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

Verified Answer :

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

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]