Question

The critical angle for glass-water surface is (refractive index of glass = 1.5 and refractive index of water = 4/3)?

Answer 1

Explanation:

Refraction at an interface is described by Snell’s law.

n1sinθ1=n2sinθ2

If n1 is the higher refractive index medium, we will have a critical angle. If θ1 is set to be exactly at the critical angle θc , then θ2 will be exactly at 90 ∘ . Substituting all of this into Snell’s law give us

n1sinθc=n2sin90∘

Solving this for θc yields

θc=sin−1(n2/n1)

In this case, n1=1.33 because it is water and n2=1.0 because it is air. This gives us a critical angle of

θc=sin−1(1.0/1.33)=48.8∘

Answer 2

Given:

Refractive Index of glass = 1.5 and Refractive Index of water is 4/3.

To find:

Critical angle of glass-water interface.

Calculation:

Critical angle can be calculated only when the light travels from denser medium (glass) to rarer medium (water).

For critical angle of incidence, the angle of refraction is 90°

Applying Snell's Law ;

$\therefore \: \mu_{glass} \times \sin( \theta) = \mu_{water} \times \sin( {90}^{ \circ} )$

$= > \: 1.5 \times \sin( \theta) = \dfrac{4}{3} \times 1$

$= > \: \dfrac{3}{2} \times \sin( \theta) = \dfrac{4}{3} \times 1$

$= > \: \sin( \theta) = \dfrac{4}{3} \times \dfrac{2}{3}$

$= > \: \sin( \theta) = \dfrac{8}{9}$

$= > \: \theta = { \sin}^{ - 1} ( \dfrac{8}{9} )$

$= > \: \theta \approx {62.73}^{ \circ}$

So, final answer is:

$\boxed{ \sf{ \: \theta = { \sin}^{ - 1} ( \dfrac{8}{9} ) \approx {62.73}^{ \circ} }}$