Question

Light passes from air into water at an angle of incidence of 42.3°. Determine the angle of refraction in the water.

Answer 1

Answer:
`0.517\°`

Step-by-step explanation:

According to Snell’s Law we have the following:

`n_(1)sin(\theta_(1))=n_(2)sin(\theta_(2))` (1)

Where:

`n_(1)=1` is the first medium index of refraction (air)

`n_(2)=1.33` is the second medium index of refraction (water)

`\theta_(1)=42.3\°` is the angle of the incident ray

`\theta_(2)` is the angle of the refracted ray

So, we need to find from (1):

`\theta_(2)=sin^(-1)((n_(1))/(n_(2))sin\theta_(1))` (2)

`\theta_(2)=sin^(-1)((1)/(1.3)sin(42.3\°))` (3)

Finally:

`\theta_(2)=0.517\°` This is the angle of refraction