Question

The critical angle for a special type of glass in air is 35.5 ? . the index of refraction for water is 1.33. what is the critical angle if the glass is immersed in water?

Answer 1

To solve this problem, we must recall Snell's Law:

`n_(1)\alpha= n_(2)\beta`

For us to solve for the critical angle
`\alpha`,
`\beta` is 90°. Thus leaving the equation with

`n_(1)\alpha = n_(2)// sin\alpha = (n_(1))/(n_(2))`

Given that the glass in air forms a critical angle of 35.5° and that a glass' index of refraction is 1.00, we have


`sin35.5 = (1)/(n_(glass)) \\ n_(glass) = (1)/(sin35.5) \\ n_(glass) = 1.72`

Now, to solve for the critical angle between glass and water, we have

`sin\alpha = (n_(air))/(n_(glass)) \\ sin\alpha = (1.33)/(1.72) \\ \alpha = sin^(-1)((1.33)/(1.72)) \\ \alpha= 50.65`

Answer: θ= 50.65°