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 -

`(sini)/(sinr)=\ n_(w)^a`

⇒
`(sin45)/(sinr) =\ 1.33`

⇒
`sinr=\ (sin45)/(1.33)`

⇒
`sinr=\ (1/√(2))/(1.33)`

⇒
`sinr=\ (1)/(1.33* √(2))`

⇒
`sinr=\ 0.53167`

⇒
`r=\ 32.1^0` [ans]