We will now have

θ_{c} = sin^{-1} (0.75) = 48.6^{0}

**Therefore, the critical angle for light traveling from water to air is 48.6 degrees.**

### Problem 2

A ray of light strikes from a medium with n = 1.67 on a surface of separation with the air with n = 1. Find the value of the critical angle.

**Data**

The refractive index of air, n_{a} = 1.67

We also have the refractive index of glass, n_{g} = 1

Additionally, we have a refractive index of the medium is, n = n_{a} / n_{g} = 1 / 1.67

**Unknown**

Critical angle, θ_{c} = ?

**Formula**

The equation for Critical angle, θ_{c} = sin^{-1} (n_{a} / n_{g})

#### Solution

We will add our data to the formula

Critical angle, θ_{c} = sin^{-1} (n_{a} / n_{g}) = sin^{-1} (1 / 1.67)

The above expression will give us

θ_{c} = sin^{-1} 0.599

And our final result for the critical angle will become

θ_{c} = 36.8^{0}

### Problem 3

The refractive index of a medium relative to air is 1.8. Calculate to the nearest degree, the critical angle for the medium.

**Data:** The information from the question that will help us solve the problem

The refractive index, n = 1.8

**Unknown:** What we need to find

Critical angle, θ_{c} = ?

**Formula:** The equation that will help us solve the problem

Critical angle, **θ _{c} = sin^{-1} (1 / n)**

#### Solution

To solve the problem, we will apply the critical angle formula above:

**θ _{c} = sin^{-1} (1 / n)** =

**sin**

^{-1}(1 / 1.8) =**sin**= 33.75^{-1}(0.55)^{0}

**Therefore, we can abbreviate the final answer as 34 ^{0} **

### Problem 4

The refractive index for a given transparent medium is 1.4. What is the minimum angle for total internal reflection to take place in the medium?

**Data:** The information from the question

The refractive index for the medium, n = 1.4

**Unknown:** Unavailable data from the question

Critical angle, θ_{c} = ?

**Formula:** The equation that will help us solve the problem

Critical angle, θ_{c} = sin^{-1} (1 / n)

#### Solution

Critical angle, **θ _{c} = sin^{-1} (1 / n)** =

**sin**

^{-1}(1 / 1.4) =**sin**=^{-1}(0.714) = 45.6^{0}**46**^{0}**Therefore, the critical angle of the medium is 46 ^{0}**

### Problem 5

If the refractive index of glass is 1.5, what is the critical angle at the air-glass interface?

**Data:** The information from the question

Refractive index, n = 1.5 = 3 / 2

**Unknown:** What we need to find

Critical angle, θ_{c} = ?

**Formula:** The equation to solve the problem

Critical angle, θ_{c} = sin^{-1} (1 / n)

#### Solution

**Critical angle, θ _{c} = sin^{-1} (1 / n)** = sin

^{-1}(1 / 1.5) = sin

^{-1}(1 / (3/2))

**Therefore, the critical angle, θ _{c}= sin^{-1} (2 / 3) = sin^{-1} (0.667) = 41.8^{0} = 42^{0} **

### Problem 6

The refractive indices of glass and water are 1.5 and 1.3 respectively. What will be the critical angle when the angle of refraction in the water medium is 90^{0}?

**Data:**

The refractive index of water, n_{w} = 1.3

Refractive index of glass, n_{g} = 1.5

Refractive angle, **θ** = 90^{0}

**Unknown:**

Critical angle, θ_{c} = ?

**Formula:**

**θ _{c}** = sin

^{-1}(n

_{w}sin90

^{0})/ n

_{g}

#### Solution

sin **θ _{c}** = (n

_{w}sin90

^{0})/ n

_{g}= (1.3 x 1) / 1.5 = 0.866

θ_{c} = sin^{-1} (0.866) = 60^{0}

## Table for Substances Their Critical Angles and Refractive Indexes

substance | Refractive index (n) | Calculation θ _{c} = sin^{-1} (1 / n) | Critical angle (θ_{c}) |

Vacuum | 1.00 | θ_{c} = sin^{-1} (1 / 1) | 90^{0} |

Pure Water | 1.33 | θ_{c} = sin^{-1} (1 / 1.33) | 48.75^{0} |

Carbondioxide | 1.0005 | θ_{c} = sin^{-1} (1 / 1.0005) | 88.19^{0} |

Ice | 1.31 | θ_{c} = sin^{-1} (1 / 1.31) | 49.76^{0} |

Ethyl Alcohol | 1.36 | θ_{c} = sin^{-1} (1 / 1.36) | 47.33^{0} |

Quartz | 1.46 | θ_{c} = sin^{-1} (1 / 1.46) | 43.23^{0} |

Vegetable oil | 1.47 | θ_{c} = sin^{-1} (1 / 1.47) | 42.86^{0} |

Olive oil | 1.48 | θ_{c} = sin^{-1} (1 / 1.48) | 42.51^{0} |

Acrylic | 1.49 | θ_{c} = sin^{-1} (1 / 1.49) | 42.16^{0} |

Table Salt | 1.51 | θ_{c} = sin^{-1} (1 / 1.51) | 41.47^{0} |

Glass | 1.52 | θ_{c} = sin^{-1} (1 / 1.52) | 41.14^{0} |

Sapphire | 1.77 | θ_{c} = sin^{-1} (1 / 1.77) | 34.40^{0} |

Zircon | 1.92 | θ_{c} = sin^{-1} (1 / 1.92) | 31.39^{0} |

Cubic zirconia | 2.16 | θ_{c} = sin^{-1} (1 / 2.16) | 27.58^{0} |

Diamond | 2.42 | θ_{c} = sin^{-1} (1 / 2.42) | 24.41^{0} |

Gallium phosphide | 3.50 | θ_{c} = sin^{-1} (1 / 3.50) | 16.60^{0} |

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