# For 589-nm light, calculate the critical angle for the following materials surrounded by air. (a) fluorite (n = 1.434) ° (b) crown glass (n = 1.52) ° (c) ice (n = 1.309)

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 Dutch 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 $40.36 ^ { \circ }$

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