Question

As light shines from air to water (n = 1.33) at an incident angle of 45.0°, what is its angle of refraction? r = _____° 0.500 1.60 148 32.1

Answer 1

Note: • is angle as I don't have a theta symbol.

n = sin•air / sin•water

1.33 = sin45 / sin•water

•water = sin^-1 (sin45 / 1.33)

•water = 32.1 degrees

Answer 2

The correct answer of this question is : $32.1^0$

EXPLANATION :

As per the question, the light is moving from air to water.

The angle of incidence is given as i = 45 degree.

The refractive index of water with respect to air is given as n = 1.33.

We are asked to calculate the angle of refraction.

From Snell's law, we know that -

                                $\frac{sini}{sinr}=\ n_{w}^a$

                               ⇒ $\frac{sin45}{sinr} =\ 1.33$

                               ⇒ $sinr=\ \frac{sin45}{1.33}$

                               ⇒ $sinr=\ \frac{1/\sqrt{2}}{1.33}$

                               ⇒ $sinr=\ \frac{1}{1.33\times \sqrt{2}}$

                              ⇒  $sinr=\ 0.53167$

                              ⇒ $r=\ 32.1^0$          [ans]