This **article aims** to find the **critical angle** for the given **materials surrounded** by air. This **article uses the concept** of the **Snell law** to solve the **critical angle**. **Snell’s law** is used to explain the relationship between angles of **incidence and refraction** when referring to light or other waves passing through an **interface** between two different isotropic media, such as air, water or glass. This law was named after D**utch astronomer and mathematician Willebrand Snellius** (also called **Snell**).

**Snell’s law** states that for a given pair of the media, ratio of the sines of **angle of incidence** $\theta_{1}$ and **angle of refraction** $ \theta _{ 2 } $ is equal to the **ratio of the phase velocities** $ ( \dfrac {v_{ 1 } } { v_{ 2 } } ) $ in the two media, or equivalently to the **refractive indices** $ (\dfrac{n_{ 2 } } { n_{ 1 } } ) $ of the two media.

\[ \dfrac{ \sin \theta_{ 1 } } { \sin \theta_{ 2 } } = \dfrac { v_{ 1 } }{ v_{ 2 } } = \dfrac{n_{2}}{n_{1}}\]

**Expert Answer**

The **critical angle is given** by

\[\sin(\theta) = \dfrac{n_{ 2 }}{n_{1}} \]

**For air**

\[n_{2} = 1\]

So

\[\sin (\theta) = \dfrac{1}{n_{1}}\]

**Part (a)**

**Fluorite** $ n_{1}=1.434^{\circ} $

\[\sin(\theta) = \dfrac{1}{1.434^{\circ}}\]

\[\sin (\theta) = 0.697 \]

\[\theta _{c} = 44.21^{\circ}\]

The value of the** critical angle for Fluorite** is $44.21^{\circ}$

**Part (b)**

**Crown glass** $ n_{1}=1.52^{\circ} $

\[\sin(\theta) = \dfrac{1}{1.52^{\circ}}\]

\[\sin(\theta) = 0.657\]

\[\theta _{c} = 41.14^{\circ}\]

The value of the **critical angle for Crown glass** is $41.14^{\circ}$

**Part (c)**

Ice $ n_{1}=1.309^{\circ} $

\[\sin(\theta) = \dfrac{1}{1.309^{\circ}}\]

\[\sin(\theta) = 0.763\]

\[\theta _{c} = 49.81^{\circ}\]

The value of the **critical angle for Ice** is $49.81^{\circ}$

**Numerical Result**

– The value of the** critical angle for Fluorite** is $44.21^{\circ}$

– The value of the **critical angle for Crown glass** is $41.14^{\circ}$

– The value of the **critical angle for Ice** is $49.81^{\circ}$

**Example**

**For $589\: nm$ light, calculate the critical angle for the following materials surrounded by air.**

**(a) Cubic zirconia $(n_{1} = 2.15^{\circ})$**

**(b) Sodium chloride $ ( n_{ 1 } = 1.544 ^ { \circ } ) $**

**Solution**

The **critical angle is given** by

\[ \sin ( \theta ) = \dfrac { n_{ 2 } } { n_{ 1 } } \]

**For air**

\[ n_{ 2 } = 1 \]

**So**

\[ \sin ( \theta ) = \dfrac { 1 }{ n_{ 1 } } \]

**Part (a)**

**Cubic zirconia** $ n_{ 1 } = 2.15 ^ { \circ } $

\[ \sin ( \theta ) = \dfrac { 1 } { 2.15 ^ { \circ } } \]

\[\sin (\theta) = 0.465 \]

\[\theta _{ c } = 27.71 ^ { \circ } \]

**Part (b)**

**Sodium chloride **$ n_{ 1 }=1.544 ^ { \circ } $

\[ \sin( \theta ) = \dfrac{ 1 } { 1.544 ^ { \circ } } \]

\[ \sin( \theta ) = 0.647\]

\[ \theta _{ c } = 40.36 ^ { \circ } \]

The **critical angle for the sodium chloride** $ 40.36 ^ { \circ } $