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

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

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

**For air**

\[n_{2} = 1\]

So

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

**Part (a)**

**Fluorite**Â $ 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**Â critical angle for Fluorite**Â is $44.21^{\circ}$

**Part (b)**

**Crown glass**Â $ 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**Â 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**Â is $49.81^{\circ}$

**Numerical Result**

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

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

– The value of the **critical angle for Ice**Â 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**Â by

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

**For air**

\[ n_{ 2 } = 1 \]

**So**

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

**Part (a)**

**Cubic zirconia**Â $ 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**Â $ 40.36 ^ { \circ } $