The drawing shows a rectangular block of glass (n = 1.55) surrounded by liquid carbon disulfide (n = 1.68). A ray of light is incident on the glass at point A with a θ1 = 43.0° angle of incidence. At what angle of refraction does the ray leave the glass at point B?
Given , μ1 = 1.68 , μ2 = 1.55 ,θ1 = 43°
Accordingto Snell law, we have
μ1sinθ1 =μ2sinθ2
1.68*sin(43°) = 1.55*sin(θ2)
sin(θ2) = 1.68sin(43°)/1.55 = 0.74
θ2 = sin^-1(0.74) = 47.663°
Therefore the refraction angle at A is θ2 = 47.663°
The angle of incidence at B = 90 - θ2 = 90 - 47.663° = 42.336°
Again using the snells law,
1.55*sin(42.336°) = 1.68*sin(θ3)
sin θ3 =1.55*sin(42.336°)/1.68
sin θ3 = 0.621
θ3 = sin^-1(0.621) = 38.416°
Thus , the angl of refraction is 38.416°
Get Answers For Free
Most questions answered within 1 hours.