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
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]